P1-Allens Made Physics Theory {PART-1}

ALTERNATING CURRENT 3 ATOMIC STRUCTURE 39 C A PAC I TA N C E 85 Current electricity 111 DIFFRACTION polarisation OF LIGHT with exe 149 ELECTROMAGNETIC INDUCTION 165 ELECTROMAGNETIC WAVES with exe 197 ELECTROSTATICS 207 GRAVITATION 257 INTERFERENCE OF LIGHT with exe 281 KINEMATICS 329 LOGIC GATES 367 MAGNETIC EFFECT OF CURRENT 373 MAGNETISM 417 MATHEMATICAL TOOLS with exe 435 Newton’s Laws of Motion 481 PRACTICAL PHYSICS with exe 531 PRINCIPLES OF COMMUNICATION SYSTEMS 561 PROPERTIES OF MATTER 573 RAY OPTICS 1 635 RAY OPTICS 2 FROM BACK 667 ROTATIONAL MOTION 691 SEMI CONDUCTOR - ELECTRONICS 757 SYSTEM OF PARICLES 785 Thermal physics 835 WAVE THEORY, SOUND WAVES _ DOPPLER_S EFFECTS 925 WORK POWER AND ENERGY 965



JEE-Physics ALTERNATING CURRENT ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction with time. It can be represented by a sine curve or cosine curve I = I sin  t or I = I cos  t 0 0 where I = Instantaneous value of current at time t, I = Amplitude or peak value 0  = Angular frequency  = 2 = 2f T = time period f = frequency T I I I0 I0 T 34T T T 2 2T T t T 34T t 4 4 –I0 –I0 I as a sine function of t I as a cosine function of t AMPLITUDE OF AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I . Peak to peak value = 2I 00 PERIODIC TIME The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. FREQUENCY The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : cycle/s ; (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING • Amplitude is constant • Alternate half cycle is positive and half negative The alternating current continuously varies in magnitude and periodically reverses its direction. I sinosudial AC I triangular AC ++ –t – I square wave AC I saw tooth wave t t Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 I0 I0 I0 t t t mixture of AC and DC Not AC (not periodic) Not AC (direction not change) 32 E

JEE-Physics AVERAGE VALUE OR MEAN VALUE The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. T/2  Idt average value of current for half cycle < I > = 0 T/2  dt 0 Average value of I = I sin t over the positive half cycle : 0 < sin  > = < sin 2 >=0 T < cos  >= < cos 2 >= 0   Iav  2 I0 dt 0 sin t 2 I0 T  2 I0 < sin  cos  > = 0 T  1 T =  cos t 2 < sin2  > = < cos2 >= 2 0  2 dt 0 • For symmetric AC, average value over full cycle = 0, Average value of sinusoidal AC Full cycle (+ve) half cycle (–ve) half cycle As the average value of AC over a 0 complete cycle is zero, it is always 2I0 –2I0 defined over a half cycle which must   be either positive or negative MAXIMUM VALUE  I =a • I = a + b sin  I = a + b ( if a and b > 0) • I = a sin Max. • Max. • I = a sin + b cos   I= a2  b2 I = a sin2   I = a (a > 0) Max. Max. ROOT MEAN SQUARE (rms) VALUE It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. Irms   T I2dt rms value = Virtual value = Apparent value 0 T 0 dt rms value of I = I sin t : 0 T (I0 sin t)2 dt I20 T sin2t dt = I0 1 T 1  cos 2t  1 t sin 2t T I0 0 0  2  T  2 2  2 2 Irms  T0 T dt  I0   T= 0 = 0 dt Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 • If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS. Current Average Peak RMS Angular fequency I= I sin t 0I I0  0 2 2 1 0 I= I sin t cos t = I0 sin 2t 0 I0 I0 2 2 22 2 0 I= I s int + I0cost 0 2 I0 I  0 3 0 Peak value • For above varieties of current rms = 2 E 33

JEE-Physics Example If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s Solution 4  < I > = 2 t.dt 3 4 t )2 dt 4  t2 4 4 ( t 2 )24 2 (2 2 4t dt = 2  2 3A 3 ( t )24 3 2 2   8  2 2  and I = 4 2  2  4 rms  2 2 dt dt 2 Example Find the time required for 50Hz alternating current to change its value from zero to rms value. Solution  I0  I0 sin t  sin t  1   2  t   t T1  2.5 ms I = I sin  t  2 2 t    T  4   8 8  50 0 4 Example 1 If E = 20 sin (100 t) volt then calculate value of E at t = s 600 Solution 1 E = 20 Sin 100  1  = 20 sin  1 At t = s 600  6  = 20 × = 10V 600 2 Example A periodic voltage wave form has been shown in fig. Determine. (a) Frequency of the wave form. (b) Average value. Solution After 100 ms wave is repeated so time period is (a) 1 T = 100 ms.  f = = 10 Hz (b) T Average value = Area/time period = (1 / 2)  100  10 = 5 volt (100) Example E +E0= 311.08 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Explain why A.C. is more dangerous than D.C. ? O 0.01 sec t Solution There are two reasons for it : -E  A.C. attracts while D.C. repels. -E0= -311.08  A.C. gives a huge and sudden shock. for 220 V ac V = 220 V rms Hence, V = 2.Vrms = 12.414 × 220 = 311.08 V 0 Voltage change from +V0 (positive peak) to –V0 (negative peak) = 311.08 – (–311.08) = 622.16 V This change takes place in half cycle i.e., in 1 s (for a 50 Hz A.C.) 100 A shock of 622.16 within 0.01 s is huge and sudden, hence fatal. 34 E

JEE-Physics Example I a DC + I AC If a direct current of value a ampere is superimposed on t b an alternating current I = b sint flowing through a wire, what is the effective value of the resulting current in the circuit ? =? t Solution As current at any instant in the circuit will be, I = I + I = a + b sint DC AC 1 T I2dt  1 T 1 T (a2  2ab sin t  b2 sin2 t)dt T0 T0 (a  b sin t)2 dt     Ieff  T0 1T 1 T tdt 1 a2  1 b2  sin tdt 0 and  Ieff  2 but as sin2  0 T0 2 T SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE Nature of Wave–form RMS Value Average or mean wave form Value + I0 Sinusoidal 0  – 2 2 2I0 = 0.707 I  = 0.637 I 0 0 Half wave 0  2 I0 I0 rectifired 0  2 2  Full wave rectifired = 0.5 I = 0.318 I 0 0 I0 2I0 2  = 0.707 I 0.637 I 0 0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Square or + I I Rectangular – 0 0 Saw Tooth wave 0  2 I0 I0 3 2 E 35

JEE-Physics MEASUREMENT OF A.C. Alternating current and voltages are measured by a.c. ammeter and a.c. voltmeter respectively. Working of these instruments is based on heating effect of current, hence they are also called hot wire instruments. Te r m s D.C. meter A.C. meter Name Based on moving coil hot wire Reads magnetic effect of current If used in average value heating effect of current A.C. circuit then they reads zero Deflection r.m.s. value  average value of A.C. = zero Scale A.C. or D.C. then meter works  = Number deflection  current of divisions    (linear) properly as it measures rms value Uniform Seperation - 1 2 3 4 5 deflection  heat -1 2 3 4 5    (non linear) rms Non uniform sepration - 1 2 3 4 5  - 1 4 9 16 25 • D.C meter in AC circuit reads zero because < AC > = 0 ( for complete cycle) • AC meter works in both AC and DC PHASE AND PHASE DIFFERENCE (a) Phase I = I sin (t + ) 0 Initial phase =  (it does not change with time) Instantaneous phase = t +  (it changes with time) • Phase decides both value and sign.  UNIT: radian (b) Phase difference Voltage V = V sin ( t +  ) Current I  I sin (t + ) 01 02 • Phase difference of I w.r.t. V =  –  21 • Phase difference of V w.r.t. I  = 1 – 2 LAGGING AND LEADING CONCEPT I=I0sin (t ) t ( a ) V leads I or I lags V  It means, V reach maximum before I V,I V=V0sint Let if V = V sin t then I = I sin (t –  and if 0 0 V = V0 sin (t+ ) then I = I0 sin t ( b ) V lags I or I leads V  It means V reach maximum after I I=I0sin (t +) t Let if V = V sin t then I = I sin (t +  V,I and if 0 V=V0sint 0 V = V sin (t –  ) then I = I sin t 0 0 PHASOR AND PHASOR DIAGRAM A diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase angle between them is called phasor diagram. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Let V = V sin t and I = I sin (t +) Y Y 0 V X 0 In figure (a) two arrows represents phasors. The length of phasors I V0 I0 represents the maximum value of quantity. The projection of a phasor I0  V0 X on y-axis represents the instantaneous value of quantity  fig (b) t ADVANTAGES OF AC fig (a) • A.C. is cheaper than D.C • It can be easily converted into D.C. (by rectifier) • It can be controlled easily (choke coil) • It can be transmitted over long distance at negligible power loss. • It can be stepped up or stepped down with the help of transformer. 36 E

JEE-Physics GOLDEN KEY POINTS • AC can't be used in (a) Charging of battery or capacitor (as its average value = 0) (b) Electrolysis and electroplating (Due to large inertia, ions can not follow frequency of A.C) Minimum, at that instant when they are near their peak values • The rate of change of A.C. Maximum, at that instant when they change their direction. • For alternating current I > I > I + 0 rms av. – • Average value over half cycle is zero if one quarter is positive and the other quarter is negative. • Average value of symmetrical AC for a cycle is zero thats why average potential difference on any element in A.C circuit is zero. • The instrument based on heating effect of current are works on both A.C and D.C supply and also provides same heating for same value of A.C (rms) and D.C. that's why a bulb bright equally in D.C. and A.C. of same value. • If the frequency of AC is f then it becomes zero 2f times in one second and the direction of current changes 2f times in one second. Also it become maximum 2f times in one second. Example The Equation of current in AC circuit is I = 4sin 100 t   Calculate. 3  A. (i) RMS Value (ii) Peak Value (iii) Frequency (iv) Initial phase (v) Current at t = 0 Solution (i) I = I0 = 4 = 2 2A (ii) Peak value I = 4A rms 2 2 0 (iii)   = 100  rad/s  frequency f = 100 = 50 Hz 2 (iv)  (v) At t = 0, I = 4sin 100  0   4 × 3 Example Initial phase = 3  = 2 =2 3 A 3 If I = I sin t, E = E cos t   . Calculate phase difference between E and I 0 3  0 Solution I = I sin t and E = E sin   t     5 0 3   phase difference = + = 0  2 23 6 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Example If E = 500 sin (100 t) volt then calculate time taken to reach from zero to maximum. Solution  = 100 T = 2 = 1 time taken to reach from zero to maximum = T 1 100 s, = s 50 4 200 Example  If Phase Difference between E and I is and f = 50 Hz then calculate time difference. 4 Solution  Phase difference time difference T T 1 2  T   Time difference = ×== = 2.5ms 2 T 2 4 8 50  8 E 37

JEE-Physics Example Show that average heat produced during a cycle of AC is same as produced by DC with I = I . rms Solution For AC, I = I0 sint, the instantaneous value of heat produced (per second) in a resistance R is, H = I2R = I 2s i n2t × R the average value of heat produced during a cycle is : 0 T T ( I20 sin2 t R )dt   Hav  0  1 I20  T 1   I0 2 H dt 2 0 2    2 0 T  R I 2 sin2 t dt  I20 T  H av  R  I2 R .....(i) 0  rms T dt 0 dt 0 However, in case of DC, H = I2 R...(ii)  I = I so from equation (i) and (ii) H = H DC rms DC av AC produces same heating effects as DC of value I = I . This is also why AC instruments which are based rms on heating effect of current give rms value. DIFFERENT TYPES OF AC CIRCUITS In order to study the behaviour of A.C. circuits we classify them into two categories : (a) Simple circuits containing only one basic element i.e. resistor (R) or inductor (L) or capacitor (C) only. (b) Complicated circuit containing any two of the three circuit elements R, L and C or all of three elements. AC CIRCUIT CONTAINING PURE RESISTANCE R Let at any instant t the current in the circuit =  V =IR Potential difference across the resistance =  R. with the help of kirchoff’s circuital law E –  R = 0 E = E0sint s E0 sin t =  R    E0 sin t  0 sin t ( 0  E0 R R s = peak or maximum value of current) E and I E = E0 sint t O I=I0 sint 2 X Alternating current developed in a pure resistance is also of sinusoidal  3/2 nature. In an a.c. circuits containing pure resistance, the voltage and /2 current are in the same phase. The vector or phasor diagram which represents the phase relationship between alternating current and al- ternating e.m.f. as shown in figure. In the a.c. circuit having R only, as current and voltage are in the same phase, hence in fig. both phasors E and 0 are in the same direction, Y 0 E making an angle t with OX. Their projections on Y-axis represent the t PE0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 instantaneous values of alternating current and voltage. i.e.  = 0 sint and E = E sint. I I0 0 E0 0  E0 E rms t X R 2 R2 R O Since 0  , hence  rms  AC CIRCUIT CONTAINING PURE INDUCTANCE L E = E0sin t A circuit containing a pure inductance L (having zero ohmic resistance) connected with a source of alternating emf. Let the alternating e.m.f. E = E sin t 0 When a.c. flows through the circuit, emf induced across inductance  L d dt 38 E

JEE-Physics Negative sign indicates that induced emf acts in opposite direction to that of applied emf. Because there is no other circuit element present in the circuit other then inductance so with the help of Kirchoff’s circuital law E   L d  0  d so we get   E0 sin  t    dt  E L L  2  dt Maximum current 0  E0 1  E0 , Hence,   0 sin  t   E and I E = E0 sint L L  2   /2  3/2 t In a pure inductive circuit current always lags behind the emf by . O 2 2  I = I0 sin (t -/2) or alternating emf leads the a. c. by a phase angle of . 2 Expression 0  E0 resembles the expression E R. L  Y P E0 This non-resistive opposition to the flow of A.C. in a circuit is called the E inductive reactance (X ) of the circiut. L X = L = 2  f L where f = frequency of A.C. t X L O Unit of X : ohm /2 – t L L) = Unit of L × Unit of  = henry × sec–1  Volt  sec1  Volt  ohm  0 Ampere / sec Ampere Q E = E0 sint I=I0 cost Inductive reactance X  f L Higher the frequency of A.C. higher is the inductive reactance offered by an inductor in an A.C. circuit. XL For d.c. circuit, f = 0  X = L = 2  f L = 0 L Hence, inductor offers no opposition to the flow of d.c. whereas a resistive path to a.c. f AC CIRCUIT CONTAINING PURE CAPACITANCE C A circuit containing an ideal capacitor of capacitance C connected with a source of alternating emf as shown in fig. The alternating e.m.f. in the circuit E = E sin t 0 When alternating e.m.f. is applied across the capacitor a similarly varying alternating current flows in the circuit. s The two plates of the capacitor become alternately positively and negatively charged E = E0sint and the magnitude of the charge on the plates of the capacitor varies sinusoidally with time. Also the electric field between the plates of the capacitor varies sinusoidally with time.Let at any instant t charge on the capacitor = q E = E0sint Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 q E and  Instantaneous potential difference across the capacitor E = /2 3/2 t O  2 C q = C E  q = CE sin t 0 The instantaneous value of current dq  d CE0 sin t  CE 0 cos t I= I0sin(t + /2)  dt Y dt E E0  t  Q P 2  I0 I E0 / C     1 sin   0 sin  t    where I = CV0  2  0 In a pure capacitive circuit, the current always leads the e.m.f. by a phase 90° angle of . The alternating emf lags behinds the alternating current by a t O X phase angle of . E 39

JEE-Physics IMPORTANT POINTS f E XC is the resistance R when both E and  are in phase, in present case they  1 is not the resistance of the capacitor, differ in phase by , hence 2 C the capacitor offer opposition to the flow of A.C. This non-resistive opposition 11 to the flow of A.C. in a pure capacitive circuit is known as capacitive reactance X. X C  C  2fC C Unit of X : ohm C Capacitive reactance X is inversely proportional to frequence of A.C. X decreases as the frequency increases. CC This is because with an increase in frequency, the capacitor charges and discharges rapidly following the flow of current. For d.c. circuit f = 0 1  X C  2fC   but has a very small value for a.c. This shows that capacitor blocks the flow of d.c. but provides an easy path for a.c. INDIVIDUAL COMPONENTS (R or L or C) TERM R L C Circuit R L C Supply Voltage V = V sin t V = V sin t V = V sin t Current 0 0 0 Peak Current I = I sin t I = I sin (t –  ) I = I sin (t+  ) Impedance ( ) 0 0 2 0 2 Z  V0  Vrms I0  V0 I0  V0 I0  V0 = V 0C I0 Irms R L 1 C V0  R V0  L  XL V0  1  XC I0 I0 I0 C R = Resistance X =Inductive reactance. X =Capacitive reactance. L C Phase difference zero (in same phase)  (V leads I)  (V lags I) 2 2 V I Phasor diagram I V I V R XL I Variation of Z with f XL f XC  1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 f G,S ,S f f LC V R does not depend on  S = 1/X (mho, seiman) G=1/R = conductance. LL S = 1/X Behaviour of device CC in D.C. and A.C Same in Inductive susceptance A C and D C L passes DC easily Capacitive susceptance (because X = 0) while C - blocks DC L (because X =) while gives a high impedance C for the A.C. of high provides an easy path for the A.C. of high frequency (XL f) frequency  X C  1 V = IX  f  Ohm's law V = IR LL V = IX R CC 40 E

JEE-Physics GOLDEN KEY POINTS • Phase diference between capacitive and inductive reactance is  • Inductor called Low pass filter because it allow to pass low frequency signal. • Capacitor is called high pass filter because it allow to pass high frequency signal. Example What is the inductive reactance of a coil if the current through it is 20 mA and voltage across it is 100 V. Solution   V = IX  X = VL 100 = 5 k LL L I  20  103 Example The reactance of capacitor is 20 ohm. What does it mean? What will be its reactance if frequency of AC is doubled? What will be its, reactance when connected in DC circuit? What is its consequence? Solution The reactance of capacitor is 20 ohm. It means that the hindrance offered by it to the flow of AC at a specific XC 11 frequency is equivalent to a resistance of 20 ohm. The reactance of capacitance   C 2 fC Therefore by doubling frequency, the reactance is halved i.e., it becomes 10 ohm. In DC circuit f = 0. Therefore reactance of capacitor =  (infinite). Hence the capacitor can not be used to control DC. Example A capacitor of 50 pF is connected to an a.c. source of frequency 1kHz Calculate its reactance. Solution 1 1 107 X= = =  L C C 2  103  50  1012  Example In given circuit applied voltage V = 50 2 sin 100t volt and A V ammeter reading is 2A then calculate value of L Solution  Reading of ammeter = I V =I X rms rms rms L X = Vrms = V0 50 2 = 25  L = XL = 25 1 IL = =H 2 Irms 2 2  100 rms 4 Example A 50 W, 100 V lamp is to be connected to an AC mains of 200 V, 50 Hz. What capacitance is essential to be put in series with the lamp ? Solution Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65  resistance of the lamp R  Vs2 (1 0 0 )2 = 200  and the maximum curent V 100 1  I  A W 50 R 200 2  when the lamp is put in series with a capacitance and run at 200 V AC, from V = IZ Z  V  200  400 Now as in case of C–R circuit Z  R2  1 , I1 (C )2 2  R2 1  (400)2  1  16 104  (200)2  12 104  1  12 102  (C )2 (C )2 C 1 100 C  F  F  9.2F 100  12 102  12 E 41

JEE-Physics RESISTANCE AND INDUCTANCE IN SERIES (R-L CIRCUIT) A circuit containing a series combination of a resistance R and an inductance L, connected with a source of alternating e.m.f. E as shown in figure. RL E = E0sint s s PHASOR DIAGRAM FOR L-R CIRCUIT Let in a L-R series circuit, applied alternating emf is E = E sint. As R Y R 0 Q and L are joined in series, hence current flowing through both will be same VL E at each instant. Let  be the current in the circuit at any instant and V and L V the potential differences across L and R respectively at that instant. R Then V = XL and V = R L R  Now, V is in phase with the current while V leads the current by . R L2 X VR P So V and V are mutually perpendicular (Note : E  V + V ) RL RL The vector OP represents VR (which is in phase with ), while OQ represents VL (which leads  by 90°). The resultant of V and V= the magnitude of vector OR E VR2  VL2 R L Thus E² = V² + V² = ² (R² + X ²)    E R L L R2  X 2 L XL The phasor diagram shown in fig. also shows that in L-R circuit the applied ZL emf E leads the current  or conversely the current  lags behind the e.m.f. E. by a phase angle  tan   VL  X L  X L  L       tan 1  L  R VR R R R  R  Inductive Impedance Z : L In L-R circuit the maximum value of current 0  E0 Here R2  2L2 represents the effective R 2  2 L2 opposition offered by L-R circuit to the flow of a.c. through it. It is known as impedance of L-R circuit and is represented by Z. ZL  R 2  2 L2  R 2  (2fL )2 The reciprocal of impedance is called admittance L 11 CR YL  Z L  R 2  2 L2 RESISTANCE AND CAPACITOR IN SERIES (R-C CIRCUIT) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 A circuit containing a series combination of a resistance R and a capacitor C, connected with a source of e.m.f. of peak value E as shown in fig. 0 PHASOR DIAGRAM FOR R-C CIRCUIT E = E0sint Current through both the resistance and capacitor will be same at every instant VR P and the instantaneous potential differences across C and R are O  X VC V =  XC and V =  R Q E= E C R where X = capacitive reactance and  = instantaneous current. V +2 V C R Now, V is in phase with , while V lags behind  by 90°. C 2 R C The phasor diagram is shown in fig. The vector OP represents V (which is in phase with ) S R E (applied emf)  and the vector OQ represents V (which lags behind  by 2 ). C The vector OS represents the resultant of V and R V = the applied e.m.f. E. C 42

JEE-Physics Hence V 2 + V 2 = E2 E  VR2  VC2 O P X RC Q XC R  E² = ² (R² + X² )     E Z = R +2 X C 2 R2  X 2 C C The term (R2  X 2 ) represents the effective resistance of the R-C S C circuit and called the capacitive impedance Z of the circuit. C Hence, in C-R circuit ZC  R2  X 2  R2   1 2 C  C  Capacitive Impedance ZC : In R-C circuit the term R2  X 2 effective opposition offered by R-C circuit to the flow of a.c. through C it. It is known as impedance of R-C circuit and is represented by Z C The phasor diagram also shows that in R-C circuit the applied e.m.f. lags behind the current  (or the current  leads the emf E) by a phase angle  given by tan   VC  XC  1 / C  1 , tan   XC  1    tan 1 1 VR R R C R R CR  CR  COMBINATION OF COMPONENTS (R-L or R-C or L-C) TERM R-L R-C L-C RL RC LC Circuit I is same in R & L I is same in R & C I is same in L & C VL VR VL I I Phasor diagram V I VC VC VR V2 = VR2  VC2 V = V – V (V >V ) V2 = VR2  VL2 L CLC   V = VC – VL (VC>VL) Phase difference V leads I ( = 0 to 2 ) V lags I ( = – 2 to 0 )  V lags I ( =  2 ,if XC>XL) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65  R2 2 in between V & I V leads I ( = ,if X >X ) Impedance Z  X L 2 LC Z = R2  X C 2 Z = XL  Xc as f, as f, Z first  then as f,Z  Variation of Z Z Z Z with f R R At very low f f Z~X f f At very high f C Z ~ R (X  0) Z~X L C Z~X Z ~ R (X  0) Z~X L C L E 43

JEE-Physics Example 30 40 Calculate the impedance of the circuit shown in the figure. Solution Z = R2  (Xc )2 = (30)2  (40)2 = 2500 = 50  Example If X = 50  and X = 40  Calculate effective value of current in given circuit. L C Solution Z = X – X = 10  XL=50 XC=40 L C  = V0  = 40 = 4 A  I= 4 =2 2 A Z 10 2 rms Example V=40sin 100 volt In given circuit calculate, voltage across inductor VR=60 VL=? Solution  V 2  VR2  VL2  V 2 = V2 – V 2 LR V V 2  VR2 = (100)2  (60)2 = 6400 = 80 V V=100 2 sint volt L 6 8 Example (i) impedance of circuit (ii) current in circuit In given circuit find out Solution (i) Z= R2  X 2  (6)2  (8)2 = 10  C (ii) V = IZ  I  V0  20  2 A so I = 2 = 2A V=20sint volt Z 10 rms 2 Example When 10V, DC is applied across a coil current through it is 2.5 A, if 10V, 50 Hz A.C. is applied current reduces to 2 A. Calculate reactance of the coil. Solution For 10 V D.C. V = IR  Resistance of coil R = 10  4 For 10 V A.C.  V = I Z  Z  V 20  5 2.5 I  10  Z  R2  X 2 5  R2  X 2  25  X 2 = 52 – 42 X= 3  L L L L Example When an alternating voltage of 220V is applied across a device X, a current of 0.5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Y, the same current again flows through the circuit but it leads the applied voltage by /2 radians. (a) Name the devices X and Y. (b) Calculate the current flowing in the circuit when same voltage is applied across the series combination of X and Y. Solution (a) X is resistor and Y is a capacitor (b) Since the current in the two devices is the same (0.5A at 220 volt) When R and C are in series across the same voltage then 220 Vrms = 220 220 R = XC = 0.5 = 440   Irms = = = 0.35A R2  X 2 C (440)2  (440)2 440 2 44 E

INDUCTANCE, CAPACITANCE AND RESISTANCE IN SERIES JEE-Physics (L-C-R SERIES CIRCUIT) LCR A circuit containing a series combination of an resistance R, a coil of inductance L and a capacitor of capacitance C, connected with a source of alternating e.m.f. A.C. source E = E0 sint of peak value of E , as shown in fig. 0 PHASOR DIAGRAM FOR SERIES L-C-R CIRCUIT Let in series LCR circuit applied alternating emf is E = E sin t. 0 As L,C and R are joined in series, therefore, current at any instant through the three elements has the same amplitude and phase. However voltage across each element bears a different phase relationship with the current. Y VL Q Let at any instant of time t the current in the circuit is  Let at this time t the potential differences across L, C, and R O X VR P V =   X, V =  X and V =  R VC L L C C R Now, V is in phase with current  but V leads  by 90° R L While V legs behind  by 90°. C The vector OP represents V (which is in phase with ) the vector R OQ represent VL (which leads  by 90°) and the vector OS represents VC (which legs behind  by 90°) Y VL Q K T V and V are opposite to each other. (VL- VC) LC E (applied emf) If V > V (as shown in figure) the their resultant will be (V – V ) which LC LC is represented by OT. O PX VR Finally, the vector OK represents the resultant of V and (V – V ), that VC R LC is, the resultant of all the three = applied e.m.f. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 E Y Thus E  VR2  (VL  VC )2 = I R 2  (X L  X C )2    R 2  (X L  X C )2 XL Q 2 T K  R Impedance Z  R2  (XL  XC )2 = R2   L  1 (XL-XC) Z  C PX The phasor diagram also shown that in LCR circuit the applied e.m.f. O leads the current by a phase angle tan = XL  XC XC R E 45

JEE-Physics PARALLEL L-C-R CIRCUIT SERIES LCR AND PAR ALLEL LCR COMBINATION R SERIES L-C-R CIRCUIT L C 1. Circuit diagram R LC I same for R, L & C V same for R,L and C 2. Phasor diagram V same for R, L & C VL I IC V VR VC IR IL (i) If VL > V C then (i) if IC > IL then VL–VC I IC-IL V VR IR (ii) If V > V then (ii) if I > I then CL LC VC–VL VR I IL-IC IR V (iii) V = VR2  (VL  VC )2 (iii) I = I2R  (IL  IC )2 Impedance Z = R2  (XL  XC )2 Admittance Y = G2  (SL  SC )2 tan = X L  X C = VL  VC tan = SL  SC IL  IC R VR G = IR (iv) Impedance triangle (iv) Admittance triangle Z X=XL–XC G R SL-SC Y GOLDEN KEY POINTS Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 • Series Parallel (a) if X > X then V leads I, (positive) (a) if S > S (X < X) then V leads I, (positive) L C L C L C circuit nature inductive circuit nature inductive (b) if X > X then V lags I, (negative) (b) if S > S ( X < X ) then V lags I,  (negative) CL CL C L circuit nature capacitive circuit nature capacitive • In A.C. circuit voltage for L or C may be greater than source voltage or current but it happens only when circuit contains L and C both and on R it never greater than source voltage or current. • In parallel A.C.circuit phase difference between I and I is  LC 46 E

JEE-Physics Example 4 9 6 Find out the impedance of given circuit. Solution Z  R2  (XL  XC )2  42  (9  6)2 = 42  32  25  5 ( X > X  Inductive) L C Example Find out impedance of given circuit. 3 Solution 6 Y2 = G2 + (SL – SC)2 1   1  1 2 6 36  6 3   2 6 Y=   Z= 2   (capacitive, because X > X) 6 L C Example Find out reading of A C ammeter and also calculate the potential difference across, resistance and capacitor. Solution Z  R2  (XL  XC )2  10 2   I0  V0 100  10 A 10 20 10 Z  2 A 10 2 E=100sint100t volt  ammeter reads RMS value, so its reading = 10 = 5A 22 so V = 5 × 10 = 50 V and V = 5 × 10 = 50 V RC Example In LCR circuit with an AC source R = 300  , C = 20 F, L = 1.0 H, E = 50V and f = 50/ Hz. Find RMS rms current in the circuit. Solution Irms  E rms  E rms  50 Z L 1 2 2  50 1 2 R2   C 3002     1   10 6  2 50    20      Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65  Irms  50 50 1    0.1A 2 100 9 16 10  103  (300)2  100  2   Example Calculate impedance of the given circuit : 5V 12V 4V 0.2F 1H 3 V=100sint volt Z 240 (ii) I=2A (iii) 120 (i) Z=? E 47

JEE-Physics Solution (i) It is parallel circuit so Y is evaluated Y  SL  SC  11  1    Z = 240  (inductive) 120 240 240 (ii) V 2 = 52 + 122 = 169  V = 13 volt therefore Z  Vs  13  6.5 s s I2 (iii) R = 3, XL = L = 1 as ( = 1) 11 Z2 = R2 + (X – X )2 = 32 + (1 – 5)2 = 25  Z = 5  X= = = 5 so LC C C (0.2).1 RESONANCE A circuit is said to be resonant when the natural frequency of circuit is equal to frequency of the applied voltage. For resonance both L and C must be present in circuit. There are two types of resonance : (i) Series Resonance (ii) Parallel Resonance SERIES RESONANCE (ii) VL = VC (iii)  = 0 (V and I in same phase) (a) At Resonance V (i) XL = XC (v) I = (current maximum) max R (iv) Z = R (impedance minimum) min (b) Resonance frequency  X =X 1  2r  1  11 LC  rL = r C LC r  LC  fr  2 LC (c) Variation of Z with f (i) If f< fr then XL < XC circuit nature capacitive,  (negative) (ii) At f = fr then XL = XC circuit nature, Resistive,  = zero Z R (iii) If f > f then X > X circuit nature is inductive, (positive) r LC Variation of I with f as f increase, Z first decreases then increase fr f Imax Imax Imax= V 2 R (d) I f as f increase, I first increase then decreases f1 fr f2 f Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 • At resonance impedance of the series resonant circuit is minimum so it is called 'acceptor circuit' as it most readily accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning we receive the desired station by making the frequency of the circuit equal to that of the desired station. Half power frequencies The frequencies at which, power become half of its maximum value called half power frequencies Band width =f = f – f 2 1 Quality factor Q    Q-factor of AC circuit basically gives an idea about stored energy & lost energy. Q  2 maximum energy stored per cycle maximum energy loss per cycle (i) It represents the sharpness of resonance. (ii) It is unit less and dimension less quantity (iii) Q = (X L )r = (XC )r = 2frL 1 L = fr = fr = RR R R C f band width 48 E

JEE-Physics Magnification I At resonance So at resonance V or V = QE (where E = supplied voltage) R1 LC R1 < R2 < R3 Magnification factor = Q-factor R2 Sharpness Sharpness  Quality factor  Magnification factor R decrease   Q increases   Sharpness increases PARALLEL RESONANCE R3 f fr (a) At resonance C LR (i) S = S (ii) I = I (iii) = 0 LC LC (iv) Z = R (impedance maximum) max V (v) I = (current minimum) min R 1 (b) Resonant frequency f = r 2 LC (c) Z Variation of Z with f as f increases , Z first increases then decreases  If f < f thenS > S,  (positive), circuit nature is inductive I r L C  If f > f then S > S ,  (negative), circuit nature capacitive. r CL fr V Imin= R ( d ) Variation of I with f as f increases , I first decreases then increases 1 1 R2 L 1 R2 fr f LC  L2  Zmax  RC  Note : For this circuit fr  For resonance 2 LC L2 GOLDEN KEY POINTS • Series resonance circuit gives voltage amplificaltion while parallel resonance circuit gives current amplification. • At resonance current does not depend on L and C, it depends only on R and V. • At half power frequencies : net reactance = net resistance. • As R increases , bandwidth increases • To obtain resonance in a circuit following parameter can be altered : (i) L (ii) C (iii) frequency of source. • Two series LCR circuit of same resonance frequency f are joined in series then resonance frequency of series combination is also f • The series resonance circuit called acceptor whereas parallel resonance circuit called rejector circuit. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 • Unit of LC is second 1F 1 H Example  R For what frequency the voltage across the resistance R will be maximum. Solution It happens at resonance f  1  1  500 Hz 2 LC 2 1  106  1  E 49

JEE-Physics Example R=220 A capacitor, a resistor and a 40 mH inductor are connected in series to an AC source of frequency 60Hz, calculate the capacitance of the capacitor, if V1 V2 V3 the current is in phase with the voltage. Also calculate the value of X and I. 300V 300V XV I Solution 110V, 60Hz At resonance L  1 , 1 1 1  176F C C  4 2 f2 L  42  (60)2  40  103 2 L V 110 V =V  X = 110 V and I    0.5 A R R 220 Example A coil, a capacitor and an A.C. source of rms voltage 24 V are connected in series, By varying the frequency of the source, a maximum rms current 6 A is observed, If this coil is connected to a bettery of emf 12 V, and internal resistance 4 , then calculate the current through the coil. Solution V V 24 At resonance current is maximum. I =    Resistance of coil R = = 6 =4 R I E 12 = 1.5 A When coil is connected to battery, suppose I current flow through it then I = R  r = 4  4 Example Radio receiver recives a message at 300m band, If the available inductance is 1 mH, then calculate required capacitance Solution Radio recives EM waves. ( velocity of EM waves c = 3 x 108 m/s)  c = f  f = 3 108 = 106 Hz Now f= 1 106 C = 1 = 25 pF =1× 42L  1012 300 2 LC Example C Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 In a L–C circuit parallel combination of inductance of 0.01 H and a AL capacitor of 1 F is connected to a variable frequency alternating current source as shown in figure. Draw a rough sketch of the current variation as the frequency is changed from 1kHz to 3kHz. Solution L and C are connected in parallel to the AC source, 104 I  so resonance frequency f  1  1  1.6kHz 2 LC 2 0.01  106 2 In case of parallel resonance, current in L–C circuit at resonance is zero, 1.6kHz f so the I-f curve will be as shown in figure. E 50

JEE-Physics POWER IN AC CIRCUIT The average power dissipation in LCR AC circuit Let V = V sint and I = I sin (t – ) 0 0 Instantaneous power P = (V sint)(I0 sin(t – ) = VI sint (sintcos – sincost) 0 00 Average power <P> = 1 T (V0 I0 sin2 t cos   V0 I0 sin t cos t sin )dt T 0 1 T 1 T  1   = 0 T 0 dt   2 V0 I0  T sin2 t cos dt  sin t cos t sin  V0 I0 cos   0  sin    <P> = V0 I0 cos  = Vrms Irm,s cos 2 Instantaneous Average power/actual power/ Virtual power/ apparent Peak power power P=V I P = VI dissipated power/power loss Power/rms Power 00 P = V I cos  P=V I rms rms rms rms • I cos is known as active part of current or wattfull current, workfull current. It is in phase with voltage. rms • I sinis known as inactive part of current, wattless current, workless current. It is in quadrature (90) with voltage. rms Power factor : Average power P  Erms Irms cos   r m s power  cos  Average power R Power factor (cos ) r m s Power and cos= Z Power factor : (i) is leading if I leads V (ii) is lagging if I lags V GOLDEN KEY POINTS • P <P . av rms • Power factor varies from 0 to 1 • Pure/Ideal  V Power factor = cos Average power V .I R 0 V, I same Phase 1 (maximum) rms rms L  V leads I 0 2 0  00 C  V lags I 2 Choke coil  V leads I 00  2 • At resonance power factor is maximum ( = 0 so cos = 1) and P =V I av rms rms Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Example 1 A voltage of 10 V and frequency 103 Hz is applied to F capacitor in series with a resistor of 500. Find  the power factor of the circuit and the power dissipated. Solution  XC 11  500 Z = R2 2 (500)2  (500)2  500 2     X C  2f C 2  103  106  R 500 1 =Vrms Irms cos  V2 cos (1 0 )2 1  1 W Power factor cos= = =  Power rms =  2 10 Z 500 2 dissipated Z 2 500 2 E 51

JEE-Physics Example  If V = 100 sin 100 t volt and I = 100 sin (100 t + ) mA for an A.C. circuit then find out 3 (a) phase difference between V and I (b) total impedance, reactance, resistance (c) power factor and power dissipated (d) components contains by circuits Solution  (a) Phase difference    (I leads V) 3 (b) Total impedance Z= V0 100  1k Now resistance R  Z cos 60  1000  1  500 I0  100  103 2 reactance X  Z sin 60  1000  3 500 R  60° 23 X Z (c)  = – 60°  Power factor = cos = cos (–60°) = 0.5 (leading) Power dissipated P  Vrms Irms cos   100  0.1  1  2.5 W 2 2 2 (d) Circuit must contains R as   and as  is negative so C must be their, (L may exist but X > X ) Example 2 CL 1 If power factor of a R-L series circuit is 2 when applied voltage is V = 100 sin 100t volt and resistance of circuit is 200 then calculate the inductance of the circuit. Solution cos = R  1R Z = 2R R2  X 2 = 2R  XL 3 R = L Z 2Z L = 3 R  L= 3R 3  200 = 2 3 = H  100  Example A circuit consisting of an inductance and a resistacne joined to a 200 volt supply (A.C.). It draws a current of 10 ampere. If the power used in the circuit is 1500 watt. Calculate the wattless current. Solution Apparent power = 200 × 10 = 2000 W True power 1500 3 Power factor cos = Apparent = = power 2000 4  3  2 10 7  4   Wattless current = rms sin  = 10 1  A 4 Example Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 A coil has a power factor of 0.866 at 60 Hz. What will be power factor at 180 Hz. Solution Given that cos  = 0.866,  = 2f = 2 × 60 = 120 rad/s, ’ = 2f’ = 2 × 180 = 360 rad/s Now, cos  = R/Z  R = Z cos  = 0.866 Z But Z = R 2  (L )2  L = Z2  R 2 = Z2  (0.866 Z )2 = 0.5 Z  L = 0.5Z 0.5Z =  120 When the frequency is changed to ’ = 2 × 180 = 3 × 120 = 300 rad/s, then inductive reactance ’ L = 3 L = 3 × 0.5 Z = 1.5 Z  New impedence Z’ = [R ' ( 'L )2 ] = (0.866 Z)2  (1.5 Z)2 = Z [(0.866 )2  (1.5 )2 ] = 1.732Z  New power factor R 0.866 Z = 0.5 == 52 E Z ' 1.732 Z

JEE-Physics CHOKE COIL In a direct current circuit, current is reduced with the help of a resistance. tube light rod Hence there is a loss of electrical energy I2 R per sec in the form of heat in the resistance. But in an AC circuit the current can be reduced by choke coil which involves very small amount of loss of energy. Choke starter coil is a copper coil wound over a soft iron laminated core. This coil is put in series with the circuit in which current is to be reduced. It also known as ballast. choke coil Circuit with a choke coil is a series L-R circuit. If resistance of choke coil = r (very small) The current in the circuit E with Z  (R  r)2  (L )2 So due to large inductance L of the coil, the I Z current in the circuit is decreased appreciably. However, due to small resistance of the coil r, rr r  cos     0 The power loss in the choke P= V I cos  0 Z r2  2 L2 L av rms rms GOLDEN KEY POINT • Choke coil is a high inductance and negligible resistance coil. • Choke coil is used to control current in A.C. circuit at negligible power loss • Choke coil used only in A.C. and not in D.C. circuit • Choke coil is based on the principle of wattless current. • Iron cored choke coil is used generally at low frequency and air cored at high frequency. • Resistance of ideal choke coil is zero Example A choke coil and a resistance are connected in series in an a.c circuit and a potential of 130 volt is applied to the circuit. If the potential across the resistance is 50 V. What would be the potential difference across the choke coil. Solution V = VR2  VL2  VL  V 2  VR2  (130)2  (50)2 = 120 V Example An electric lamp which runs at 80V DC consumes 10 A current. The lamp is connected to 100 V – 50 Hz ac source compute the inductance of the choke required. Solution Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Resistance of lamp V 80  8 R  I 10 Let Z be the impedance which would maintain a current of 10 A through the Lamp when it is run on V 100 Z = R 2  (L )2 100 Volt a.c. then. Z = I = 10 = 10  but  (L)2 = Z2 – R2 = (10)2 – (8)2 = 36L = 6 L = 6 = 6 = 0.02H  2  50 Example Calculate the re sistance or i nductance required to operate a lamp (60V, 10W) from a source of (100 V, 50 Hz) E 53

JEE-Physics Solution Maximum voltage across lamp = 60V R (a) 100V, 50Hz  V + V = 100  V = 40V Lamp R R Now current througth Lamp is = Wattage = 10 1 = A voltage 60 6 But V = IR  1  R = 240  L R 40 = (R) 6 (b) Now in this case (V )2 + (V )2 = (V)2 Lamp L (60)2 + (VL)2 = (100)2  VL = 80 V 100V, 50Hz 1 X = 80 × 6 = 480  = L (2f)  L = 1.5 H Also V = IX = X so L L L 6L A capacitor of suitable capacitance replace a choke coil in an AC circuit, the average power consumed in a capacitor is also zero. Hence, like a choke coil, a capacitor can reduce current in AC circuit without power dissipation. Cost of capacitor is much more than the cost of inductance of same reactance that's why choke coil is used. Ex am p l e A choke coil of resistance R and inductance L is connected in series with L, R C a capacitor C and complete combination is connected to a.c. ~ voltage, Circuit resonates when angular frequency of supply is  = 0. (a) Find out relation betwen 0, L and C V=V0 sinwt(volt) (b) What is phase difference between V and I at resonance, is it changes when resistance of choke coil is zero. Solution 11 (a) At resonance condition XL = XC  0L = 0C     0 = LC (b)   cos  = R = R = 1  = 0° No, It is always zero. ZR LC OSCILLATION Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called LC Oscillation. UNDA MPED OSCILLATION When the circuit has no resistance, the energy taken once from the source and given to capacitor keeps on oscillating between C and L then the oscillation produced will be of constant amplitude. These are called undamped oscillation. I CL t 54 E

JEE-Physics After switch is closed Q di Q d2Q d2Q 1 L 0  L 0   Q0 C dt2 C dt dt2 LC By comparing with standard equation of free oscillation d2x  2 x    dt2 0  2  1 Frequency of oscillation f  1 LC 2 LC Charge varies sinusoidally with time q = q cos t m dq  current also varies periodically with t I = dt = q  cos (t  ) m 2 If initial charge on capacitor is q then electrical energy strored in capacitor is U = 1 q 2 m m E 2C At t = 0 switch is closed, capacitor is starts to discharge. As the capacitor is fully discharged, the total electrical energy is stored in the inductor in the form of magnetic energy. U= 1 L I 2m where I = max. current B 2 m (U ) = (U ) 1 q 2  1 L I 2m m 2 max EPE max MPE 2C DA MPED OSCILLATION Practically, a circuit can not be entirely resistanceless, so some part of energy is lost in resistance and amplitude of oscillation goes on decreasing. These are called damped oscillation. RL C Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65  1 R2 Angular frequency of oscillation  LC 4L2 I frequency of oscillation f  1 1 R2  2 LC 4L2 t 1 R2 0 oscillation to be real if  LC 4L2 1 R2 Hence for oscilation to be real  LC 4L2 E 55

JEE-Physics GOLDEN KEY POINTS • In damped oscillation amplitude of oscillation decreases exponentially with time. • At t  T , 3T , 5T ..... energy stored is completely magnetic. 44 4 • At t  T , 3T , 5T ..... energy is shared equally between L and C 88 8  when charge is maximum, current minimum • Phase difference between charge and current is when charge is minimum,current maximum 2 Example An LC circuit contains a 20mH inductor and a 50F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed to be t = 0. (a) What is the total energy stored initially. (b) What is the natural frequency of the circuit. (c) At what time is the energy stored is completely magnetic. (d) At what times is the total energy shared equally between inductor and the capacitor. Solution 1 q2 1 (10  103 )2 (a) U= =  = 1.0J 2 50  106 E 2C (b)  = 1 = 1 = 103 rad/sec  f = 159 Hz LC 20  10 3  50  10 6 (c)  q = q cos t 0 Energy stored is completely magnetic (i.e. electrical energy is zero, q = 0) at t  T , 3T , 5T ......... where 1 = 6.3 ms 44 4 T f (d) Energy is shared equally between L and C when charge on capacitor become q0 2 T 3T 5T Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 so, at t  , , ...... energy is shared equally between L and C 88 8 56 E

JEE-Physics SOME WORKED OUT EXAMPLES Example#1 For the given circuit (A) The phase difference between I & IR1 is 0° L (B) The phase difference between V & VR2 is 90° C (C) The phase difference between I & IR1 is 180° L (D) The phase difference between V & VR2 is 180° C Solution Ans. (B) The phase difference between V and VR 2 is  rad or 90° C Example#2 A periodic voltage V varies with time t as shown in the figure. T is the time period. The r.m.s. value of the voltage is :- V V0 T/4 T t (A) V0 (B) V0 (C) V (D) V0 8 2 0 4 Solution Ans. (B) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 T/4 V02  T  V02 T  V02  V0  4   4  42  V02 dt   Root mean square value <V> = T T 0 T  dt 0 Example#3 The potential difference V and current I flowing through the AC circuit is given by V = 5 cos(t – /6) volt and I = 10sint ampere. The average power dissipated in the circuit is 25 3 (B) 12.5 W (C) 25 W (D) 50 W (A) W 2 E 57

JEE-Physics Solution Ans. (B) V = 5 cos (t - /6); i = 10 sin t = 10 cos (t - /2)      P  VI cos   5 10  1  12.5W 2 6 3; 2 22 Example#4 The radius of a coil decreases steadily at the rate of 10–2 m/s. A constant and uniform magnetic field of in- duction 10–3 Wb/m2 acts perpendicular to the plane of the coil. The radius of the coil when the induced e.m.f. in the coil is 1V, is :- 2 3 4 5 (A)  cm (B)  cm (C)  cm (D)  cm Solution Ans. (D) e  d  d dr e 1 0 6 5 dt dt (r2B) = 2rB r  r   cm dt  dr  2 103 102   2 B dt  Example#5 . .2H A time varying voltage V = 2t volt is applied across an ideal inductor of inductance L = 2H as shown in figure. Then select incorrect statement V=2t (A) current versus time graph is a parabola (B) energy stored in magnetic field at t = 2 s is 4J Ans. (D) (C) potential energy at time t = 1 s in magnetic field is increasing at a rate of 1 J/s (D) energy stored in magnetic field is zero all the time Solution di = 2t  2 × di di t2  i – t graph parabola dt V = 2t  L dt = 2t  dt  i = 2 11 dU di t2 U= Li2 = × 2 × 4 = 4J and dt =Li dt =2× × t = t3 = 1 J/s 2 2 2 Example#6 A circular coil of 500 turns encloses an area of 0.04 m2. A uniform magnetic field of induction Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 0.25 Wb/m2 is applied perpendicular to the plane of the coil. The coil is rotated by 90° in 0.1 second at a constant angular velocity about one of its diameters. A galvanometer of resistance 25 was connected in series with the the coil. The total charge that will pass through the galvanometer is - (A) 0.4 C (B) 1 C (C) 0.2 C (D) Zero Solution Ans. (C) Induced current I  e  dq  e   d  1  q    NBA R dt R  dt  R R R Total charge 500  0.25  0.04 q = 25 = 0.2 C 58 E

JEE-Physics Example#7 A condenser of capacity 6 µF is fully charged using a 6-volt battery. The battery is removed and a resistanceless 0.2 mH inductor is connected across the condenser. The current which is flowing through the inductor when one-third of the total energy is in the magnetic field of the inductor is :- (A) 0.1 A (B) 0.2 A (C) 0.4 A (D) 0.6 A Solution Ans. (D) 11 Total energy = Intial energy on capacitor = CV2, Magnetic field energy = LI2 22  1 1 1 CV2  CV 2 6 106  6  6 LI2 = 32 I=  = 0.6 A 3L 3  2.0 103 2 Example#8 For the circuit shown, which of the following statement(s) is(are) correct? (A) Its time constants is 2 second. Ans. (B) (B) In steady state, current through inductance will be 1A. (C) In steady state, current through 4 resistance will be 2/3 A. (D) In steady state, current throgh 8 resistance will be zero. Solution Time constant   L  12  1s R 12 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 In steady state current through 4 resistance = 6 1  1 A 6 12 3 E 59

JEE-Physics Example#9 In the circuit shown the capacitor has charge Q. At t = 0 sec the key is closed. The charge on the capacitor at the instant potential difference across the inductor L is zero, is 1 (A) Q Q 2Q (D) 0 (B) 3 (C) 3 Solution Ans. (D) When V = 0  i = i  q = 0 across L max c Example#10 The figure shows a rod of length  with points A and B on it. The rod is moved in a uniform magnetic field (B ) in different ways as shown. In which case potential difference (V –V ) between 0 AB A & B is minimum? B0 B B0 B0 B A /2 B B0   2 /2 A  v= /2 /2 B  A (A) (B) (C) (D)  A Solution Ans. (C) For (A): VB  VA  1 B 2  VA  VB  1 B2 2 2 For (B) : VB  VA  Bv  B      1 B 2  VA  VB   1 B2  2  2 2 For (C) : V _ V = 0 AB 1 B2 1    2 3 B2  3 B2 2 2  2  8 8 For (D) : VB  VA   B   VA  VB  Example#11 A bent rod PQR (PQ = QR = ) shown here is rotating about its end P with angular speed  in a region of transverse magnetic field of strength B. Q (A) e.m.f. induced across the rod is B2 60° Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 (B) e.m.f. induced across the rod is B2/2 B (C) Potential difference between points Q and R on the rod is B2/2  R (D) Potential difference between points Q and R on the rod is zero P Solution Ans. (B,D) The rod in equivalent to a rod joining the ends P and R of the rod rotating is the same sense. 60° B P R R VR  VP  B2 ; VQ  VP  B 2 ; VR  VP  B2 ; V –V =0 2 2 2 Q R 60 E

Example#12 JEE-Physics A conducting loop is kept so that its center lies at the origin of the coordinate system. A magnetic field has the induction B pointing along Z-axis as shown y B x in the figure (A) No emf and current will be induced in the loop if it rotates about Z-axis (B) emf is induced but no current flows if the loop is a fiber when it rotates about y-axis. (C) emf is induced and induced current flows in the loop if the loop is made of copper & is rotated about y-axis. (D) If the loop moves along Z-axis with constant velocity, no current flows in it. Solution Ans. (A,C,D) If the loop rotates about Z axis, the variation of flux linkage will be zero. Therefore no emf is induced. Consequently no current flows in the loop. When it rotates about y axis, its flux linkage changes. However, in insulators there can not be motional emf. If the loop is made of copper, it is conductive therefore induced current is set up. If the loop moves along the Z axis variation of flux linkage is zero. Therefore the emf and current will be equal to zero.n Example#13 Initially key was placed on (1) till the capacitor got fully charged. Key is placed on (2) at t=0. The time when the energy in both capacitor and inductor will be same- (1) (2) L E C  LC  LC 5 LC 5 LC (A) (B) (C) (D) 4 2 4 2 Solution. Ans. (A,C) For given situation q L di 0  d2q  q 0  d2q  2 q 0  q = q cos t & i = –q sint C dt dt2 LC dt2 0 0 According to given conditions q2  1 Li2  q0 cos2 t  1 L q 2 2 sin2 t 2C 2 2C 2 0  cot2t = 1  t =  , 3 , 5 , 7 .........  t   LC , 3 LC , 5 LC , 7 LC ..... 44 4 4 44 4 4 Example#14 For an LCR series circuit, phasors of current i and applied voltage V = V sint are shown in diagram 0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 at t =0. Which of the following is/are CORRECT?  i0 V0 (A) At t = 2 , instantaneous power supplied by source is negative.  /3 2 (B) From 0 < t < 3 , average power supplied by source is positive. 5 (C) At t = 6 , instantaneous power supplied by source is negative. (D) If  is increased slightly, angle between the two phasors decreases. E 61

JEE-Physics Solution i/v Ans. (BCD) current The graph shows V & I as function of time. Current leads the voltage by /3 Power is positive if V & I are of same sign. t Voltage Power is negative if V & I are of opposite sign If  1  thus angle decreases. C Example#15 to 17 Consider a conducting circular loop placed in a magnetic field as shown in figure. When magnetic field changes with time, magnetic flux also changes and emf e   d is induced. dt If resistance of loop is R then induced current is i  e . For current, charges must have non–zero average R velocity. Magnetic force cannot make the stationary charges to move. Actually there is an induced electric field in the conductor caused by changing magnetic flux, which makes the charges to move,       d .This E d dt induced electric field is non conservative by nature. 1 5 . A cylindrical space of radius R is filled with a uniform magnetic induction B dB parallel to the axis of the cylinder. If dt = constant, the graph, showing the variation of induced electric field with distance r from the axis of cylinder, is (A) (B) (C) (D) 16. A square conducting loop is placed in the time varying magnetic field  dB  ve constant  . The centre of Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65  dt  square coincides with axis of cylindrical region of magnetic field. The directions of induced electric field at point a, b and c. 62 E

JEE-Physics (A) (B) (C) (D) 1 7 . A line charge  per unit length is pasted uniformly onto the rim of a wheel of mass m and radius R. The wheel has light non–conducting spokes and is free to rotate about a vertical axis as shown in figure.A uniform magnetic field B exist as shown in figure. What is the angular velocity of the wheel when the field is suddenly switched off? 2a2 B a2 B 3 a2B a2 B (A) (B) (C) (D) mR mR mR 2mR Solution 15. Ans. (B) For r  R; E 2r   r2 dB  E  r For r  R; E 2r   R 2 dB  E  1 dt dt r 16. Ans. (A) Induced electric field is circular. 17. Ans. (B)   L I mR 2 but    2R  ER    a 2 B  R  a2 BR m R 2 a2BR a2 B t   t  t   t t t t mR Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Example#18 to 20 A conducting square wire frame ABCD of side  is pulled by horizontal force so that it moves with constant velocity v. A uniform magnetic field of strength B is existing perpendicular to the plane of wire. The resistance per unit length of wire is  and negligible self inductance. If at t= 0, frame is just at the boundary of magnetic field. Then E 63

JEE-Physics  1 8 . The emf across AB at t = 2v is Bv (B) zero (C) Bv (D) None of these (A) (D) None of these (D) None of these 4  1 9 . Potential difference across BC at time t = 2v is Bv (B) Bv 3Bv (A) (C) 4 4  2 0 . Find the applied horizontal force on BC, (F) as a function of time 't' (t < v ) B2 v B2 v B2 v (A) t (B) (C)  2 4 Solution 18. Ans. (B) e   .    as  |    = 0° e=0 B ( v) e| v 19. Ans. (C) A B Bv 3 B v 4() 4 V = E – ir = Bv – []  DC 20. Ans. (C) F  Bi  B (B v )  B22 v  B 2 v 4() 4() 4 Example#21 Magnetic flux in a circular coil of resistance 10 changes with time as shown in figure.  direction indicates a direction perpendicular to paper inwards. (Wb) ××× 10 × × × Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 8 10 14 16 26 t(s) × × × –10 ××× Column–I Column–II (A) At 1 second induced current is (P) Clockwise (B) At 5 second induced current is (Q) Anticlockwise (C) At 9 second induced current is (R) 0.5 A (D) At 15 second induced current is (S) 5 A (T) None of these 64 E

JEE-Physics Solution Ans. (A) Q,R (B) T (C) P,R (D) Q,R For (A) d  5  i 5 1 A, Anticlockwise For (B) d  0 i=0 = zero dt  dt 10 2 For (C) d = –5  i  5  1 A, Clockwise For (D) d =5  i 5  1 A , Anticlockwise dt   dt  2 10 2 10 Example#22 In the circuit shown in figure E=25V, L=2H, C=60 F, R1 = 5 and R2 = 10. Switch S is closed at t = 0. S E CL R1 Column-I (A) Current through R1 at t = 0 R2 (B) Current through R2 at t = 0 Column-II (C) Current through R1 at t =  (P) 0 (D) Current through R2 at t =  (Q) 5A (R) 2.5 A Solution (S) 7.5 A (T) None of these Ans. (A)(Q), (B)(P), (C)(P), (D)(R) At t = 0 At t =  EE R1 R2 E 25 I1 = R1 = = 5A R1 R2 5 I1 = 0 I2 = 0 E 25 = 2.5 A I2 = R 2 = 10 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 Example#23 A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region is proportional to 1/rn. Find the value of n. r xx P x x ax Solution Ans. 1 d  E 2r   E  1  n  1 dt r E 65

JEE-Physics Example#24 An inductor (X = 2) a capacitor (X = 8) and a resistance (8) is connected in series with an ac source. The LC voltage output of A.C source is given by V = 10 cos 2 50t. Find the instantaneous p.d. between A and B when the voltage output from source is half of its maximum. XL=2 XC= 8 8 A B ~ Solution Ans. 3 VAB  6   10  3   R AB  X C  X C and Z  X C  X L 2  R2 10  2  Example#25 Figure shows a uniform circular loop of radius ‘a’ having specific resistance  placed in a uniform magnetic field B perpendicular to plane of figure. A uniform rod of length 2a & resistance R moves with a velocity v as shown. a Find the current in the rod when it has moved a distance from the centre of circular loop. 2 [Given B = 3, a =2, v =3,   9 , R = 2/3 all in SI units] 4 x xx x B xa v x x x a/2 x xx x xx x Solution Ans. 6 6 3 1 3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\03 AC.p65 a3 × B × v = 18 V 18  i  18  6 A 3 66 E



JEE-Physics ATOMIC STRUCTURE VARIOUS MODELS FOR STRUCTURE OF ATOM • Dalton's Theory Every material is composed of minute particles known as atom. Atom is indivisible i.e. it cannot be subdivided. It can neither be created nor be destroyed. All atoms of same element are identical physically as well as chemically, whereas atoms of different elements are different in properties. The atoms of different elements are made up of hydrogen atoms. (The radius of the heaviest atom is about 10 times that of hydrogen atom and its mass is about 250 times that of hydrogen). The atom is stable and electrically neutral. • Thomson's Atom Model The atom as a whole is electrically neutral because the positive charge present on the atom (sphere) is equal to the negative charge of electrons present in the sphere. Atom is a positively charged sphere of radius 10–10 m in which electrons are embedded in between. The positive charge and the whole mass of the atom is uniformly distributed throughout the sphere. electron positively charged matter • Shortcomings of Thomson's model (i) The spectrum of atoms cannot be explained with the help of this model (ii) Scattering of –particles cannot be explained with the help of this model RUTHERFORD ATOM MODEL • Rutherford experiments on scattering of  – particles by thin gold foil The experimental arrangement is shown in figure. –particles are emitted by some radioactive material (polonium), kept inside a thick lead box. A very fine beam of –particles passes through a small hole in the lead screen. This well collimated beam is then allowed to fall on a thin gold foil. While passing through the gold foil, –particles are scattered through different angles. A zinc sulphide screen was placed out the other side of the gold foil. This screen was movable, so as to receive the –particles, scattered from the gold foil at angles varying from 0 to 180°. When an –particle strikes the screen, it produces a flash of light and it is observed by the microscope. It was found that : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65vacuumZnS lead box N()leadgold foil screenmicroscope screen 10–7m source of -particle beam of N() cosec4  -particle 2  most -pass through about 1 in 8000 is some are deviated 90° 180° repelled back through large angle   E1

JEE-Physics • Most of the  – particles went straight through the gold foil and produced flashes on the screen as if there were nothing inside gold foil. Thus the atom is hollow. • Few particles collided with the atoms of the foil which have scattered or deflected through considerable large angles. Few particles even turned back towards source itself. • The entire positive charge and almost whole mass of the atom is concentrated in small centre called a nucleus. • The electrons could not deflected the path of a  – particles i.e. electrons are very light. • Electrons revolve round the nucleus in circular orbits. So, Rutherford 1911, proposed a new type of model of the atom. According to this model, the positive charge of the atom, instead of being uniformly distributed throughout a sphere of atomic dimension is concentrated in a very small volume (Less than 10–13m is diameter) at it centre. This central core, now called nucleus, is surrounded by clouds of electron makes. The entire atom electrically neutral. According to Rutherford scattering formula, the number of  – particle scattered at an angle  by a target are given by N  16 N 0nt(2Ze2 )2 2 )2  1  (40 )2 r2 (m v 0 sin4 2 Where N = number of  – particles that strike the unit area of the scatter 0 Number of target atom per m3 n= Thickness of target t= Charge on the target nucleus Ze = Charge on  – particle 2e = Distance of the screen from target r= Velocity of  – particles at nearer distance of approach the size of a v= nucleus or the distance of nearer approach is given by 0 r0 1 (2Ze)2 1 (2Ze)2    4 0 1  4 0 EK where E = K.E. of particle  2 2  K m v 0 r0 + nucleus Ze electronnucleus+ -particle  NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 b target nucleus area = b2 Bohr's Atomic Model In 1913 Neils Bohr, a Danish Physicist, introduced a revolutionary concept i.e., the quantum concept to explain the stability of an atom. He made a simple but bold statement that \"The old classical laws which are applicable to bigger bodies cannot be directly applied to the sub–atomic particles such as electrons or protons. Bohr incorporated the following new ideas now regarded as postulates of Bohr's theory. 1 . The centripetal force required for an encircling electron is provided by the electrostatic attraction between the 1 Zee mv2 v nucleus and the electron i.e. 40 r2  r ...(i) r Electron 0 = Absolute permittivity of free space = 8.85 × 10–12 C2 N–1 m–2 + m = Mass of electron v = Velocity (linear) of electron Nucleus r = Radius of the orbit in which electron is revolving. Z = Atomic number of hydrogen like atom. +Ze 2E

JEE-Physics 2 . Electrons can revolve only in those orbits in which angular momentum of electron about nucleus is an integral h nh multiple of . i.e., mvr = ...(ii) 2 2 n = Principal quantum number of the orbit in which electron is revolving. 3 . Electrons in an atom can revolve only in discrete circular orbits called stationary energy levels (shells). An electron in a shell is characterised by a definite energy, angular momentum and orbit number. While in any of these orbits, an electron does not radiate energy although it is accelerated. 4 . Electrons in outer orbits have greater energy than those in inner orbits. The orbiting electron emits energy when it jumps from an outer orbit (higher energy states) to an inner orbit (lower energy states) and also absorbs energy when it jumps from an inner orbit to an outer orbit. En – Em = hn,m E3 E2 where, En = Outer energy state E1 Em = Inner energy state + n,m = Frequency of radiation 5 . The energy absorbed or released is always in the form of electromagnetic radiations. Nucleus MATHEM ATICAL ANALYSIS OF BOHR'S THEORY 1 Zee mv2 nh From above equation (i) and (ii) i.e.,  and mvr = ...(ii) 4 0 r2 r 2 We get the following results. 1 . Velocity of electron in nth orbit : By putting the value of mvr in equation (i) from (ii) we get 1 Ze2   nh  v v Z  e2  Z v0 4 0  2     ....(iii) n  2  h  n  0   1.6  1019 2 c = 2.2 × 106 m/s Where, v0 = = 2.189 × 106 ms–1 = 2  8.85  10 12  6.625  10 34 137 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 where c = 3 × 108 m/s = speed of light in vacuum 2 . Radius of the nth orbit : From equation (iii), putting the value of v in equation (ii), we get  Z e2  nh n2  0h2   n2 r0 ...(iv) m  n  20h  r  2  r= Z  m e 2  Z    8.85  1012  6.625  1034 2 2 = 0.529 × 10–10 = 3.14 where  r0 m  0.53Å  9.11  1031 1.6  1019  E3

JEE-Physics 3 . Total energy of electron in nth orbit : From equation (i) KE  1 mv2  Ze2 and PE  1 Zee  |PE| = 2 KE 2 80r  2K.E. 40 r Ze2 Total energy of the system E = KE + PE = –2KE + KE = – KE = 8 0 r By putting the value of r from the equation (iv), we get Z2  me4  Z2 ...(v) E  n2   8 20h2   n2 E0     9.11  103 1.6  1019 4    where  13.6 eV E0  2 2 8 8.85  1012 6.625  1034  2 r 4 . Time period of revolution of electron in nth orbit : T  v n3  4 20 h 3  n3 Z2  me  Z2 By putting the values of r and v, from (iii) and (iv) T   4   T0 where, T0     4  8.85  1012 2  6.625  1034 3 = 1.51 × 10–16 second  9.11  1031  1.6  1019 4 5 . Frequency of revolution in nth orbit : 1 Z2 me4 Z2  9.11  1031  1.6  1019 4    f  T  n3  420h3  n3  f0 f0  = 6.6 × 1015 Hz where, 4 8.85  10 12 2 6.625  1034 3 6 . Wavelength of photon E  E n2  E n1  me4 1  1  Z2  1 3 .6 1  1  Z2  E  hc  1   me4  1  1  Z2 8 20 h2          n 2 n 2   n12 n 2  8 20 h 3 c  n 2 n 2  1 2 2 1 2  R 1  1  Z2 where,  is called wave number. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65    n 2 n 2   1 2 R   R H = Rydberg constant  9.11  1031  1.6  1019 4    = 8  8.85  1012 2 6.625  1034 3  3  10 = 1.097 × 107 m–1 = 1.097 × 10–3 Å–1 (for stationary nucleus). If nucleus is not stationary (i.e., mass of nucleus is not much greater than the mass of the revolving particle like electron), then R = R where, m = mass of revolving particle and M = mass of nucleus 1m/M 4E

JEE-Physics SPECTRAL SERIES OF HYDROGEN ATOM It has been shown that the energy of the outer orbit is greater than the energy of the inner ones. When the Hydrogen atom is subjected to external energy, the electron jumps from lower energy state i.e. the hydrogen atom is excited. The excited state is unstable hence the electron return to its ground state in about 10–8 sec. The excess of energy is now radiated in the form of radiations of different wavelength. The different wavelength constitute spectral series. Which are characteristic of atom emitting, then the wave length of different members of series can be found from the following relations   1  R 1  1      n12 n 2  2 This relation explains the complete spectrum of hydrogen. A detailed account of the important radiations are listed below. • Lyman Series : The series consist of wavelength which are emitted when electron jumps from an outer orbits to the first orbit i. e., the electronic jumps to K orbit give rise to lyman series. Here n = 1 & n = 2, 3, 4, ....... 1 2 The wavelengths of different members of Lyman series are : • First member : In this case n = 1 and n = 2 hence 1  R 1  1   3R 12  12 22  4  4 4  3R   = 1216 × 10–10 m = 1216Å 3 10.97 106 • Second member : In this case n = 1 and n = 3 hence 1  R 1  1   8R 12  12 32  9 9 9 = 1026 × 10–10 m =1026Å     8R 8 10.97 106 Similarly the wavelength of the other members can be calculated. • Limiting members : In this case n = 1 and n =  , hence 1  R 1  1   R 12  12 2     1    1 = 912 × 10–10m = 912Å R 10.97 106 This series lies in ultraviolet region. • Balmer Series : This series is consist of all wave lengths which are emitted when an electron jumps from an outer orbit to the second orbit i. e., the electron jumps to L orbit give rise to Balmer series. Here n = 2 and n = 3, 4, 5 ......... The wavelength of different members of Balmer series. 12 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 • First member : In this case n = 2 and n = 3, 1  R  1 1   5R 1 2 hence  22   36  32    36 36 = 6563 × 10–10m = 6563Å 5R   5 10.97 106 • Second member : In this case n = 2 and n = 4, hence 1  R  1  1   3R 1 2   22 42  16 16 16 = 4861 × 10–10m = 4861Å     3R 3 10.97 106 • Limiting members: In this case n = 2 and n =  , hence 1  R 1  1  R  4 = 3646Å 1 2   2 2   4 R This series lies in visible and near ultraviolet region. E5

JEE-Physics • Paschen Series : This series consist of all wavelength are emitted when an electron jumps from an outer orbit to the third orbit i. e., the electron jumps to M orbit give rise to paschen series. Here n =3 & n= 4, 5, 6 ..... 1 2 1 1 1  The different wavelengths of this series can be obtained from the formula  R      3 2 n 2  2 where n = 4, 5, 6 ... 2 For the first member, the wavelengths is 18750Å. This series lies in infra–red region. • Bracket Series : This series is consist of all wavelengths which are emitted when an electron jumps from an outer orbits to the fourth orbit i. e., the electron jumps to N orbit give rise to Brackett series. Here n1 = 4 & n2 = 5, 6, 7, ...... 1 1 1  The different wavelengths of this series can be obtained from the formula  R      4 2 n 2  2 where n = 5, 6, 7 .......... 2 This series lies in infra–red region of spectrum. • Pfund series : The series consist of all wavelengths which are emitted when an electron jumps from an outer orbit to the fifth orbit i. e., the electron jumps to O orbit give right to Pfund series. Here n = 5 and n = 6, 7, 8 ......... 1 2 1 1 1  The different wavelengths of this series can be obtained from the formula  R      5 2 n 2  2 where n = 6, 7, 8 ....... 2 This series lies in infra–red region of spectrum. The result are tabulated below S. No. Series Value of n Value of n Position in the 1. Observed 12 Spectrum 1 2,3,4... Lym an Series Ultra Violet 2. Balm er Series 2 3, 4 , 5. . . V isible 3. Paschen Series 3 4, 5, 6 . .. .  Infra–red 4. Brackett Series 4 5, 6, 7 . .. .  Infra–red 5. Pfund Series 5 6, 7, 8 . .. .  Infra–red 5thexcited state n = 6 Continuum –0.38 eV NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 4thexcited state n = 5 –0.54 eV 3rdexcited state n = 4 Pfund Series –0.85 eV 2ndexcited state n =3 Brackett Series –1.5 eV Paschen Series 1st excited state n = 2 –3.40 eV Balmer Series Ground state n = 1 Lyman Series –13.6 eV 6 E

JEE-Physics EXCITATION AND IONISATION OF ATOMS Consider the case of simplest atom i. e., hydrogen atom, this has one electron in the innermost orbit i.e., (n = 1) and is said to be in the unexcited or normal state. If by some means, sufficient energy is supplied to the electron. It moves to higher energy states. When the atom is in a state of a high energy it is said to be excited. The process of raising or transferring the electron from lower energy state is called excitation. When by the process of excitation, the electron is completely removed from the atom. The atom is said to be ionized. Now the atom has left with a positive charge. Thus the process of raising the atom from the normal state to the ionized state is called ionisation. The process of excitation and ionisation both are absorption phenomena. The excited state is not stationary state and lasts in a very short interval of time (10–8 sec) because the electron under the attractive force of the nucleus jumps to the lower permitted orbit. This is accompanied by the emission of radiation according to BOHR'S frequency condition. photon of photon of wavelength  wavelength    + + Ze Ze spectrum spectrum origin of emission spectra origin of absorption spectra The energy necessary to excite an atom can be supplied in a number of ways. The most commonly kinetic energy (Wholly or partly) of the electrons is transferred to the atom. The atom is now in a excited state. The minimum potential V required to accelerate the bombarding electrons to cause excitation from the ground state is called the resonance potential. The various values of potential to cause excitation of higher state called excitation potential. The potential necessary to accelerate the bombarding electrons to cause ionisation is called the ionization potential. The term critical potential is used to include the resonance, excitation and ionisation potential. We have seen that the energy required to excite the electron from first to second state is 13.6 – 3.4 = 10.2 eV from first to third state is 13.6 – 1.5 = 12.1 eV., and so on. The energy required to ionise hydrogen atom is 0 – (– 13.6) = 13.6 eV. Hence ionization potential of hydrogen atom is 13.6 volt. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 SUCCESSES AND LIMITATIONS Bohr showed that Planck's quantum ideas were a necessary element of the atomic theory. He introduced the idea of quantized energy levels and explained the emission or absorption of radiations as being due to the transition of an electron from one level to another. As a model for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple, rational ordering of the electrons in larger atoms and quantitatively helps to predict a greater deal about chemical behavior and spectral detail. Bohr's theory is unable to explain the following facts : • The spectral lines of hydrogen atom are not single lines but each one is a collection of several closely spaced lines. • The structure of multielectron atoms is not explained. • No explanation for using the principles of quantization of angular momentum. • No explanation for Zeeman effect. If a substance which gives a line emission spectrum is placed in a magnetic field, the lines of the spectrum get splitted up into a number of closely spaced lines. This phenomenon is known as Zeeman effect. E7

JEE-Physics Example A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de–excitation. Ground state energy of hydrogen atom is –13.6 eV. Solution The energy released during de–excitation in hydrogen like atoms is given by : E n2  E n1  me4 1  1  Z2 8 02 h2   n 2 n 2   1 2 Energy released in de–excitation will be maximum if transition takes place from nth energy level to ground state i.e., 1 1  Z2 1 1  E2n  E1  1 3 .6     2 12 2n  = 204 eV ...(i) & also E2n  En  1 3 . 6   2n  Z = 40.8 eV...(ii) 2  n 2 2 Taking ratio of (i) to (ii), we will get 4n2 1  n2 =4  n=2 5 3 Putting n=2 in equation (i) we get Z2 = 204 16  Z=4 13.6 15  E = – 13.6 Z2  E = –13.6 × 42 = – 217.6 eV = ground state energy n n2 1 12 E is minimum if transition will be from 2n to 2n–1 i.e. between last two adjacent energy levels.  E min  E 2n  E 2n 1 = 13.6 1  1  4 2 = 10.57 eV  3 2 42  is the minimum amount of energy released during de–excitation. Example A single electron orbits around a stationary nucleus of charge +Ze where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from the second orbit to third orbit. Find (i) The value of Z. (ii) The energy required to excite the electron from the third to the fourth Bohr orbit. (iii) The wavelength of electronic radiation required to remove the electron from first Bohr orbit to infinity. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 (iv) Find the K.E., P.E. and angular momentum of electron in the 1st Bohr orbit. [ The ionization energy of hydrogen atom = 13.6 eV, Bohr radius = 5.3 × 10–11 m, Velocity of light = 3 × 108 m/s, Planck's constant = 6.6 × 10–34 J–s ] Solution The energy required to excite the electron from n to n orbit revolving around the nucleus with charge +Ze is 12 me4 1 1  1 1    8 20 h2  E n2 given by  E n1  n 2  n 2 Z2  E n2  E n1  Z2  13.6  n12  n 2   1 2  2  (i) Since 47.2 eV energy is required to excite the electron from n = 2 to n =3 orbit 12 8E

JEE-Physics 47.2 = Z2 × 13.6 1  1   Z2 = 47.2  36 = 24.988  25  Z=5  2 2 32  13.6  5 (ii) The energy required to excite the electron from n =3 to n =4 is given by 12 E – E = 13.6 1  1  25 13.6  7 43 Z2=5 32 42  = =16.53 eV 144 (iii) The energy required to remove the electron from the first Bohr orbit to infinity  is given by E  E3  13.6  Z2 1  1   13.6  25eV =340eV 12 2  In order to calculate the wavelength of radiation, we use Bohr's frequency relation  hc 6.6  1034  108  3 hf   36.397Å   = 13.6 × 25 × (1.6 × 10–19)J    1.6  10 19 13.6  25  (iv) K.E. = 1 m v 2  1  Ze2 r1 = 543.4 × 10–19 J P.E. = – 2 × K.E. = – 1086.8 × 10–19 J 2 1 2 4 g 0 h Angular momentum = mv r = =1.05 × 10–34 Js 1 1 2 The radius r of the first Bohr orbit is given by 1 r= 0h2 1 0.53  1010  0 h2  0.53  10 10 m  = 1.106 × 10–10 m = 0.106 Å 1   m e2  me2 Z 5 Example An isolated hydrogen atom emits a photon of 10.2 eV. (i) Determine the momentum of photon emitted (ii) Calculate the recoil momentum of the atom (iii) Find the kinetic energy of the recoil atom [Mass of proton= m = 1.67 × 10–27 kg] p Solution E 10.2  1.6  1019 (i) Momentum of the photon is p =  = 5.44 × 10–27 kg–m/s 1c 3 108 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 (ii) Applying the momentum conservation p = p = 5.44 × 10–27 kg–m/s 21 p2 p1 atom Photon  1  p 2 p2 (iii) K = mv2 (v = recoil speed of atom, m = mass of hydrogen atom) K= m  m  = 2 2 2m Substituting the value of the momentum of atom, we get K =  5.44  1027 2 = 8.86 × 10–27J 2  1.67  10 27 E9

JEE-Physics Physical quantity Formula Ratio Formulae of hydrogen atom Radius of Bohr orbit (rn) rn  n2h2 ; rn = n2 Å r1: r2:r3...rn =1:4:9...n2 4 2 m kZ e 2 0.53 Z 2 k Z e 2 Z 11 1 Velocity of electron in nth Bohr v = ; v =2.2 × 106 v :v :v ...v = 1 : : ... orbit (v ) n nh n n 123 n 23 n n p= 2 m ke 2 z ; p  Z p: p: p ... p = 1: 1 : 1 ... 1 n n 1 2 3 n 2 3n Momentum of electron (p ) nh n n 83k2Z2mc4 Z2 1:2:3...n = 1 : 1 : 1 ... 1 Angular velocity of electron(n) n= n3h 3 ; n  n3 8 27 n3 Time Period of electron (T ) T= n3h3 ; T  n3 T :T :T ...T = 1:8:27: ... :n3 n n n 4 2 k 2 Z 2 m e 4 Z2 123 n f= 4 2 k 2 Z 2 e 4 m ; f Z2 111 Frequency (f )  f : f : f ... f = 1 : : ... n n n3h3 n n3 123 n 8 27 n3 Orbital current (I ) I= 4 2 k 2 Z 2 m e 5 ; I Z2 11 1 n  I : I : I ... I =1 : : ... n n3h3 n n3 123 n 8 27 n3 Angular momentum (J ) J= nh ; Jn n J :J :J ...J = 1 : 2 : 3...n n n 123 n 2 164k3Z3me6 Z3 11 1 Centripetal acceleration (a ) a= ; an n4 a : a : a ... a = 1 : : ... n n n4h4 123 n 16 81 n4 Kinetic energy (E kn ) RchZ2 Z2 E K1 : E K2 ...E Kn = 1 : 1 : 1 ... 1 EKn  n2 ; EKn  n2 4 9 n2 2RchZ2 Z2 11 1 Potential energy (U ) U= ; U  U : U : U ...U = 1 : : ... n n n2 n n2 123 n 4 9 n2 RchZ2 Z2 11 1 Total energy (E ) E= ; E  E : E : E ... E = 1 : : ... n2 n n n2 n n2 123 n 4 9 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 10 E

JEE-Physics X–RAYS ROENTGEN EXPERIMENT Roentgen discovered X–ray. While performing experiment on electric discharge tube Roentgen observed that when pressure inside the tube is 10–3mm of Hg and applied potential is kept 25 kV then some unknown radiation are emitted by anode. These are known as X–ray. X–rays are produced by bombarding high speed electrons on a target of high atomic weight and high melting point. evacuated tube + electron target cathode X-rays V To Produce X–ray T hree T hi ngs are Required (i) Source of electron (ii) Means of accelerating these electron to high speed (iii) Target on which these high speed electron strike COOLIDGE METHOD Coolidge developed thermoionic vacuum X–ray tube in which electron are produced by thermoionic emission method. Due to high potential difference electrons (emitted due to thermoionic method) move towards the target and strike from the atom of target due to which X–ray are produced. Experimentally it is observed that only 1% or 2% kinetic energy of electron beam is used to produce X–ray. Rest of energy is wasted in form of heat. V(in kV) anode e T water e F target filament C X-ray W NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 window Characteristics of target (a) Must have high atomic number to produce hard X–rays. (b) High melting point to withstand high temperature produced. (c) High thermal conductivity to remove the heat produced (d) Tantalum, platinum, molybdenum and tungsten serve as target materials • Control of intensity : The intensity of X–ray depends on number of electrons striking the target and number of electron depend on temperature of filament which can be controlled by filament current. Thus intensity of X–ray depends on current flowing through filament. E 11

JEE-Physics • Control of Penetrating Power: The Penetrating power of X–ray depends on the energy of incident electron. The energy of electron can be controlled by applied potential difference. Thus penetrating power of X–ray depend on applied potential difference. Thus the intensity of X–ray depends on current flowing through filament while penetrating power depends on applied potential difference Soft X–ray Hard X–ray Wavelength 10 Å to 100 Å 0.1 Å – 10 Å Energy 12400 12400  eV–Å  eV–Å Penetrating power Use Less More Radio photography Radio therapy • Continuous spectrum of X–ray : 12mv12 When high speed electron collides from the atom of target and passes e– v1 h close to the nucleus. There is coulomb attractive force due to this electron pXh-roatyon is deaccelerated i.e. energy is decreased. The loss of energy during e– v2 12mv22 deacceleration is emitted in the form of X–rays. X–ray produced in this + way are called Braking or Bremstralung radiation and form continuous spectrum. In continuous spectrum of X–ray all the wavelength of X–ray K are present but below a minimum value of wavelength there is no X–ray. It L M is called cut off or threshold or minimum wavelength of X–ray. The minimum wavelength depends on applied potential. • Loss in Kinetic Energy 11  if I mv 2 – mv 2 = h + heat energy v = 0 , v = v (In first collision, heat = 0) 21 2 2 2 1 1 ...(i) V3 2 mv2 = hmax V2 V1 1 ...(ii) [here V is applied potential] mv2 = eV V3 > V2 > V1  2 hc 12400 12400 volt from (i) and (ii) hmax = eV   min = eV   min = V × volt = V × 10–10 m × Continuous X–rays also known as white X–ray. Minimum wavelength of these spectrum only depends on applied NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 potential and doesn' t depend on atomic number. • Characteristic Spectrum of X–ray e– e– pXh-roatyon When the target of X–ray tube is collide by energetic electron it emits two h type of X–ray radiation. One of them has a continuous spectrum whose + wavelength depend on applied potential while other consists of spectral lines e– whose wavelength depend on nature of target. The radiation forming the K line spectrum is called characteristic X–rays. When highly accelerated electron L strikes with the atom of target then it knockout the electron of orbit, due to M this a vacancy is created. To fill this vacancy electron jumps from higher energy level and electromagnetic radiation are emitted which form E characteristic spectrum of X–ray. Whose wavelength depends on nature of target and not on applied potential. 12