DC Pandey Electricity And Magnetism

590 — Electricity and Magnetism Solution For lamp, (Vrms )R = (irms ) (R) = 10 × 5 = 50 V L R Choke Lamp VL VR V = V0 sin ωt In series, (Vrms )2 = (Vrms )R2 + (Vrms )L2 ∴ (Vrms )L = (Vrms )2 – (Vrms )R2 = (160)2 – (50)2 = 152 V As, (Vrms)L = (irms)XL = (irms) (2πfL) ∴ L = (Vrms)L (2πf ) (irms) Substituting the values, we get L = 152 (2π ) (50) (10) = 4.84 × 10–2 H Ans. Now, when the lamp is operated at 160 V, DC and instead of choke let an additional resistance R′ is put in series with it, then V = i (R + R′ ) or 160 = 10 (5 + R′ ) ∴ R′ = 11 Ω Ans. In case of AC, as the choke has no resistance, power loss in choke is zero. In case of DC, the loss in additional resistance R′ is P = i2R′ = (10)2(11) = 1100 W Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : In an AC circuit, potential difference across the capacitor may be greater than the applied voltage. Reason : VC = IXC , whereas V = IZ and XC can be greater than Z also. 2. Assertion : In series L-C-R circuit, voltage will lead the current function for frequency greater than the resonance frequency. Reason : At resonance frequency, phase difference between current function and voltage function is zero. 3. Assertion : Resonance frequency will decrease in L-C-R series circuit if a dielectric slab is inserted in between the plates of the capacitor. Reason : By doing so, capacity of capacitor will increase. 4. Assertion : Average value of current in the given graph is 3 A. i (A) 4 2 34 56 t (s) Reason : - Average value can’t be greater than the peak value of any function. 5. Assertion : In series L-C-R circuit, if a ferromagnetic rod is inserted inside an inductor, current in the circuit may increase or decrease. Reason : By doing so XLwill increase. 6. Assertion : Potential difference across, resistor, capacitor and inductor each is 10 V. Then, voltage function and current functions should be in phase. Reason : At this condition current in the circuit should be maximum. 7. Assertion : At some given instant I1 and I2 both are 2 A each. Then, I at this instant should be zero.

592 — Electricity and Magnetism I1 I2 I Reason : There is a phase difference of π between I1 and I2 functions. 8. Assertion : Peak value of current in AC through a resistance of 10 Ω is 2 A. Then, power consumed by the resistance should be 20 W. Reason : Power in AC is P = I 2 R rms 9. Assertion : An inductor coil normally produces more current with DC source compared to an AC source of same value of rms voltage. Reason : In DC source, applied voltage remains constant with time. 10. Assertion : In an L-R series circuit in AC, current in the circuit will decrease with increase in frequency. Reason : Phase difference between current function and voltage function will increase with increase in frequency. 11. Assertion : In series L-C-R, AC circuit, current and voltage are in same phase at resonance. Reason : In series L-C-R, AC circuit, resonant frequency does not depend on the value of resistance. Hence, current at resonance does not depend on resistance. Objective Questions 1. The term cos φ in an AC circuit is called (a) form factor (b) phase factor (c) power factor (d) quality factor 2. A DC ammeter cannot measure alternating current because (a) AC changes its direction (b) DC instruments will measure the average value (c) AC can damage the DC instrument (d) AC produces more heat 3. As the frequency of an alternating current increases, the impedance of the circuit (a) increases continuously (b) decreases continuously (c) remains constant (d) None of these 4. Phasor diagram of a series AC circuit is shown in figure. Then, V I (a) The circuit must be containing resistor and capacitor only (b) The circuit must be containing resistor and inductor only 30° (c) The circuit must be containing all three elements L, C and R (d) The circuit cannot have only capacitor and inductor 5. The rms value of an alternating current (a) is equal to 0.707 times peak value (b) is equal to 0.636 times peak value (c) is equal to 2 times the peak value (d) None of the above

Chapter 28 Alternating Current — 593 6. In an AC circuit, the applied potential difference and the current flowing are given by V = 200 sin 100 t volt, I = 5 sin 100 t − π amp 2 The power consumption is equal to (a) 1000 W (b) 40 W (c) 20 W (d) zero 7. The impedance of a series L-C-R circuit in an AC circuit is (a) R + (XL − XC ) (b) R2 + (XL2 − XC2 ) (c) R (d) None of these 8. If V0 and I0 are the peak current and voltage across the resistor in a series L-C-R circuit, then the power dissipated in the circuit is (Power factor = cos θ) (a) V0I0 (b) V0I0 (c) V0I0 cos θ (d) V0I0 cos θ 2 2 2 9. A generator produces a time varying voltage given by V = 240 sin 120 t, where t is in second. The rms voltage and frequency are (a) 170 V and 19 Hz (b) 240 V and 60 Hz (c) 170 V and 60 Hz (d) 120 V and 19 Hz 10. An L-C-R series circuit has a maximum current of 5 A. If L = 0.5 H and C = 8 µF, then the angular frequency of AC voltage is (a) 500 rad / s (b) 5000 rad / s (c) 400 rad / s (d) 250 rad / s 11. The current and voltage functions in an AC circuit are i = 100 sin 100 t mA,V = 100 sin 100t + π3 V The power dissipated in the circuit is (a) 10 W (b) 2.5 W (c) 5 W (d) 5 kW 12. A capacitor becomes a perfect insulator for (a) alternating current (b) direct current (c) both (a) and (b) (d) None of these 13. For an alternating voltage V = 10 cos 100 πt volt, the instantaneous voltage at t = 1 s is 600 (a) 1 V (b) 5 V (c) 5 3 V (d) 10 V 14. In a purely resistive AC circuit, (b) voltage lags current (d) nothing can be said (a) voltage leads current (c) voltage and current are in same phase 15. Identify the graph which correctly represents the variation of capacitive reactance XC with frequency Xc Xc Xc Xc (a) (b) (c) (d) fO f fO f O O

594 — Electricity and Magnetism 16. In an AC circuit, the impedance is 3 times the reactance, then the phase angle is (a) 60° (b) 30° (c) zero (d) None of these 17. Voltage applied to an AC circuit and current flowing in it is given by V = 200 2 sin ωt + π and i = − 2 cos ωt + π 4 4 Then, power consumed in the circuit will be (b) 400 W (d) None of these (a) 200 W (c) 200 2 W 18. When 100 volt DC source is applied across a coil, a current of 1 A flows through it. When 100 V AC source of 50 Hz is applied to the same coil, only 0.5 A current flows. Calculate the inductance of the coil. (a) (π / 3 ) H (b) ( 3 /π ) H (c) (2/π ) H (d) None of these 19. In the circuit shown in figure, the reading of the AC ammeter is 1 µF A V = 200√2 sin 100t (a) 20 2 mA (b) 40 2 mA (c) 20 mA (d) 40 mA 20. An AC voltage is applied across a series combination of L and R. If the voltage drop across the resistor and inductor are 20 V and 15 V respectively, then applied peak voltage is (a) 25 V (b) 35 V (c) 25 2 V (d) 5 7 V 21. For wattless power in an AC circuit, the phase angle between the current and voltage is (a) 0° (b) 90° (c) 45° (d) Not possible 22. The correct variation of resistance R with frequency f is given by RR R R (a) (b) (c) (d) ff f f 23. If L and R be the inductance and resistance of the choke coil, then identify the correct statement. (a) L is very high compared to R (b) R is very high compared to L (c) Both L and R are high (d) Both L and R are low

Chapter 28 Alternating Current — 595 24. When an AC signal of frequency 1 kHz is applied across a coil of resistance 100 Ω, then the applied voltage leads the current by 45°. The inductance of the coil is (a) 16 mH (b) 12 mH (c) 8 mH (d) 4 mH 25. The frequency of an alternating current is 50 Hz. The minimum time taken by it in reaching from zero to peak value is (a) 5 ms (b) 10 ms (c) 20 ms (d) 50 ms 26. An alternating voltage is applied across the R-L combination. V = 220 sin 120 t and the current I = 4 sin (120t − 60° ) develops. The power consumption is (a) zero (b) 100 W (c) 220 W (d) 440 W 27. In the AC network shown in figure, the rms current flowing through the C inductor and capacitor are 0.6 A and 0.8 A, respectively. Then, the current L coming out of the source is (a) 1.0 A (b) 1.4 A (c) 0.2 A (d) None of the above 28. The figure represents the voltage applied across a pure inductor. The diagram which correctly represents the variation of current i with time t is given by V Ot ii i i (a) O t (b) O t (c) O t (d) O t 29. A steady current of magnitude I and an AC current of peak value I are allowed to pass through identical resistors for the same time. The ratio of heat produced in the two resistors will be (a) 2 : 1 (b) 1 : 2 (c) 1 : 1 (d) None of these 30. A 50 Hz AC source of 20 V is connected across R and C as shown in figure. R C The voltage across R is 12 V. The voltage across C is (a) 8 V (b) 16 V (c) 10 V (d) Not possible to determine unless value of R and C are given

596 — Electricity and Magnetism Subjective Questions Note You can take approximations in the answers. 1. A 300 Ω resistor, a 0.250 H inductor, and a 8.00 µF capacitor are in series with an AC source with voltage amplitude 120 V and angular frequency 400 rad/ s. (a) What is the current amplitude? (b) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? (c) What are the voltage amplitudes across the resistor, inductor, and capacitor? 2. A series circuit has an impedance of 60.0 Ω and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity? 3. Voltage and current for a circuit with two elements in series are expressed as V (t) = 170 sin (6280t + π/ 3) volt i (t) = 8.5 sin (6280t + π/ 2) amp (a) Plot the two waveforms. (b) Determine the frequency in Hz. (c) Determine the power factor stating its nature. (d) What are the values of the elements? 4. A 5.00 H inductor with negligible resistance is connected across an AC source. Voltage amplitude is kept constant at 60.0 V but whose frequency can be varied. Find the current amplitude when the angular frequency is (a) 100 rad /s (b) 1000 rad /s (c) 10000 rad /s 5. A 300 Ω resistor is connected in series with a 0.800 H inductor. The voltage across the resistor as a function of time is VR = (2.50 V) cos [(950 rad/ s) t]. (a) Derive an expression for the circuit current. (b) Determine the inductive reactance of the inductor. (c) Derive an expression for the voltage VL across the inductor. 6. An L-C-R series circuit with L = 0.120 H, R = 240 Ω , and C = 7.30 µF carries an rms current of 0.450 A with a frequency of 400 Hz. (a) What are the phase angle and power factor for this circuit? (b) What is the impedance of the circuit? (c) What is the rms voltage of the source? (d) What average power is delivered by the source? (e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated ( converted to other forms) in the capacitor? (g) In the inductor?

LEVEL 2 Single Correct Option 1. A capacitor and resistor are connected with an AC source as shown in figure. Reactance of capacitor is XC = 3 Ω and resistance of resistor is 4 Ω. Phase difference between current tan−1 3 37° I and I1 is  4 =  I2 XC = 3Ω I1 R = 4Ω I V = V0 sin ωt (a) 90° (b) zero (c) 53° (d) 37° 2. A circuit contains resistance R and an inductance L in series. An alternating voltage V = V0 sin ωt is applied across it. The currents in R and L respectively will be RL AC (a) IR = I0 cos ωt, IL = I0 cos ωt (b) IR = − I0 sin ωt, IL = I0 cos ωt (c) IR = I0 sin ωt, IL = − I0 cos ωt (d) None of the above 3. In the circuit shown in figure, the AC source gives a voltage V = 20 cos (2000t). Neglecting source resistance, the voltmeter and ammeter readings will be 6Ω A 5 mH, 4Ω 50 µF I I1 V (a) 0 V, 2.0 A (b) 0 V, 1.4 A (c) 5.6 V, 1.4 A (d) 8 V, 2.0 A 4. A signal generator supplies a sine wave of 200 V, 5 kHz to the circuit shown in the figure. Then, choose the wrong statement. 1 µF π 100 Ω 200 V, 5 kHz (a) The current in the resistive branch is 0.2 A (b) The current in the capacitive branch is 0.126 A (c) Total line current is ≈ 0.283 A (d) Current in both the branches is same

598 — Electricity and Magnetism 5. A complex current wave is given by i = (5 + 5 sin 100 ωt) A. Its average value over one time period is given as (a) 10 A (b) 5 A (c) 50 A (d) 0 6. An AC voltage V = V0 sin 100 t is applied to the circuit, the phase difference between current and voltage is found to be π/ 4, then V, l VI t π/4 (a) R = 100 Ω, C = 1 µF (b) R = 1 kΩ, C = 10 µF (c) R = 10 kΩ, L = 1 H (d) R = 1 kΩ, L = 10 H 7. In series L-C-R circuit, voltage drop across resistance is 8 V, across inductor is 6 V and across capacitor is 12 V. Then, (a) voltage of the source will be leading in the circuit (b) voltage drop across each element will be less than the applied voltage (c) power factor of the circuit will be 3 /4 (d) None of the above 8. Consider an L-C-R circuit as shown in figure with an AC source of peak value V0 L and angular frequency ω. Then, the peak value of current through the AC C R source is V, ω (a) V0 2 (b) V0 1 + ωC − 1  2  R2 ωL  1 + ωL − 1  R2 ωC (c) V0 (d) None of these 2 ωL 1  R2 + − ωC 9. The adjoining figure shows an AC circuit with resistance R, inductance L and source voltage Vs. Then, RL VV 70 V 20 V Vs (a) the source voltage Vs = 72.8 V (b) the phase angle between current and source voltage is tan−1 (7/2) (c) Both (a) and (b) are correct (d) Both (a) and (b) are wrong

Chapter 28 Alternating Current — 599 10. When an alternating voltage of 220 V is applied across a device P, a current of 0.25 A flows through the circuit and it leads the applied voltage by an angle π/ 2 radian. When the same voltage source is connected across another device Q, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of P and Q? (a) 1 A lagging in phase by π /4 with voltage (b) 1 A leading in phase by π /4 with voltage 42 42 (c) 1 A leading in phase by π/4 with voltage (d) 1 A leading in phase by π/2 with voltage 2 42 11. In a parallel L-C-R circuit as shown in figure if IR , IL , IC and I represent the rms values of current flowing through resistor, inductor, capacitor and the source, then choose the appropriate correct answer. IR R IL L IC C I (a) I = IR + IL + IC (b) I = IR + IL − IC (c) IL or IC may be greater than I (d) None of these 12. In a series L-C-R circuit, current in the circuit is 11 A when the applied voltage is 220 V. Voltage across the capacitor is 200 V. If the value of resistor is 20 Ω, then the voltage across the unknown inductor is (a) zero (b) 200 V (c) 20 V (d) None of these 13. In the circuit shown in figure, the power consumed is RL V = V0 sin ωt (a) zero (b) V02 (c) V02R (d) None of these 2R 2(R2 + ω2L2) 14. In a series L-C circuit, the applied voltage is V0. If ω is very low, then the L C voltage drop across the inductor VL and capacitor VC are V0 V0 (a) VL = 2 ; VC = 2 (b) VL = 0 ; VC = V0 (c) VL = V0 ; VC = 0 (d) VL = − VC = V0 2

600 — Electricity and Magnetism 15. A coil, a capacitor and an AC source of rms voltage 24 V are connected in series. By varying the frequency of the source, a maximum rms current of 6 A is observed. If coil is connected to a DC battery of emf 12 volt and internal resistance 4 Ω, then current through it in steady state is (a) 2.4 A (b) 1.8 A (c) 1.5 A (d) 1.2 A 16. In a series C - R circuit shown in figure, the applied voltage is 10 V and the voltage across capacitor is found to be 8 V. The voltage across R, and the phase difference between current and the applied voltage will respectively be 8 V VR CR 10 V (a) 6 V, tan−1  34 (b) 3 V, tan−1  34 (c) 6 V, tan−1  43 (d) None of these 17. An AC voltage source described by V = 10 cos (π/ 2) t is connected to a 1 µF capacitor as shown in figure. The key K is closed at t = 0. The time (t > 0) after which the magnitude of current I reaches its maximum value for the first time is V = 10 cos π t C = 1 µF 2 K (a) 1 s (b) 2 s (c) 3 s (d) 4 s 18. An AC voltage source V = V0 sin ωt is connected across resistance R and capacitance C as shown in figure. It is given that R = 1/ωC. The peak current is I0. If the angular frequency of the voltage source is changed to ω/ 3, then the new peak current in the circuit is R V0 sin ωt C (a) I0 (b) I0 2 2 (c) I0 (d) I0 3 3

Chapter 28 Alternating Current — 601 More than One Correct Options 1. In a R-L-C series circuit shown, the readings of voltmeters V1 and V2 are 100 V and 120 V. Choose the correct statement(s). V2 V1 V = 130 V (a) Voltage across resistor, inductor and capacitor are 50 V, 86.6 V and 206.6 V respectively (b) Voltage across resistor, inductor and capacitor are 10 V, 90 V and 30 V respectively (c) Power factor of the circuit is 5 13 (d) Circuit is capacitive in nature 2. Current in an AC circuit is given by i = 3 sin ωt + 4 cos ωt, then (a) rms value of current is 5 A (b) mean value of this current in positive one-half period will be 6 π (c) if voltage applied is V = Vm sin ωt, then the circuit may contain resistance and capacitance (d) if voltage applied is V = Vm cos ωt, then the circuit may contain resistance and inductance only 3. A tube light of 60 V, 60 W rating is connected across an AC source of 100 V and 50 Hz frequency. Then, (a) an inductance of 2 H may be connected in series 5π (b) a capacitor of 250 µF may be connected in series to it π (c) an inductor of 4 H may be connected in series 5π (d) a resistance of 40 Ω may be connected in series 4. In an AC circuit, the power factor (a) is unity when the circuit contains an ideal resistance only (b) is unity when the circuit contains an ideal inductance only (c) is zero when the circuit contains an ideal resistance only (d) is zero when the circuit contains an ideal inductance only 5. In an AC series circuit, R = 10 Ω, XL = 20 Ω and XC = 10 Ω. Then, choose the correct options (a) Voltage function will lead the current function (b) Total impedance of the circuit is 10 2 Ω (c) Phase angle between voltage function and current function is 45° (d) Power factor of circuit is 1 2

602 — Electricity and Magnetism 6. In the above problem further choose the correct options. (a) The given values are at frequency less than the resonance frequency (b) The given values are at frequency more than the resonance frequency (c) If frequency is increased from the given value, impedance of the circuit will increase (d) If frequency is decreased from the given value, current in the circuit may increase or decrease 7. In the circuit shown in figure, 40 Ω XL = 20 Ω 2 A 100 V V = V0 sin ωt (a) VR = 80 V (b) XC = 50 Ω (c) VL = 40 V (d) V0 = 100 V 8. In L-C-R series AC circuit, (b) If L is increased, then current will decrease (d) If C is increased, then current will decrease (a) If R is increased, then current will decrease (c) If C is increased, then current will increase Comprehension Based Questions Passage I (Q. No. 1 to 3) A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4 A. He then replaced the 12 V DC source by a 12 V, (ω = 50 rad/ s ) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 µF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series). Based on the readings taken by the student, answer the following questions. 1. The value of resistance of the coil calculated by the student is (a) 3 Ω (b) 4 Ω (c) 5 Ω (d) 8 Ω 2. The power developed in the circuit when the capacitor of 2500 µF is connected in series with the coil is (a) 28.8 W (b) 23.04 W (c) 17.28 W (d) 9.6 W 3. Which of the following graph roughly matches the variations of current in the circuit (with the coil and capacitor connected in the series) when the angular frequency is decreased from 50 rad/s to 25 rad/s? ii i i (a) (b) (c) (d) 25 50 ω 25 50 ω 25 50 ω 25 50 ω

Chapter 28 Alternating Current — 603 Passage II (Q. No. 4 to 6) It is known to all of you that the impedance of a circuit is dependent on the frequency of source. In order to study the effect of frequency on the impedance, a student in a lab took 2 impedance boxes P and Q and connected them in series with an AC source of variable frequency. The emf of the source is constant at 10 V. Box P contains a capacitance of 1 µF in series with a resistance of 32 Ω. And the box Q has a coil of self-inductance 4.9 mH and a resistance of 68 Ω in series. He adjusted the frequency so that the maximum current flows in P and Q. Based on his experimental set up and the reading by him at various moment, answer the following questions. 4. The angular frequency for which he detects maximum current in the circuit is (a) 105 /7 rad /s (b) 104 rad /s (c) 105 rad /s (d) 104 /7 rad /s 5. Impedance of box P at the above frequency is (a) 70 Ω (b) 77 Ω (c) 90 Ω (d) 100 Ω 6. Power factor of the circuit at maximum current is (a) 1/2 (b) 1 (c) 0 (d) 1/ 2 Match the Columns 1. Match the following two columns for a series AC circuit. Column I Column II (a) Only C in the circuit (p) current will lead (b) Only L in the circuit (q) voltage will lead (c) Only R in the circuit (r) φ = 90° (d) R and C in the circuit (s) φ = 0° 2. Applied AC voltage is given as V = V0 sin ωt Corresponding to this voltage, match the following two columns. Column I Column II (a) I = I0 sin ωt (p) only R circuit (b) I = − I0 cos ωt (q) only L circuit (c) I = I0 sin (ωt + π /6) (r) may be C-R circuit (d) I = I0 sin (ωt − π /6) (s) may be L-C-R circuit 3. For an L-C-R series AC circuit, match the following two columns. Column I Column II (a) If resistance is increased (p) current will increase (b) If capacitance is increased (q) current will decrease (c) If inductance is increased (r) current may increase or decrease (d) If frequency is increased (s) power may decrease or increase

604 — Electricity and Magnetism 4. In the circuit shown in figure, match the following two columns. In Column II, quantities are given in SI units. XC = 30 Ω, XL = 15 Ω 2 A 40 V Column I Column II (a) Value of resistance R (p) 60 (b) Potential difference across capacitor (q) 20 (c) Potential difference across inductor (r) 30 (d) Applied potential difference (s) None of the above 5. Corresponding to the figure shown, match the two columns. 1 2 Column I Column II 3 (a) Resistance (p) 4 (b) Capacitive reactance (q) 1 4 (c) Inductive reactance (r) 2 ω (d) Impedance (s) 3 Subjective Questions Note Power factor leading means current is leading. 1. A coil is in series with a 20 µF capacitor across a 230 V, 50 Hz supply. The current taken by the circuit is 8 A and the power consumed is 200 W. Calculate the inductance of the coil if the current in the circuit is (a) leading (b) lagging 2. The current in a certain circuit varies with time as shown in figure. Find the average current and the rms current in terms of I0. I0 O τ 2τ t –I0 . 3. Two impedances Z1 and Z2 when connected separately across a 230 V, 50 Hz supply consume 100 W and 60 W at power factor of 0.5 lagging and 0.6 leading respectively. If these impedances are now connected in series across the same supply, find (a) total power absorbed and overall power factor (b) the value of reactance to be added in series so as to raise the overall power factor to unity.

Chapter 28 Alternating Current — 605 4. In the figure shown, the reading of voltmeters are V1 = 40 V, V2 = 40 V and V3 = 10 V. Find V1 V2 V3 R=4Ω L C E = E0 sin 100 πt + π 6 (a) the peak value of current (b) the peak value of emf (c) the value of L and C 5. In the circuit shown in figure power factor of box is 0.5 and power factor of circuit is 3/ 2 . Current leading the voltage. Find the effective resistance of the box. Box 10 Ω 6. A circuit element shown in the figure as a box is having either a capacitor or an inductor. The power factor of the circuit is 0.8, while current lags behind the voltage. Find C Box 1 A VC = 100 V R = 80 Ω V, 50 Hz (a) the source voltage V, (b) the nature of the element in box and find its value. 7. The maximum values of the alternating voltages and current are 400 V and 20 A respectively in a circuit connected to 50 Hz supply and these quantities are sinusoidal. The instantaneous values of the voltage and current are 200 2 V and 10 A, respectively. At t = 0, both are increasing positively. (a) Write down the expression for voltage and current at time t. (b) Determine the power consumed in the circuit. 8. An L-C circuit consists of an inductor coil with L = 5.00 mH and a 20.0 µF capacitor. There is negligible resistance in the circuit. The circuit is driven by a voltage source with V = V0 cosωt. If V0 = 5.00 mV and the frequency is twice the resonance frequency, determine (a) the maximum charge on the capacitor (b) the maximum current in the circuit (c) the phase relationship between the voltages across the inductor, the capacitor and the source.

606 — Electricity and Magnetism 9. A coil having a resistance of 5 Ω and an inductance of 0.02 H is arranged in parallel with another coil having a resistance of 1 Ω and an inductance of 0.08 H. Calculate the power absorbed when a voltage of 100 V at 50 Hz is applied. 5 Ω 0.02 H 1 Ω 0.08 H i 100 V 50 Hz 10. A circuit takes a current of 3 A at a power factor of 0.6 lagging when connected to a 115 V – 50 Hz supply. Another circuit takes a current of 5 A at a power factor of 0.707 leading when connected to the same supply. If the two circuits are connected in series across a 230 V, 50 Hz supply, then calculate (a) the current (b) the power consumed and (c) the power factor Answers Introductory Exercise 28.1 2. 0.036 H, 111.8 V 3. 7.7 H, 6 A 1. (a) 628 Ω (b) 6.37 mH (c) 1.59 kΩ (d) 1.59 mF Introductory Exercise 28.2 1. 650 Hz, 0 2. 0.2 Exercises LEVEL 1 Assertion and Reason 1. (a) 2. (b) 3. (a) 4. (b) 5. (a or b) 6. (b) 7. (a) 8. (a,b) 9. (b) 10. (b) 11. (c) Objective Questions 1. (c) 2. (b) 3. (d) 4. (d) 5. (a) 6. (d) 7. (d) 8. (d) 9. (a) 10. (a) 11. (b) 12. (b) 13. (c) 14. (c) 15. (b) 16. (d) 17. (d) 18. (b) 19. (c) 20. (c) 21. (b) 22. (a) 23. (a) 24. (a) 25. (a) 26. (c) 29. (a) 27. (c) 28. (c) 30. (b)

Chapter 28 Alternating Current — 607 Subjective Questions 1. (a) 0.326 A (b) 35.3°, lagging (c) 97.8 V, 32.6 V, 102 V 2. (a) Inductor (b) 0.133 H 3 3. (b) 1000 Hz (c) , leading (d) R = 17.32 Ω ,C = 15.92µF 2 4. (a) 0.12 A (b) 1.2 × 10–2 A (c) 1.2 × 10–3 A 5. (a) (8.33 mA)cos (950 rad/s) t (b) 760 Ω (c) – (6.33 V) sin (950 rad/s)t 6. (a) 45.8° , voltage leads the curren,t 0.697 (b) 343 Ω (c) 155 V (d) 48.6 W (e) 48.6 W (f) 0 (g) 0 LEVEL 2 Single Correct Option 1. (c) 2. (d) 3. (c) 4. (b) 5. (b) 6. (b) 7. (d) 8. (b) 9. (a) 10. (b) 14. (b) 15. (c) 16. (a) 17. (a) 18. (b) 11. (c) 12.(b) 13.(c) More than One Correct Options 1.(a,c,d) 2.(c,d) 3.(c,d) 4.(a,d) 5.(a,b,c,d) 6.(b,c,d) 7.(a,b,c) 8.(a) Comprehension Based Questions 1. (a) 2. (c) 3. (b) 4. (a) 5. (b) 6. (b) Match the Columns 1. (a) → p,r (b) → q,r (c) → s (d) → p (c) → r,s (d) → s 2. (a) → p,s (b) → q (c) → r,s (d) → r,s (c) → r (d) → s 3. (a) → q,s (b) → r,s (c) → r (d) → q 4. (a) → q (b) → p 5. (a) → s (b) → p Subjective Questions 1. (a) 0.416 H (b) 0.597 H 2. zero, I0 3. (a) 99 W, 0.92 leading (b) 194.2 Ω 3 4. (a) 10 2A (b) 50 2V (c) 1 H, 1 5. 5 Ω (d) F 25 π 100 π 1.6 6. (a) 100 V (b) inductor, L = H π 7. (a) V = 400 sin (100 πt + π /4), i = 20 sin (100 πt + π /6) (b) P = 3864 W 8. (a) 33.4 nC (b) 0.211 mA (c) Source and inductor voltages in phase. Capacitor voltage lags by 180°. 9. 797 W 10. (a) 5.5 A (b) 1.188 kW (c) 0.939 lag





23. Current Electricity INTRODUCTORY EXERCISE 23.1 or vd = × 1028 1 × 10−4 × 1.6 × 10−19 8.5 1. Q i = q = ne = 0.735 × 10−6 m/s tt ∴ = 0.735 µ m/s = it = (0.7) (1) n e 1.6 × 10− 19 t= l vd = 4.375 × 1018 2. Q q = it = 10 × 103 s 0.735 × 10−6 = (3.6) (3 × 60 × 60) = 38880 C = 10 × 103 yr × 60 × 60 3. (a) Q q = it = (7.5) (45) = 337.5 C 0.735 × 10−6 × 24 × 365 (b) n= q = 337.5 19 = 431.4 yr e 1.6 × 10− = 2.1 × 1021 INTRODUCTORY EXERCISE 23.4 4. f = v = 2.2 × 106 1. R =ρ l = (1.72 × 10− 8 ) (35) = 0.18 Ω 2πr (2π ) (5.3 × 10− 11) A (π /4) (2.05 × 10− 3)2 = 6.6 × 1015 Hz 2. ρ = 1 ⇒ ρσ = constant I = qf σ = (1.6 × 10 − 19 ) (6.6 × 1015) 3. R = ρl = 1.06 × 10− 3 A A = 1.06 mA ∴ A = ρl 10 10 R idt = (10 + 4t) dt ∫ ∫5. ∆q = m = (Vd), where V = volume and d = density 00 = 300 C ∴ m = (Ald) = ρl2 d 6. Current due to both is from left to right. So, the R = (1.72 × 10− 8 ) (3.5)2 (8.9 × 103) two currents are additive. 0.125 = 15 × 10− 3 kg INTRODUCTORY EXERCISE 23.2 1. False. Only under electrostatic conditions (when = 15 g i = 0 ) all points of a conductor are at same 4. R = ρ (L) = ρ L = ρ potential. A tL t INTRODUCTORY EXERCISE 23.3 i.e. R is independent of L. 1. i = neAvd Hence, the correct option is (c). i.e. vd ∝ i INTRODUCTORY EXERCISE 23.5 When current has increased from i = 1.2 A to 1. Copper is metal and germanium is semiconductor. i = 6.0 A, i.e five times, then drift velocity will also increase to five times. Resistance of a metal decreases and that of a semiconductor increases with decrease in 2. From i = neAvd vd = i temperature. neA ∴ Correct option is (d). We have

Chapter 23 Current Electricity — 611 2. 4.1[1 + 4.0 × 10− 3 (θ − 20)] INTRODUCTORY EXERCISE 23.7 = 3.9 [1 + 5.0 × 10− 3 (θ − 20)] 1. Applying loop law equation in upper loop, we Solving we get, have θ ≈ 85° C E + 12 − ir − 1 = 0 …(i) INTRODUCTORY EXERCISE 23.6 Applying loop law equation in lower loop, we have where 1. PD across each resistance is 10 V. i = 1 + 2 = 3A 10 E + 6−1= 0 …(ii) 2 ∴ i2Ω = = 5A Solving these two equations, we get E = − 5 V and r = 2Ω 10 i4Ω = 4 = 2.5A 2. Power delivered by a battery = Ei 2. VA = 0 V (as it is earthed) = 12 × 3 = 36 W VC − VA = 5 V Power dissipated in resistance ∴ VC = 5 V = i2R = (3) (2)2 VB − VA = 2 V = 12 W ∴ VB = 2 V VD − VC = 10 V INTRODUCTORY EXERCISE 23.8 ∴ VD = 10 + VC = 15 V i1 Ω = VC − VB = 3A from C to B as VC > VB 1. (a) Equivalent emf (V) of the battery 1 PD across the terminals of the battery is equal to i2 Ω = VD − VA = 7.5A from D to A as VD > VA its emf when current drawn from the battery is 2 zero. In the given circuit, 3. VA = VB r2 V2 i ∴ VAB = 0 i=0 i=0 or E − ir = 0 A r1 V1 B ∴ E −  15 + E (2) = 0 8 Solving this equation, we get Current in the internal circuit, E = 5V i = Net emf = V1 + V2 Total resistance r1 + r2 4. Net emf = (n − 2m) E = (10 − 2 × 2) (1) Therefore, potential difference between A and =6V B would be i = Net emf Net resistance = 6 = 0.5 A VA − VB = V1 − ir1 10 + 2  V1 + V2  V1r2 − V2r1 5. VR 1 = 0 ∴ VA − VB = V1 −  r1 + r2  r1 = r1 + r2 ∴ iR 1 = 0 So, the equivalent emf of the battery is V = V1 r2 − V2 r1 ∴ VR 2 = VR 3 = 10 V r1 + r2 iR 2 = iR 3 = 10 =1A Note that if V1 r2 = V2 r1 : V = 0 10

612 — Electricity and Magnetism If V1r2 > V2r1 : VA − VB = Positive i.e. A side of 3. V = ig G the equivalent battery will become the positive ∴ ig = V terminal and vice-versa. Now, G (b) Internal resistance (r) of the battery V G r1 and r2 are in parallel. Therefore, the internal nV = ig (G + R) = (G + R) resistance r will be given by ∴ R = (n − 1) G 1/r = 1/r1 + 1/r2 or r = r1r2 INTRODUCTORY EXERCISE 23.10 r1 + r2 2. 1. r = R  l1 − 1  l2 4 V 0.5 Ω E r E = Σ (E/r) = (6/1) − (2/1) = 2 V = 5  0.52 − 1 Σ (1/r) (1/1) + (1/1) 0.4 Now, net emf of E and 4 V is 2V as they are = 1.5 Ω oppositely connected. 2. (a) VAJ = E or emf of lower battery Σ (E/r) 2 3. Eeq = Σ (1/r) E = (10 / 1) + (4 /2) + (6/2) ∴ i RAJ = 2 (1/1) + (1/2) + (1/2) or E   61050r  (l) = E  15r + r 2 = 7.5V 1=1+ 1+ 1 Solving this equation, we get r12 2 l = 320 cm ∴ r = 0.5 Ω (b) Resistance of 560 cm =  6105r0 (560) INTRODUCTORY EXERCISE 23.9 = 14r Now the circuit is as under, 1. V = ig (G + R) r ∴ R = V − G = series resistance connected with E ig i1 galvanometer 14r r =  5 −1 = 999 Ω i1–i2  5 × 10− 3 i2 G S r E/2 2. i –ig Applying loop law in upper loop ig G we have, ig = S E − 14r (i1 − i2) − i1r − i1r = 0 …(i) i − ig G Applying loop law in lower law loop …(ii) we have, ∴  ig  − E − i2r + (i1 − i2 ) (14 r) = 0 S = G 2  i − ig  (50 × 10− 6)  Solving these two equations  10− 3 − 50 × 10− =  (5 × 6 )  (100) ≈ 1.0 Ω we get, i2 = 3E  22r

Chapter 23 Current Electricity — 613 INTRODUCTORY EXERCISE 23.11 ∴ X = 10 × 53 = 10.6 Ω 50 1. R > 2 Ω ⇒ 100 − x > x ∴ Correct option is (b). 2Ω R 3. Slide wire bridge is most sensitive when the G resistance of all the four arms of bridge is same. x 100 – x Hence, B is the most accurate answer. INTRODUCTORY EXERCISE 23.12 1. P = R ⇒ X =  Q  R =  110 R P QX R 2Ω R lies between 142 Ω and 143 Ω. G Therefore, the unknown resistance X lies between 14.2 Ω and 14.3 Ω. x + 20 80 – x 2. Experiment can be done in similar manner but now K2 should be pressed first then K1. 3. BC, CD and BA are known resistances. The unknown resistance is connected between A and D. Applying P =R INTRODUCTORY EXERCISE 23.13 QS We have 2= x …(i) 1. Yellow → 4 R 100 − x Red → 2 R = x + 20 …(ii) Orange → 103 2 80 − x Gold → 5 Solving Eqs. (i) and (ii), we get R = 3 Ω ∴ R = (4.2 × 103 ± 5%) Ω ∴ Correct option is (a). 2. 2 → Red 2. Using the concept of balanced, Wheatstone bridge, 4 → Yellow we have, 106 → Blue P =R ⇒ X = 10 5% → Gold QS (52 + 1) (48 + 2) Exercises LEVEL 1 3. R = ρl or R = resistance per unit length = ρ ∝ 1 Al AA Assertion and Reason 1. If PD between two terminals of a resistance is Near A, area of cross-section is less. Therefore, zero, then current through resistance is zero, this is resistance per unit length will be more. Hence confirmed. But PD between any two points of a from the equation, H = i2Rt, heat generation near circuit is zero, this does not mean current is zero. A will be more. (as i is same) 2. In parallel, V = constant Current density, J = i or J ∝ 1 ∴ From the equation AA P=V2 ⇒ P∝ 1 RR 4. Since net resistance decreases, therefore main current increases. Hence, net potential difference across voltmeter also increases.

614 — Electricity and Magnetism 5. Even if ammeter is non-ideal, its resistance should 4. R= l σA be small and net parallel resistance is less than the ∴ smallest individual resistance. σ = l = m m2 = ohm− 1 -m− 1 RA ohm - ∴ Rnet < resistance of ammeter in the changed situation. Hence, net resistance of the circuit will 6. 0.5 = E …(i) decrease. So, main current will increase. But maximum percentage of main current will pass r + 3.75 through ammeter (in parallel combination) as its resistance is less. Hence, reading of ammeter will 0.4 = E …(ii) increase. r + 4.75 Initial voltmeter reading = emf of battery Solving these two equations, we get Final voltmeter reading = emf of battery E = 2V − potential drop across shown resistance. 7. In parallel current distributes in increase ratio of Hence, voltmeter reading will decrease. resistance 6. If current flows from a to b, then equation will ∴ IG = S IS G become ∴ G =  IS  (S ) Va − ir − E = Vb or Va − Vb = E + ir  IG  So, Va − Vb is always positive. Hence, Va is always greater than Vb. =  502−020 (12) 7. Current in the circuit will be maximum when = 18 Ω R = 0. 8. IG = S 8. Resistance will increase with temperature on IS G ∴ heating. Hence current will decrease. Further P = V 2 or P ∝ 1 S =  IG  G RR  IS  Resistance is increasing. Hence, power consumed =  928 G = G across R should decrease. 49 V = IR is just an equation between P D across a 9. P = V 2 or P ∝ 1 resistance current passing through it and its R resistance. This is not Ohm’s law. R ∴ 9. Electrons get accelerated by the electric. Then, P2 = R1 = l1 (as R ∝ l ) P1 R2 l2 suddenly collision takes place. Then, again accelerated and so on. 11. Eeq = E1 /r1 + E2 /r2 P2 =  l1  P1 =  l l P1 = 1.11 P1 (1/ r1 ) + (1/ r2 )  l2  0.9  (1/r1) + (E2 /E1) (1/r2) So, power will increase by 11%.  (1/r1) + (1/r2)  = E1 10. By symmetry, VA = VB So, Eeq may be greater than E1 also, if E2/E1 > 1 or VAB = 0 r1 and r2 are in parallel. Hence, req is less than both 11. r = R  l1 − 1 = 10  75 − 1 r1 and r2 individually.  l2 60 Objective Questions = 2.5 Ω 3. H = I 2 Rt 12. Let V0 = V  H  Now, I AO + I BO = I OC  I 2t  ∴ [R ] = ∴ 6−V + 3−V =V −2 6 32  ML2T− 2   I2T  ML2T− 3I− =   = [ 2 ] Solving this equation, we get V = 3V

Chapter 23 Current Electricity — 615 13. vd = i = i ⇒ vd ∝ i or P + 15 = 2 …(ii) neA (πr2) r2 Q3 ne 14. For making voltmeter of higher range, more Solving these two equations, we get P = 9Ω resistance is required. 15. V20Ω = VTotal 23. Total potential of 10V equally distributes between ∴ (20) (0.3) = (RTotal ) (0.8) 50 Ω and other parallel combination of 100 Ω and voltmeter. Hence, their net resistance should be ∴ RTotal = 30 Ω same. Or 4 100 × R = 50 ∴ 4=1+ 1+1 100 + R 30 R1 20 15 ∴ R = 100 Ω = resistance of voltmeter Solving we get R1 = 60Ω 24. VAC = VDE 16. Net resistance will decrease by increasing the 2V 1Ω parallel resistors. Therefore, main current i will increase, further, (PD)Voltmeter = VTotal − (PD)Ammeter i = VTotal − i (resistance of ammeter) iCi B Since, i has increased. Hence, PD across voltmeter A will decrease. G IG = 0 17. P = V 2 or P ∝ 1 …(i) D E RR 1.2V 4Ω Resistor is cut in n equal parts. Therefore, each i (RAC ) = E = 1.2 resistance will become R . Now, these are ∴  2  4 × l = 1.2 n  4 + 1 100 connected in parallel. Therefore, net resistance will become 1 times R . or R . Solving this equation, we get n n n2 l = 75 cm Now, from Eq. (i), power will become n2 times. 18. If A is fused, then complete circuit is broken. 1A 2A 19. E − ir = 0 25. 2 Ω 2 Ω 4A 6A ∴ 3−  3 + 15  (1) = 0 3A A 3V 1Ω 2V 1Ω B  1 + 2 + R Solving this equation, we get VA − 3 × 2 − 3 − 1 × 4 + 2 − 1 × 6 = VB R = 3Ω ∴ VA − VB = 17 V 20. =  (2500 R)  2V 0.5Ω  2500 + R  100 (5) Solving this equation, we get 26. R = 20Ω 21. Five parallel combination, each of value R R + R =R Equivalent simple circuit is given as 10 10 5 maximum power across R is obtained 22. P = 20 or Q = 4P …(i) When R = r = 0.5 Ω i = 2 = 2A Q 100 − 20 R+r ∴ P<Q Now, P + 15 = 40 ∴ = i2R = (2)2 (0.5) = 2W Q 100 − 40

616 — Electricity and Magnetism 27. r = R  E − 1 = 5  2.2 − 1 = 10 Ω 36. R0 = RAB = R + (R + R0) R V 1.8 9 (R + R0) + R 28. i = 10 − 5 = 1 A (clockwise) Solving this equation, we get R = R0 3 2.5 + 2.5 + 40 9 VB − 15 i − 25 i = VA 37. Simple circuit is as shown in figure, 40 P 9 ∴ VA − VB = 40 i = − V 29. Potential drop across potentiometer wire Q = (0.2 × 10− 3) (100) = 0.02 V Each → R Now given resistance and potentiometer wire are 38. Wheatstone (balanced) between A and B. So, in series with given battery. So, potential will drop in direct ratio of resistance. resistance between C and D can removed. ∴ 0.02 = R A 2 − 0.02 490 CD ∴ R = 4.9Ω 30. When K is open Rnet = 3R/2 B ∴ i1 = E / (3R / 2) = 2E 3R When K is closed 2R RB Rnet = 2  R × 2R  = 4 R 39.  R + 2R  3 A ∴ i2 = E / (4 R / 3) = 3E 4R 2R ∴ i1 = 8 i2 9 R =  24πr (2r) = 4 π 31. IG = S Now 2Ω, 2Ω and R are in parallel. IS G ∴ S =  IG  G R R/2  IS  40. R/2 D = (1/34) × 3663 = 111 Ω R/2 ⇐ R/2 R/2 (33/ 34 ) C R/2 32. Simple series and parallel grouping of resistors. A B AB R R 33. Two balanced Wheatstone bridges in parallel. (3 × 15) R R 3 + 15 A 34. Rab = = 2.5 Ω As R60° =  31680° (60° ) = 3Ω 41. R R R R 35. RAB = 2 [Net resistance of infinite series] + 1 R R R In parallel net resistance is always less than the smallest one. Hence, net resistance of infinite R series is less than 1Ω. B ∴ 1Ω < RAB < 3Ω RR Connection can be removed from centre. 3R and 3R from two sides of AB are in parallel.

Chapter 23 Current Electricity — 617 Subjective Questions 9. i = 150 − 50 = 20 A (anti-clockwise) 1. Under electrostatic conditions (when no current 2+ 3 flows), E = 0. When current is non-zero, then VQ + 150 − 20 × 2 = VP electric field is also non-zero. ∴ VQ = VP − 110 = − 10 V 2. There is random or thermal motion of free 10. ρ = 8.89 × 103 kg/m3 electrons in the absence of potential difference. Mass of 1m3 = 8.89 × 103 kg 3. i = qf = q  2πvR = 8.89 × 103 kg = 8.89 × 106g = (1.6 × 10−1 9 ) (2.2 × 106) ∴ Number of gram moles = 8.89 × 106 = 1.4 × 105 (2π ) (5 × 10− 11) 63.54 = 1.12 × 10− 3 A = 1.12 mA Number of atoms = 1.4 × 105 × 6.02 × 1023 4. P = V 2 = 8.42 × 1028 R One atom emits one conduction electron. ∴ R=V2 Therefore, number of free electrons in unit volume P (or 1m3 volume) (120)2 n = 8.42 × 1028 per m3 40 R1 = = 360 Ω Now, i = neA vd ∴ R2 = (120)2 = 240 Ω vd = i = i 60 neA ne πr2 R3 = (120)2 = 192 Ω = (8.42 × 1028 ) (1.6 × 2.0 (0.5 × 10− 3 )2 75 10− 19 )(π ) Now, all these resistors are in parallel. = 1.9 × 10− 4 m/s 5. (a) i = 12 − 6 = 0.5 A 11. (a) In 1 m, potentials difference, 4+8 V = 0.49 ∴ V = iR (b) PR1 = i2R1 = 1W ∴ i = 0.49 = (0.49) A ⇒ PR2 = i2R2 = 2W R ρl = (0.49) (π /4) (0.84 × 10− 3 )2 (2.75 × 10− 8 ) (1) (c) Power supplied by E1 = E1 i = 6 W and power consumed by E2 = E2i = 3W = 9.9 A 6. 8 = l (b) PD between two points, 12 m apart 12 40 − l = (0.49 V/m) (12m) = 5.88 V Solving this equation, we get (c) R = V = 5.88 = 0.6 Ω i 9.9 l = 16 cm 7. (a) Ideal voltmeter means infinite resistance. 12. Radius at distance x from end P, ∴ i=0 =a+  b − a l (b) V = E (if i = 0) r x =5V Q (c) Reading of voltmeter = E = 5 V 8. (a) E1 > E2 P b a Therefore, net current is anti-clockwise or from B to A. x (b) Current through E1 is normal. Hence, it is dx doing the positive work. l (c) Current flows from B to A ∴ VB > VA Resistance of element of thickness dx is

618 — Electricity and Magnetism dR = ρ (dx) (Using R = ρl)  32××00.0.02187××76.5 (1 mm) πr2 A = X=l  ∫∴ R = dR = 0.569 mm X=0 13. i = E ⇒ P = power across R = i2R 17. (a) E = V = 0.938 = 1.25 V/m R+r l 0.75 (b) E = Jρ  E  2 P =  R + r …(i) ∴ ρ = E = 1.25 R J 4.4 × 107 For power to be maximum, = 2.84 × 10− 8 Ω-m dP = 0 18. (a) J= i = V = V dR A RA  ρAl By putting dP = 0 we get, R = r A dR Further, by putting R = r in Eq. (i) J =V …(i) ρl We get, Pmax = E2 4r or J ∝ 1 l 14. As derived in the above question, Pmax = E2 lmin = d. So, J is maximum. Hence, potential 4r difference should be applied across the face Here, E = net emf= 2 + 2 = 4 V (2d × 3d) and r = net internal resistance From Eq. (i), = 1+ 1= 2Ω J max = V ρd ∴ Pmax = (4 )2 = 2W (4) (2) (b) i = V = V = VA R (ρl/ A) ρl 15. In series, or i ∝ A α eq = R01 α 1 + R02 α2 l R01 + R 02 Across face (2d × 3d), area of cross-section is = (600) (0.001) + (300) (0.004) 600 + 300 maximum and l is minimum. Hence, current is maximum. = 0.002 per°C imax =V (2d × 3d) = 6Vd ρ (d) ρ Now, Rt = R0 [1 + α ∆θ ] 19. (a) ρ = RA = (0.104) (π /4) (2.5 × 10− 3)2 = (600 + 300) [1 + 0.002 × 30] = 954Ω l 14 16. In parallel current distributes in inverse ratio of = 3.65 × 10− 8 Ω-m resistance 1→ Aluminium 2 → Copper R1 = i2 (b) V = El = 1.28 × 14 = 17.92 V R2 i1 ∴ i = V = 17.92 = 172.3 A ρ1l1 / A1 = 2 R 0.104 ρ2 l2 / A2 3 (c) vd = i neA ρ1l1d22 2 ∴ ρ2l2d12 = 3 = 172.3   (8.5 × 1028 ) (1.6 × 10− 19 )  π  (2.5 × 10− 3)2  4 ∴ d2 =  2ρ2l2  d1  3ρ1l1 = 2.58 × 10− 3 m/s

Chapter 23 Current Electricity — 619 20. R1 + R2 = 20 …(i) Now, this is a balanced Wheatstone bridge in parallel with 12 Ω resistance. R1α 1 + R2α 2 α eq = R1 + R2 (in series) 25. First case 0 = R1 (− 0.5 × 10− 3) + R2 (5.0 × 10− 3) i = 12 + 6 = 3A (clockwise) 20 1+ 2+ 3 ∴ R1 = 10R2 …(ii) Now, VA − VG = 12 V ∴ VA = 12 V, as VG = 0 Solving Eqs. (i) and (ii), we get VA − VB = 1 × 3 = 3 V R2 = RFe = 20 Ω ∴ VB = VA − 3 = 9 V 11 VB − VC = 2 × 3 = 6 V = 1.82Ω ∴ VC = VB − 6 = 3 V R1 = RCu VG − VD = 6 V = 10R2 ∴ VD = − 6 V, as VG = 0 = 18.18 Ω In the second case, 21. 8Ω and 12Ω resistors are in parallel. i = 12 − 6 = 1 A 1+ 2+ 3 ∴ Rnet = 8 × 12 = 4.8 Ω 8 + 12 ∴ i = 24 Rest procedure is same. (anti-clockwise) 4.8 26. i = 200 = 5 A = 5A 5 + 10 + 25 22. All four resistors are in parallel ∴ V3 − VG = 25 × 5 = 125 V3 = 125 V as VG = 0 ∴ 1=1+ 1+ 1+ 1 ∴ R 8 4 6 12 VG − V2 = 10 × 5 = 50 R=8Ω ∴ V2 = −50V 5 Now, V2 − V1 = 5 × 5 = 25 V ∴ i = 24 V1 = V2 − 25 = − 75 V 8/5 V3 − 2 = V3 − V2 = 15 A 23. All these resistors are in parallel. 4.3V 1.0Ω 24. The given network is as shown below. 27. (a) 6 50 Ω 2.0 Ω V1 V3 4 200 Ω 12 6 Rnet = 1.0 + 2.0 + 50 × 200 50 + 200 3 V2 V4 = 43 Ω 2 ∴ i = 4.3 = 0.1 A 43 The simple circuit is as shown below. 6 V3 3 = Readings of ammeter Readings of voltmeter 6 = (i) net resistance of 50 Ω and 200 Ω V1 4 V4 2 V2 = (0.1)  50 × 200  50 + 200 =4V 12

620 — Electricity and Magnetism i 4.3 V 1.0 Ω 30. (a) (i) When switch S is open, V1 and V2 are in 50 Ω 2.0 Ω (b) series, connected to 200 V battery. Potential will drop in direct ratio of their resistors. i1 ∴ V1 : V2 = RV 1 : RV 2 = 3000 : 2000 i2 = 3:2 200 Ω ∴ V1 = 3 × 200 = 120 V 5 Rnet = 1.0 + 52 × 200 = 42.27 Ω V2 = 2 × 200 = 80 V 52 + 200 5 ∴ i = 4.3 ≈ 0.1 A (ii) When S is closed then V1 and R1 are in Now 42.27 parallel. Similarly, V2 and R2 are also in ∴ parallel. Now, they are in series and they come i1 = 200 i2 52 out to be equal. So, 200 V will equally i1 =  220520 (0.1) = 0.08 A distribute between them. ∴ V1 = V2 = 200 = 100 V each 2 = Reading of ammeter (b) i2 = 100 = 1 A ∴ Reading to voltmeter 2000 20 = Potential difference across 50 Ω and 2.0 Ω i4 = 100 = 1 A = 0.08 × 52 ≈ 4.2 V 3000 30 5Ω A 4Ω i2 28. 42 V i1–i2 10 V 3000 Ω 2000 Ω (1) 6Ω (2) i1 i1 V1 i3 V2 E i2–i3 B 1Ω C 8Ω i2 i4 i3 (3) 16 Ω D 2000 Ω P 3000 Ω 4V 100 V 100 V Loop 1 If we apply junction law at P, then current through − 42 − 6 (i1 − i2) − 5i1 − i1 = 0 …(i) switch Loop 2 = i2 − i4 = 1 A in upward direction. 60 − 4i2 − 10 − 8 (i2 − i3) + 6 (i1 − i2) = 0 …(ii) Loop 3 31. Power absorbed by resistor is i2R or 2W. 8 (i2 − i3) = − 16 i3 + 4 = 0 …(iii) Therefore, remaining 3W is absorbed by the Solving these equations, we get battery (= Ei). Hence, E is 3 V and current of 1 A i1 = 4 A, i2 = 1.0 A and i3 = 0.5 A enters from the position terminal as shown below. 29. Net resistance of voltmeter (R = 400 Ω) and 400 Ω A 2Ω 3V B will be 200 Ω. Now, we are getting a balanced 1A Wheatstone bridge with 100 Ω and 200 Ω resistors on each side. Potential difference across each side VAB = E + ir (Here, r = R) = E + iR …(i) will be 10 V which will distribute in direct ratio of = 3 + (1) (2) = 5 V resistors 100 Ω and 200 Ω. ∴ V100 Ω = 100 = 1 32. 8.4 = E − 1.5 r V200 Ω 200 2 9.4 = E + 3.5 r …(ii) Solving these two equations, we get or V200 Ω =  23 (10) = 20 V 3 r = 0.2 Ω and E = 8.7 V

Chapter 23 Current Electricity — 621 33. During charging, where, R = resistance of voltmeter. Solving the above equation, we get V = E + ir = 2 + (5) (0.1) = 2.5 V R = 1200 Ω 34. Simple circuit is as shown below In the new situation, i 2i i 4Ω 4Ω 4Ω Rnet = 400 + (300) (1200) = 640 Ω 300 + 1200 2V ∴ i = 60 = 0.09375 A 640 Now voltage drop across 2V 2V Voltmeter = 60 − potential drop across 400 Ω resistor By symmetry, currents on two sides will be same (let i) = 60 − (400) i Now if we apply loop law in any of the closed = 60 − (400) (0.09375) loop, we will get i = 0. = 22.5 V 35. Net resistance should remain unchanged. 38. Rnet = 60 + (60) (120) = 100 Ω 60 + 120 GS ∴ R + G = R′ + G+S ∴ i = 120 = 1.2 A 100 ∴ R′ − R = G − GS = G 2 G+S G+S Now, reading of voltmeter 36. Current through voltmeter = 120 − potential drop across R1 = 120 − (60) (1.2) = 48 V r 39. In parallel current distributes in inverse ratio of 4.96 A resistance. S 5A i –ig V 0.04A 2500Ω = V = 100 = 0.04 A G R R 2500 ig In parallel current distribution in inverse ratio of ∴ i − ig = G + R resistors. Hence,   ig  S 4.96 = 2500 0.04 r ∴ R = i − ig S −G  ∴ r = 20.16 Ω  ig  37. Voltmeter reads 30 V, half of 60 V. Hence, =  20  (0.005) − 20 = 80 Ω 10− 3 resistance of 400 Ω and voltmeter is also equal to 300 Ω. Note In calculations, we have taken i − ig ≈ i. 60 V V i 400 Ω 40. 300 Ω A 2Ω 100 Ω V 120 Ω ∴ 300 =  400 × R 3.4V 3Ω  400 + R

622 — Electricity and Magnetism Reading of voltmeter = 3.4 − Voltage drop across 43. P = i2R ammeter and 3Ω resistance Pmax = 36 = 15 A = (3.4) − 0.04 × 2 − 0.04 × 3 ∴ imax = R 2.4 = 3.2 V R/2 R 3.2 V = (0.04) (100 R) Now, imax (100 + R) where, R = resistance of voltmeter Total maximum power = (imax )2  32R ∴ R = 400 Ω If voltmeter is ideal, then = (15) (1.5) (2.4) = 54 W i = 3.4 = 0.03238 A 44. V = E − ir 2 + 100 + 3 ∴ r = E − V = 2.6 − 2 Reading of voltmeter = 100i = 3.238 V i1 41. (a) V = E − ir = 0.6 Ω Now, power generated in the battery = E −  E r  r + RV  P = i2r V =  r ERV  …(i) = (1)2 (0.6) = 0.6 W  + RV  Power supplied by the battery = Ei = 2.6 W (b) V = E ∴ Net power supplied for external circuit 100 = 2.6 − 0.6 = 2.0 W Substituting in Eq. (i), we get RV = 4.5 × 10− 3 Ω 2Ω 2Ω i1 i1 – i2 (c) V =E 1 45. 3Ω (2) 1+ r (1) RV 7V 1V If RV is increased from this value, V will i2 increase. 42. (a) IA = R + ε r Loop equation in loop (1) RA+ ∴ ε = (R + RA + r) IA + 7 − 2i1 − 3 (i1 − i2) = 0 …(i) Loop equation in loop (2) …(ii) Now, I A′ = R ε r …(i) + − 1 + 3 (i1 − i2) − 2i2 = 0 Solving Eqs. (i) and (ii), we get Substituting the value of ε, we get i1 = 2A and i2 = 1 A 1 + RA  I′A = IA RA + r Power supplied by E1 = E1i1 = 14 W If RA → 0, IA′ → IA But power consumed by E2 = E2i2 = 1W (b) In Eq. (i) substituting 46. (a) i = 12 = 2 A 5+1 IA = 0.99 I′A and the given values, we get ⇒ P1 = Ei = 24 W RA = 0.0045 Ω (b) P2 = i2r = (2)2 (1) = 4W (c) IA = I ′A = I A′ (c) P = P1 − P2 = 20 W 1+ RA 1+ 1 r r 47. (b) i = V in each resistance RA + 1+ R RA (e) P = V 2 in each resistance If RA is decreased from this value, then IA will R increase from 99% of I′A.

Chapter 23 Current Electricity — 623 (f) P = V 2 Now let us find VAB across path ACB, R VAB = 2i1 + 8i2 = 61.94 V or P∝ 1 (as V is same) Now, VAB = inet Rnet = (10) Rnet R ∴ 61.94 = 10 Rnet 48. (a) P = V 2 or Rnet = 6.194 Ω (f) Two resistors in vertical middle wire can be R removed. ∴ V = PR = 5 × 15 × 103 = 273.8 V (g) Now balanced Wheatstone bridge, in parallel with 1Ω resistance between points A and B. (b) P =V2 = (120)2 = 1.6 W The encircled resistance of 2Ω can be removed R 9 × 103 from the Wheatstone bridge. 4Ω A 49. (a) 8Ω 6Ω 1Ω 2Ω 4Ω B 2Ω 2Ω A 4Ω B 2Ω 2Ω 1Ω 1Ω (b) A balanced Wheatstone bridge in parallel 2Ω with R. One resistor of 2Ω has already removed from the original circuit given the question. As its two ends 4Ω will be the same potential (by symmetry). (c) 2Ω 3Ω 2Ω 50. Ri + 5R 3Ω If connected by two equal resistors between B and D A 1Ω 1Ω B and between C and E, the combination is a balanced Wheatstone bridge and two resistors in series. 10 Ω ∴ Rf = R + R + R = 3R = 0.6 Ri (d) 10Ω 5Ω 5Ω 51. A 2Ω 15 Ω 15 Ω 8Ω B 5Ω 6Ω 40 Ω 10 Ω 5Ω AB 20Ω 30Ω 10 Ω 52. (b) Three resistors are in parallel. Then, one resistor 2Ω C 8Ω i2 i1 i1 – i2 in series with this combination. (e) A 10A (1) B (c) Balanced Wheatstone bridge. Hence, two 10 –i1 2Ω (2) resistors in vertical wire can be removed. 4Ω 10Ω 10–i2 (d) All four resistors are in parallel (e) A balanced Wheatstone bridge. 53. (a) A balanced Wheatstone bridge with one resistance in parallel. Let us take a current of 10 A between A and B Loop equation in loop (1) (b) − 2i1 − 2 (i1 − i2) + 4 (10 − i1) = 0 …(i) A B Loop equation in loop (2) − 8i2 + 10 (10 − i2) + 2 (i1 − i2) = 0 …(ii) Solving these two equations, we get i1 = 6.53 A and i2 = 6.11 A

624 — Electricity and Magnetism LEVEL 2 Further RA = R Single Correct Option or RBC = (1.5R) (3R) = R ∴ 1.5 R + 3R Because 1. No current will flow through voltmeter. As it is RA = RBC 8. VA = VBC ideal (infinite resistance). Current through two iRA = iRBC batteries 5A i = 1.5 − 1.3 = 0.2 2A r1 + r2 r1 + r2 C Now, V = E2 − ir2 B 7A ∴ 1.45 = 1.5 −  0.2  5Ω 3A  r1 + r2 (r2) 2A R Solving this equation, we get r1 = 3r2 100 V 2. In series, PD distributes in direct ratio of Applying loop equation in closed loop we have, + 100 − 30 − 35 − 2R = 0 resistance. In first case, 198 = 900 …(i) ∴ 2R = 35 V = VR VAB − 198 R1 In second case, 180 = 900 …(ii) V5Ω = 7 × 5 = 35 V VAB − 180 2R1 ∴ V5Ω = 1 Solving these two equations, we get VR VAB = 220 V 9. r = R  l1 − 1  l2 3. Maximum current will pass through A. = R  y − 1 =  y− x R P = i2R x x or P ∝ i2 (R is same) 10. Let R = at + b 4. 4 (R + RA) = 20 V At t = 10 s, R = 20 Ω ∴ 20 = 10 a + b ∴ R = 5 − RA At t = 30a + b …(i) where, RA = resistance of ammeter Solving these two equations, we get …(ii) 5. r = R  l1 − 1 = 132.4  70 − 1 ≈ 22.1 Ω a = 1.0 Ω/s  l2 60 and b = 10 Ω ∴ R = (t + 10) 6. Initial current, i1 = E1 + E2 R+ r1 + r2 i = E = 10 R t + 10 Final current, when second battery is short circuited is i2 = E1 r1 30 R+ ∫∆q = idt 10 i2 > i1 if E1 > E1 + E2 ∫= 30  10  dt R + r1 R+ r1 + r2 10  t + 10 or E1R + E1r1 + E1r2 > E1R = 10 loge (2) + E1r1 + E2R + E2 r1 11. Suppose n (< 1) fraction of length is stretched to m or E1r2 > E2 (R + r1) times. (1 − n) l + (nl) m = 1.5l 7. B and C are in parallel Then, nm − n = 0.5 or ∴ VB = VC …(i)

Chapter 23 Current Electricity — 625 R = ρl = ρl (V = volume) 15. When K1 and K2 both are closed R1 is A (V /l) short-circuited, = ρl2 Rnet = (50 + r) Ω V When K1 is open and K2 is closed, current remains or R ∝ l2 (if V = constant) half. Now, the second condition is Therefore, net resistance of the circuit becomes (1 − n) R + (nR) m3 = 4R two times. ∴ nm2 − n = 3 …(iii) or (50 + r) + R1 = 2 (50 + r) Of the given options, the above equation is Solving these two equations, we get satisfied if n= 1 r = 0 and R1 = 50 Ω 8 16. 100 Ω, 25 Ω and 20 Ω are in parallel. l1 = X = 2 12. Initially, 100 − l1 R 3 Their, net resistance is 10 Ω ∴ l = 2 × 100 ∴ Rnet = 4 Ω + 10Ω + 6Ω = 20Ω 15 V = i Rnet = 80 V 17. All these resistors are in parallel. = 40 cm ∴ Rnet = R + r = 4Ω 3 Finally, l2 = X = 12 = 3 ∴ 100 − l2 R′ 8 2 Hence, the main current l2 = 3 × 100 i = E =1A 5 Rnet = 60 cm Current through either of the resistance ∴ J is displaced by is i or 1 A 33 l2 − l1 = 20 cm ∴ V = iR =  13 (9) = 3 V 13. In parallel, current distributes in inverse ratio of S resistance. 9Ω 0.9Ω I 0.01A (I – 0.01)A 18. 0.03 – IG A 0.1 Ω B IG G ∴ 0.01 = 0.1 In parallel, current distributers in inverse ratio of I − 0.01 9 + 0.9 resistance. Solving we get, I = 1 A 0.03 − IG = G = r = 4 IG S (r/4) 14. Equivalent emf of two batteries ε1 and ε2 is Solving this equation, we get ε = ε1 /r1 + ε2 /r2 = (2/2) + (4 /6) 1/r1 + 1/r2 (1/2) + (1/6) IG = 0.006 A = 2.5 V 19. 8Ω = 4 Ω 6Ω 3Ω Now, VAN = ε ∴ VA = VB or VAB = 0 ∴ (IAN ) (RAN ) = ε 20. V = iR or  12  (4) (l) = 2.5  4 +4× 4  ∴ V ∝R (as i = constant) Solving this equation, we get ∴ VA =  ρlA   ρπlrBB2 l = 25 m VB  πrA2   24 

626 — Electricity and Magnetism or rB = VA × lB between 2kΩ and 2kΩ. Hence, reading of rA VB lA voltmeter = 3×1=1 262 = 10 = 5 V 2 21. Current decreases 20 times or 2 times. Therefore, 27. R2 = R3 as P = V2 and in parallel V is same. 30 3 R net resistance should become 3 times. i /2 2 ∴ R + 50 = 3 (2950 + 50) i R2 R1 i/2 R3 2 Solving we get, R = 4450Ω Hence, PR2 = PR3 If R2 = R3 22. E0 = VAC = (i)AC (R)AC =  1E0  10 × 0.2 Now current through R1 is double so R1 should be 1 1 4 th of R2 or R3 for same power. As P = i2R. =E …(i) More Than One Correct Options 5 …(ii) V2 V 2t1 In second case, R1 H E   10 0.3 1. H = t1 ⇒ R1 =  10 + x 1 E0 = × Solving Eqs. (i) and (ii), we get Similarly, R2 = V 2t2 H x = 5Ω  V2  23. V and V0 are oppositely connected. In series, H =   t  R1 + R2 24. Balanced Wheatstone bridge. Hence, 1.5 Ω H (R1 + R2 ) resistance can be removed from the circuit. t = V2 i1 20 Ω 4Ω Substituting the values of R1 and R2, we get t = t1 + t2 1.4 A i2 In parallel, H = V2 t = V 2t  1 + 1 50 Ω Rnet  R1 R2 10 Ω i1 = 50 + 10 = 2.5 =V 2t H + H i2 20 + 4 1  V 2t1 V 2t2 ∴ i1 =  2.5  (1.4) = 1 A Solving we get, t = t1t2  2.5 + 1 t1 + t2 6V 25. Resistance between A and B can be removed due 2. 2 Ω i to balanced Wheatstone bridge concept. Now, RDE and RGH are in series and they are connected in parallel with 10 V battery. ∴ I DE = 10 = 10 5V 3Ω RDE + RHG 2+ 2 i = 6 − 5 = 0.2 A = 2.5 A 2+ 3 26. Net resistance of 3kΩ and voltmeter is also 2kΩ. V1 = E1 − ir1 = 6 − 0.2 × 2 = 5.6 V Now, the applied 10 V is equally distributed

Chapter 23 Current Electricity — 627 3. (a) In series, current is same. Vb − Va = 2i − 10 = 2 V ∴ i = 6A ∴ IA = IB (b) VA + VB = VC Now, VC − Va = 2 × 6 = 12 V ∴ IARA + IBRB = ICRC 10. Between a and c, balanced Wheatstone bridge is (d) In parallel, current distributes in inverse ratio formed. Across all other points simple series and of resistance. ∴ IB = IA = RC parallel grouping of resistors. IC IC RA + RB Comprehension Based Questions 4. Same as above. 1 and 2. r = R  l1 − 1  l2 5. (a) V = iR ∴ 10 = R  500 − 1 In series i is same. Hence, V is also same as R 490 is given same. Solving this equation, we get (b) R = ρl R = 490 Ω A Further r = R  E − 1 R is same. Hence, A should be smaller in first V wire. Secondly, vd = i or vd ∝ 1 or 10 = 490  2 − 1 ne A A V A of first wire is less. Hence, its drift velocity Solving, we get V = 1.96 V should be more. (c) E = V or E ∝ 1 (V → same) Match the Columns ll 1. Let potential of point e is V volts. Then, 7. If switch S is open, Iae + Ibe + Ice + Ide = 0 i1λl = E2 ∴  2 − V  +  4 − V  +  6 − V  +  4 − V  =0 where, i1 = current in upper circuit and λ is 1 2 1 2 resistance per unit length of potentiometer wire. ∴ Null point length, l = E2 or V = 4 V i1λ Now current through any wire can be obtained by the equation, (a) If jockey is shifted towards right, resistance in upper circuit will increase. So, current i1 will I = PD decrease. Hence, l will increase. R (b) If E1 is increased, i1 will also increase. So, l 2. i1 = i2 or i is same at both sections. will decrease. A1 < A2 (c) l ∝ E2 (a) Current density = i ∝ 1 (d) If switch is closed, then null point will be AA obtained corresponding to (c) Resistance = ρ ∝ 1 V2 = E2 − i2 r2 length A A which is less than E2. Hence, null point length (d) and (b) E or potential difference per unit will decrease. length = (i) (Resistance per unit length) 8. By closing S1, net external resistance will = (i)  ρA ∝ 1 decrease. So, main current will increase. A By closing S2, net emf will remain unchanged but 3. By introducing parallel resistance R3 in the circuit, net internal resistance will decrease. Hence, main current will increase. total resistance of the circuit will decrease. Hence, main current i will increase. 9. a 2Ω 10 V b ci Now, VR1 = E − VR2 = E − iR2 Since, i is increasing, so VR2 will increase. Hence, Vb + 10 − 2i = Va VR1 or current passing through R1 will decrease.

628 — Electricity and Magnetism 4. (a) R = H 2. (a) Due to symmetry about the shaded plane, current i2t distribution on either side of the plane will be ∴ [R ] = [ML2T− 2 ]/ [A2 T ] = [ML2T− 3A− 2 ] identical and points E and F will be at same potential and no current will flow through it. (b) V = iR ∴ [V] = [A ][ML2 T− 3A− 2 ]= [ML2 T− 3A− 1 ] i i1 E A i1 (c) ρ = RA l D i2 ∴ [ρ ] = [ML2T− 3A− 2 ][L2 ] = [ML3T− 3A− 2 ] [L] i2 B i3 F i3 1 (d) [σ ] = ρ  = [M − 1L− 3T3A2 ] C E 5. i = 4 − 1 = 1A (anti-clockwise) A 1+ 1+ 1 D (a) VA = E − ir = 4 − 1 × 1 = 3 V BF (b) VB = E + ir = 1 + 1 × 1 = 2 V C (c) |PA | = Ei − i2r = (4 × 1) − (1)2 (1) = 3W (d) |PB | = Ei + i2r = (1) (1) + (1)2 (1) = 2W Subjective Questions ∴ RAD = 2r × 8r = 8r Ans. 3 3 15 1. (a) Points D and E are symmetrically located with 2r+ 8r 33 respect to points A and C. The circuit can be redrawn as shown in figure. (b) Redrawing the given arrangement for resistance across AB. Potentials VD = VE B E R R/2 R R/2 D, E R/2 C A D AR F This is a combination of a balanced Wheatstone bridge in parallel with a resistance R. So, the B resistance between B and D (or E) can be C removed. 1 = 1 + R 1 R + 1 VC = VF RAC R + 2R 22 ∴ No current flows through DE and CF. or RAC = 2R Ans. 3 r× r 3 5 r+ r 5 ∴ RAB = 2 = r Ans. (b) With respect to D and E, points A, B and C all are 3 symmetrically located. Hence, the simplified circuit can be drawn as shown in figure. 2 RR 3. (a) Current flowing through resistance 5Ω is 11 A Power dissipated = i2R R R = (121)5 = 605 W D E (b) VB + 8V + 3V + 12V − 12V − 5V = VC RR A, B, C VB + 11V − 5V = VC ∴ RDE = R + R = 2R Ans. 6V = VC − VB 3 3 3 (c) Both batteries are being charged. Ans.

Chapter 23 Current Electricity — 629 4. (a) VA − VB = 6 V For voltmeter, range Since, VB = 0 V = Ig (99 + 101) V = 200 Ig ∴ VA = 6 V Also resistance of the voltmeter …(ii) VA − VC = 4 V ⇒ VC = VA − 4 = 2 V = 99 + 101 = 200 Ω (b) VA − VD = VA − VC = 4 V From unitary method, we can find that, Ig 99 Ω 101 Ω AD =  1060 (4) = 66.67 cm G (c) Since, they are at same potential, no current V will flow through it. In Fig. 2, resistance across the terminals of the (d) VA − VB is still 6 V ∴ VA = 6 V Further, VA − VC = 7.5 V battery ∴ VC = − 1.5 V Since, EMF of the battery in lower circuit is R1 = r + 200 × 2 = 2.99 Ω more than the EMF of the battery in upper 200 + 2 circuit. No such point will exist. ∴ Current drawn from the battery, 5. (a) There are no positive and negative terminals I1 = 12 = 4.01 A on the galvanometer because only zero 2.99 deflection is needed. ∴ Voltmeter reading 4 V = 12 – I1 r = 12 – 4.01 × 1.01 5 G V = 7.96 × 5 = 9.95 V Ans. 4 Ans. (b) X 12 Ω 9.95 AJB C D Using Eq. (ii), Ig = 200 = 0.05 A Using Eq.(i), range of the ammeter (c) AJ = 60 cm ⇒ BJ = 40 cm I = 100 Ig = 5 A If no deflection is taking place. Then, the 7. Applying Kirchhoff’s laws in two loops we have, Wheatstone bridge is said to be balanced. R 10 V Hence, X = RBJ or X = 40 = 2 1 i1 12 RAJ 12 60 3 6V 3Ω or X = 8 Ω Ans. 2 6. For ammeter 99Ig = (I – Ig) 1 i2 or I = 100 Ig …(i) i1 + i2 I Ig 99 Ω 6Ω G 10 − i1R − 6 + 3i2 = 0 …(i) 6 − 6(i1 + i2) − 3i2 = 0 …(ii) 1Ω Solving these two equations, we get …(iii) Ig is the full scale deflection current of the i1 = R 6 2 galvanometer and I the range of ammeter. + For the circuit in Fig.1, given in the question Power developed in R, 12 V × 1 = 3 A ⇒ r = 1.01 Ω Ans. P = i12R = 36 R r + 99 + 2)2 2 + 99 + 1 (R

630 — Electricity and Magnetism For power to be maximum, PD across switch = 10 + (1)i2 = 10 + 2 = 12 V Ans. dP = 0 dR 20 V 4Ω ∴ (R + 2)2(36) − (36R)(2)(R + 2) = 0 A6 A1 2Ω 6V A5 i3 or R + 2 − 2R = 0 Ans. (b) i1 15 V i2 2Ω ∴ R=2Ω A4 A2 For maximum power from Eq. (iii), we have A3 1Ω 10 V Pmax = 4.5 W Ans. i4 8. Applying loop law in loops 1, 2 and 3, we have 2A R When switch is closed 1 A E1 3 5A −2(i1 − i2) + 15 = 0 …(i) 7A E2 6Ω 2i1 − 9i2 – 2i3 − 11 + i4 = 0 …(ii) 4Ω 1 …(iii) 3A 3Ω 2 i2 + i3 + 3 = 0 …(iv) 8A 10 − (i4 − i2) = 0 Solving these four equation, we get i1 = 12.5 A, i2 = 5.0 A , i3 = − 8.0 A i4 = 15 A E1 − 12 − 24 = 0 Ammeter A1 A2 A3 A4 ∴ E1 = 36 V Reading (amp) 12.5 2.5 10 7 −E2 + 24 + 30 = 0 or E2 = 54 V And the current through switch is 15 A. Ans. −2R − E1 + E2 = 0 or R = E2 − E1 2V 2 10. AB = 10 m RAB = 30 Ω =9Ω A CB 9. Using the loop current method, 20 V 4Ω A6 1Ω G A1 2 Ω 6 V A5 1.5V i1 i2 i3 Potential gradient across wire, A2 2 Ω A4 15 V AB = 2 = 0.2 V/m 10 A3 1Ω 10V Now, VAC = 1.5 V or (0.2)(AC ) = 1.5 ∴ S AC = 7.5 m Ans. (a) −2(i1 − i2) + 15 = 0 …(i) (a) VAB =  RAB  × 2 =  30  × 2 = 12 V  RAB + 5  30 + 5 7 −4i2 + 20 − 2(i2 − i1) − 15 − i2 − 10 − 2(i2 + i3) − 6 = 0 2V or 2i1 − 9i2 − 2i3 − 11 = 0 …(ii) − 2(i2 + i3) − 6 = 0 5Ω A or i2 + i3 + 3 = 0 …(iii) 1Ω Solving these equations, we get 1.5 V C1 B G i1 = 9.5 A, i2 = 2 A and i3 = − 5 A Ammeter A1 A2 A3 A4 Reading (amp) 9.5 9.5 2 5

Chapter 23 Current Electricity — 631 ∴ Potential gradient across ∴ Power generated in Rx is AB = 12 V/m 70 P = V12 = 900Rx V2 Rx (50Rx + 600)2 Now, VAC = 1.5 V ∴  1720 (AC1) = 1.5 For P to be constant, ∴ AC1 = 8.75 m dP = 0 2V dRx Ans. (50Rx + 600)2(900 V 2) or − 1800 × 50 × Rx V 2(50Rx + 600) = 0 (50Rx + 600)4 A C2 B or 50Rx + 600 − 100Rx = 0 Ans. 1.5 V 1Ω ∴ Rx = 12 Ω 12. The two batteries are in parallel. Thermal power generated in R will be maximum when, i total internal resistance = total external resistance 5Ω or R = R1R2 R1 + R2 (b) VAC2 = V  E1 RE22 or (0.2)( AC 2 ) =  5  (1.5) Eeq =  R1 + 1  5 + 1 + R2 1 or AC2 = 6.25 m Ans.  R1 11. V = constant  E1R2 + E2R1   R1 + R2  20 Ω = Rnet = 2R1R2 R1 + R2 V 30 Ω ∴ i = Eeq = E1R2 + E2R1 Rx Rnet 2R1R2 Maximum power through R 20 Ω Pmax = i2R = (E1R2 + E2R1)2 Ans. 4R1R2(R1 + R2) 30 Rx 13. V 2 = k (T − T0) + C  dT  30+Rx V1 R dt ⇒V or dT = dt V2 k (T C R − − T0)  30Rx  ∫ ∫or T dT = t dt   T0 V2 k (T 0 C  30 + Rx  R − − T0) V1 =  30Rx + 20 V 30 + Rx (at t = 0, temperature of conductor T = T0)  50R3x0+Rx600 Solving this equation, we get  = V T = T0 + V2 (1 − e−kt/C ) Ans. kR

24. Electrostatics INTRODUCTORY EXERCISE 24.1 F = Force between two point charges =  1  q × q  4πε0 a2 1. Due to induction effect, a charged body can attract a neutral body as shown below. Fnet = F 2 + F 2 + 2FF cos 60° ++ – + = 3F + ++ – +  3   qa 2 + 1+ 2   ++ – ++ =  4πε0 Body-1 is positively charged and body-2 is q q neutral. But we can see that due to distance factor A B attraction is more than the repulsion. r F 4. Number of atoms in 3 gram-mole of hydrogen –q 4. atom = number of electrons in it O = 3 N 0 = (3 × 6.02 × 1023) q where, N 0 = Avogadro number DC ∴ Total charge Net force on − q from the charges at B and D is = − (1.6 × 10− 19 ) (3 × 6.02 × 1023) zero. = − 2.89 × 105C So, net force on − q is only due to the charge at A. INTRODUCTORY EXERCISE 24.2 F =  1 q×q  4π ε0 r2 1. Fe = 1 q1q2 where, r = 2a = a 4π ε0 r2 ∴ 22 and Fg =G m1 m2 F =  1 q2 r2  4π ε0 (a/ 2)2 ∴ Fe = (1/4π ε0) q1q2 =  1  aq 2 Fg G m1m2  2π ε0 = (6.67 × (9 × 10+ 9 ) (1.6 × 10− 19 )2 × 10− 27 ) 5. The charged body attracts the natural body 10− 11) (9.11 × 10− 31) (1.67 because attraction (due to the distance factor) is = 2.27 × 1039 more than the repulsion. 2. F = 1 q1q2 7. F = 1 ⋅ q1q2 4π ε0 r2 4π ε0 r2 ∴ ε0 = q1q2 (q1)min = (q2)min = e2 4π Fr2 ∴ Fmin = 1 e2 4π ε0 r2 Units and dimensions can be found by above 9. Two forces are equal and opposite. equation. q INTRODUCTORY EXERCISE 24.3 3. a a q aq 60° F 1. Electric field lines are not parallel and equidistant. √3F = Fnet 2. Electric lines flow higher potential to lower F potential. ∴ VA > VB

Chapter 24 Electrostatics — 633 3. If charged particle is positive, and at rest. Electric = (9 × 109 ) (− 2 × 10− 12 )  1 − 21.0 1.0 field lines are straight then only it will move in the direction of electric field. = − 9 × 10− 3J 4. See the hint of above question. = − 9 mJ 5. Electric field lines start from positive charge and 3. Work done by electrostatic forces terminate on negative charge. W = − ∆U = U i − U f ∴ Uf =Ui −W 6. In case of five charges at five vertices of regular = (− 6.4 × 10− 8 ) − (4.2 × 10− 8 ) pentagon net electric field at centre is zero. Because five vectors of equal magnitudes from a = − 10.6 × 10− 8 J closed regular pentagon as shown in Fig. (i). 4. U ∞ = 0 D C DC E BE B Ur = 1 ⋅ q1q2 (For two charges) 4π ε0 r AA Ur ≠U∞ (i) (ii) For r ≠ ∞ Where one charge is removed. Then, one vector Ur = 1  q1q2 + q2q3 + q3q1  4 πε 0  r12 r23 r31  (let AB) is deceased hence the net resultant is equal to magnitude of one vector Now,U r can be equal to U ∞ for finite value of r. = |BA | = 1 q 4 πε 0 a2 INTRODUCTORY EXERCISE 24.5  1 7. E =  4πε0  q  (rp − rq ) 1. Wab = ∆U a − b = U b − U a = q (Vb − Va ) = q Vba r3 Here, r = (3)2 + (4)2 = 5m ∴ Vba = Wab = 12 = 1200 V q 10− 2 (9 × 109 ) (− 2 × 10−6 ) ∴ E = (5)3 (3i$ + 4 $j) (a) α = λ C/m C 2. x = m = m2 = − (4.32i$ + 5.76$j) × 102 N/C x dq INTRODUCTORY EXERCISE 24.4 (b) A x = –d x=0 dx 1. Ki + U i = K f + U f x+d ∴ 1 q1q2 = 1 mv2 + 1 q1q2 dV = 1 ⋅ dq 4πε0 ri 2 4πε0 rf 4πε0 x + d ∴v = 2  1 (q1q2 )  1 − 1  =  1  λdx  m  4πε0  ri rf   4πε0  x + d 2 109 ) 10− 12 )  1 01.5 = 1   αx x+ddx  10− 1.0  4πε0   = (9 × (− 2 × − 4 α L  ∫∴ x=L = 4 πε 0 1 + dL  = 18.97 m/s V= − d ln dV x=0 2. Work done by electrostatic forces P = − ∆U r =Ui −Uf 3. d dq =  1 (q1q2 )  1 − 1  x = –l x = 0 x dx x=l  4π ε0  ri rf 

634 — Electricity and Magnetism dq =  2ql dx Substituting in Eq. (i), we have W = Qq At point P, 2πε0L ∴ dV =  1  drq INTRODUCTORY EXERCISE 24.6  4πε0  q ⋅ dx  1.  ∂V i$ + ∂V $j  1  2l  (a) E = −  ∂x ∂y =  4π ε0    d2 + x2  = − a [(2x) i$ − (2 y) $j] = − 2a [xi$ − y$j] x=l (b) Again ∫V = 2 dV = −  ∂V i$ + ∂V $j x=0  ∂x ∂y E x dx = − a [ y$i + x$j] r 4. dq 2. E = − dV = − Slope of V -x graph. dx From x = − 2m to x = 0, slope = + 5 V/m ∴ E = − 5 V/m R From x = 0 to x = 2m, slope = 0, ∴ E=0 W = − ∆U = U apex − U ∞ From x = 2m to x = 4m, slope = + 5V/m, W = qVapex ∴ U∝ = 0 …(i) ∴ E = − 5 V/m as Vapex r=R From x = 4 m to x = 8m, slope = − 5 V/ m xL ∴ ∴ E = + 5 V/m r =  RL x Corresponding E-x graph is as shown in answer. 3. ∂V = − 50 = − 10 V/m ∂x 5 Surface charge density,  ∂V  2  ∂V  2  ∂V  2 σ= Q ∂x  ∂y  ∂z πRl Now, |E| = + + dq = (σ) (dA) No information is given about ∂V and ∂V . Hence, =  πQRl (2πr) dx ∂y ∂z =  πθRL (2π ) R x dx |E| ≥ ∂V or |E| ≥ 10 V/m L ∂x =  2Q  x dx 4. VA = VD and VB = VC as the points A and D or B L2 and C are lying on same equipotential surface (⊥ to Now, dV =  1  dxq electric field lines). Further, VA or VD > VB or VC as ∴  4 πε 0  electric lines always flow from higher potential to lower potential  2Q   πε0L2  VA − VB = VD − VC = Ed   = (20) (1) = 20 V = 4 dx INTRODUCTORY EXERCISE 24.7 ∫L Q 1. (a) Given surface is a closed surface. Therefore, we V = dV = can directly apply the result. 0 2πε0L

Chapter 24 Electrostatics — 635 φ = qin = 0 as qin = 0 B ε0 3. (a) (b) Again given surface is a closed surface. AC Hence, we can directly apply the result. Net flux entering from AB = net flux entering φ = qin = q as qin = q from BC. ε0 ε0 (b) S (c) Given surface is not closed surface. Hence, we cannot apply the direct result of Gauss’s φ = ES = E (πR2) theorem. If we draw a complete sphere, then φ through complete sphere = q 4. Given electric field is uniform electric field. Net ε0 flux from any closed surface in uniform electric ∴ φ through hemisphere = 1  q field = 0. 2  ε0 2. Net charge from any closed surface in uniform electric field = 0 ∴ Net charge inside any closed surface in uniform electric field = 0 Exercises LEVEL 1 ∫6. A VA − VB = − E ⋅ dr Assertion and Reason B 1. An independent negative charge moves from lower ∫= − (4,0) (4i$ + 4$j) ⋅ (dx$i + dy$j) (0,4) potential to higher potential. In this process, electrostatic potential energy decreases and kinetic (4, 0) energy increases. ∫= − (4dx + 4dy) = 0 (0, 4) 2. Two unlike charges come together when left ∴ VA = VB freely. 7. At stable equilibrium position, potential energy is 3. E = −  ∂V $i + ∂V $j + ∂V k$  minimum.  ∂x ∂y ∂z 8. In uniform electric field, net force on an electric  ∂V  2  ∂V  2  ∂V  2 ∂x  ∂y  ∂z dipole = 0 ∴ |E| = + + Therefore, no work is done in translational motion of the dipole. ∴ |E| ≥ ∂V ∂x Electric lines also flow from higher potential to lower potential. Electrostatical force on positive or = 10 V/m charge acts in the direction of electric field. Therefore, work done is positive. 4. V = kq or kq = VR 9. Charge on shell does not contribute in electric field R just inside the shell. But it contributes in the For inside points (r ≤ R), electric field just outside it. So, there is sudden change in electric field just inside and just outside E = kq r it. Hence, it is discontinuous. R3 10. |E | = 0 minimum at centre and |V | = 1 ⋅ q is or E ∝ r At distance r = R , 4πε0 R maximum at centre. 2 E = (VR)  R2 = V R3 2R

636 — Electricity and Magnetism Objective Questions 8. V = 1 ⋅ q1 + 1 ⋅ q2 1. φ = ES →  mV (m2) → volt-m 4πε0 3R 4πε0 3R ∴ 2. ge = g −  qmE = qnet (4πε0) (3R) R =  1  qnet   4πε0 3V l will or ge will decrease. Hence, T = 2π ge = (9 × 109 ) (3 × 10− 6) (3) (9000) increase. 3. Electric lines terminate on negative charge. =1m 4. W = ∆U i =Uf −Ui R r 2√2R 1 qq  1 qq  9.  4πε0 2l   4πε0 l  = 3 ⋅ − 3 ⋅ P =  − 32  4 1 ⋅ q2  l = − 3 Fl r = 3R  πε0 l2  2   = k qnet = k (3q) = kq Vp r 3R R 5. v = 2qV 10. 1 q2 = mg sin θ m 4 πε 0 r2 vp = 2eV  1  q2  m  4πε0   ∴ r= 2e (2V )  mg sin θ vd = 2m vα = 2 (2e) (4V ) = (9 × 109 ) (2.0 × 10− 6)2 4m (0.1) (9.8) sin 30° The ratio is 1: 1: 2. = 27 × 10− 2m 6. V = 1 ⋅ q = 27 cm 4πε0 r Q1 a Q1 V′ =  1  42qr = V q  4πε0 2 11. a –q q –q q F3 P Q1 7. ⇐ Q1 F1 F1 F2 q –q –q q Q1 = (2 2 − 1) Q W = ∆U = U f − U i F1 = Force between Q1 and Q1 at distance a = 1  − q2 − q2 + q2 + q2 − q2 − q2 F2 = Force between Q1 and Q1 at distance 2a 4 πε 0  a 2a a a 2 a a F3 = Force between Q1 and q at distance a  q2 q2 − q2 − q2 + q2 q2   2  − 2a   − + −   a 2a a a a For F3 to be in the shown direction, q and Q1 should have opposite signs. For net charge to be = q2 [4 − 2 2 ] 4 πε 0a zero on Q1 placed at P. |F3| = Resultant of F1, F2 and F1

Chapter 24 Electrostatics — 637 ∴ k Q1 q = 2F1 + F2 16. F1 = k q1 q2  = 1 (a/ 2)2 r2  k 4 πε 0  = 2 k Q1 Q1  + k Q1 Q1 F2 = k q′1 q′2  a2  ( 2a)2 r2 ∴ = Q1  + 1  =  2 2+ 1 where, q′1 = q′2 = q1 + q2 2  2   4  2 |q| 2  Q1 ∴ F2 = k (q1 + q2 /2)2 r2 =  2 2 + 1 (2 2 − 1)Q   4 + 2  q1 2 q2 > q1q2 = 7 Q or q = − 7 Q 44 This is because, 12. |a| = qE = qσ (q1 + q2)2 = (q1 − q2)2 + 4q1q2 m 2ε0 m or (q1 + q2)2 > 4q1q2 t = 2s = 4s ε0 m  q1 + q2 2 a | q| |σ| 2 or > q1q2 = 4 × 0.1 × 8.86 × 10− 12 × 1.67 × 10− 27 ∴ F2 > F1 1.6 × 10− 19 × 2.21 × 10− 9 = 4 × 10− 6 s 17. 13. +q –q 4Q B A Between A and B two forces on third charge will –4Q act in same direction. So, this charge cannot remain in equilibrium. +2Q –2Q To the right of B or left of A forces are in opposite +8Q directions but their magnitudes are different. Because charges have equals magnitudes but 18. 1000  4 πr3 = 4 πR3 ⇒ R = 10r distances are different. 3 3 14. Between 2q and − q, two electric fields are in same i.e. radius has become 10 times. direction. So their resultant can’t be zero. To the Charge will become 1000 times. right of 2q left of − q they are in opposite directions. So, net field will be zero nearer to V = 1 ⋅ (Charge) or V ∝ Charge charge having small magnitude. 4πε0 (Radius) Radius Hence, potential will become 100 times. q q 19. PD = qin  1 − 21R or PD ∝ qin 2 3 4 πε 0 R 15. E5 E1 q 20. E4 4 q 5 Q 1 E3 2a E2 q –Q E1 and E4 are cancelled. 3Q –3Q E2 and E5 are cancelled. +2Q ∴ Enet = E3 = 1⋅ q = q The desired ratio is − 3Q = − 3 4 πε 0 (2a)2 16πε0a2 2Q 2

638 — Electricity and Magnetism 21. S = (1) i$ ⇒ φ = B ⋅S = 5 V-m θ T θ q Q 26. qE 22. F1 mg Q q √2F1 T sin θ = qE F1 F2 or T = qE sin θ F2 = Force between q and q Similarly, T cos θ = mg = 1 q×q 27. E = σ E1 = E2 4 πε 0 ( 2 a)2 σ1 = σ2 ε0 V = σR F1 = Force between Q and q. For net force on q to ∴ be zero . ε0 or V ∝R F2 = 2 F1 ∴ V1 = R1 = a V2 R2 b ∴ 1 q2 = 2 1 ⋅ qQ  (as σ → same) 4 πε 0 2a2  4πε0 a2  ∴ |q| = 2 2Q with sign, q=−2 2Q y E 23. VB = 0 28. + –x q ∴ kqA + kqB = 0 q rB rB E ∴ qB = − qA = − q Potential is zero at infinite and at origin. Charge distribution is as shown below. Therefore, PD = 0. Hence, the work done asked in part (c) is also zero. –q q 29. According to principle of generator PD in this case only depends on the charge on inner shell. 30. 500 = k |q| …(i) r2 − 3000 = k (− q) …(ii) r From Gauss’s theorem, electric field at any point is given by Solving these two equations, we get E = kqin r=6m r2 ∴ |q| = 500 r2 qin inside A and outside B is zero. Therefore, E = 0 k 24. kq = 3  kRq 2 or r = 4 R = (500) (6)2 r 2 3 9 × 109 ∴ Distance from surface = r = R = 2 × 10− 6C =R 3 = 2 µC 25. φ = q in 31. 1⋅ q1q2 = 1 ⋅ q1q2 4 πε 0 r12 4 πε 0 k r22 ε0 ∴ r2 = r1 = 50 qin = 0 k5 ∴ φ=0 = 22.36 m

Chapter 24 Electrostatics — 639 32. WA→ B = q (VB − VA) 38. E = 0, inside a hallow charged spherical ABEdr = q A conducting shell. ∫ ∫= q − E dr 39. W = F ⋅ r B ∫= q a λ dr = qλ ln  21 = (QE) ⋅ r = Q (E ⋅ r) 2a 2πε0r 2πε0 = Q (E1 a + E2 b) +λ Subjective Questions 1. F = kq (Q − q)  where, k = 1 r2  4 πε 0  33. – + p For F to be maximum, dF = 0 dq Negative charge of dipole is near to positively By putting dF = 0, we get charge line charge. Hence, attraction is more. dq 34. r = (4 − 1)2 + (2 − 2)2 + (0 − 4)2 q= Q 2 =5m 3. E = σ V = 1 ⋅q σ = (2E ε0) 2ε0 = 2 × 3.0 × 8.86 × 10− 12 4πε0 r ∴ = (9 × 109 ) (2 × 10− 8 ) = 36 V = 5.31 × 10− 11 C/m2 5 Field is in the direction of r = rp − rq 4. C R CP 35. V = kq R ∴ kq = VR S Q E E = Kq = VR (i) (ii) r2 r2 36. W = Fs cos θ = qEs cos θ If loop is complete, then net electric field at centre C is zero. Because equal and opposite pair of ∴ E= W electric field vectors are cancelled. qs cos θ =4 If PQ portion is removed as shown in figure, then 0.2 × 2 × cos 60° electric field due to portion RS is not cancelled. Hence, electric field is only due to the option RS. = 20 N/C 1 qRS  1 (q/2πR) x 4 πε 0 R2  4 πε 0  R2 Q –Q ∴ E= = 37. k = 1 = qx 8π 2 ε0R3 C1 d C2 4πε0 = kQ − kQ 5. Let q1 and q2 are the initial charges. After they are R R2 + d2 VC1 connected by a conducting wire, final charge on them become. VC2 = kQ − kQ q′1 = q′2 =  q1 + q2 R2 + d2 R 2   Now, given that   ∴ VC1 − VC2 = Q 1 − 1 (9 × 109 ) (q1) (q2) 2πε0 (0.5)2  R R2 + d2  0.108 = …(i)