Electricity Shobhit Nirwan

ELECTRICITY HANDWRITTEN NOTES \\- - - - - ll Designed with Baa . Snobhit Nisman

② Electric charge [ Symbol of ] A physical entity which is defined by excess or deficiency of electron on a body . ° It is a scalar Quantity . charge° SI Unit of is Coulomb K) . chargedMagnitude of PROTON e= = -11.6×10-19 c on One \\ ELECTRON = e = - 16×10-19 C ° As on one electron = 1.6×10- MC chcahargrgeeo: on n electron = n x 1.6×10 -19C T e- . k3B Hitt question# no. of electrons symbol /q=NxeT[ Italia at THAT IT G- formula ] [- click that tree , Electric Current of electrons . p no. flowingchargeThe amount of O' ' through a particular area of cross- section in unit time 't' . ° It is a scalar Quantity ° Sz unit of current is Ampere CA) [ LA = I Esoeucbotnnbd] . 34T dldt definition at WH tf 9¥ ¥ current AT formula FT :- IE=¥T - K'B Direction -of electric current is taken as opposite to the flow of electrons and 3-HIT circuit ITT EFF current ④ve terminal of battery tf Ove terminal AT 4Th 01¥ I ° Electric current is measured by a device called Ammeter . LP : A current of LA is drawn by a filament of an electric bulb Find . - throughpassingfilament in 16 seconds I a cross-section of the the number of electrons set : Given I LA: - - t= 16 seconds fifty 419 3114T ? ? ) gHenoemdinoidthI but yet q To find : ofno electrons ? ? .

② IfWe know, E- #IA = 19=16-07 ¥611049Now, see D= ne x( ) 46C = n 1→n=Q Electric Potential [ Symbol V] The amount of work done 1W) when a unit positive charge (g) is moved 1y=wqTfrom infinity to a point . ° It is a scalar Quantity . - ° SI unit is volt CV) . SymbolElectric Potential Difference [ DV) It measures the work don e(wnp) er unit ( q) is defined as the difference in edloencetricperpoutennittialqubeatnwteiteyn two points charger. It ectric fiel d , equal to the work in an electrostatic field it from one poi of charge i in an el nt to another n moving Wapotential Difference - AV = or, Vb- Va = WAI Q ° It is a scalar Quantity . HIT tret Iit # formulated af acted I ° SI unit is my ) , → ° The electric potential difference between two points in a circuit is usingmeasured a device called Voltmeter. IP:- How much work is done in moving a charge of 2C from a point of 118 V to a point at 128 V ? SEL : Given : Q 2C- ( AspsouinmtingCBth! a)t charge is taken from point 'A'to - , VB =L 28V VA = 118 V To find : we ? ! Now , DV = Vb - VA = 428 - 118)V LOV Potential T ' DEF ht @blog) 20J diff . = GOV) ( 2 C) = K3B DV atoll formula UET thaw ? = line # , HHT TIFFT I Anse:- get question of line 4-64 stint Dr # definition dial

③ Electric circuit A closed and continuous path through which electric current flows is known includingas electric circuit . It has various components a source of current ( a cell or battery) g a load ( bulb appliance) , key Ito open or close a circuit) fuse owr iraensy , etc , all connected through . These wires are generally made of copper. ° When keyIswitch is closed, then circuit is called closed circuit lie . current will flow) . key° When I switch is open , then circuit is called open circuit (i.e. current would not flow) Circuittsiagram It is the pictorial representation of a circuit in which different electrical components of the circuit are presented by their symbols . Jes e #RattaMaarto :&::ammo:#no:÷÷÷÷:÷÷: ** -Ohm 's1Law . According to this law , the electric current flowing through a c ondu ctor is g directly proportional to the potential difference applied across it s ends ,prov idin the physical conditions (s uch as temperature) remains unchanged . Mathematically : If V is the pd. applied across the ends of a conductor through which current I flows , then according to Ohm baba:- V** d 2 Cat const temp) . org 111=171 Instant of proportionality called resistance .

④ K3B ° The conductors which obeys Ohm's law are called ohmic conductors while the obeyconductors which do not Ohm's Law are called non- ohmic conductors . from Ohm's law → f-IR f-Vz or Ratz ( R is inversely proportional to I) ( If V→ const ) . so if resistance is doubled then current gets halted and if resistance is halved , , , current doubled then gets . . 42MV7I,T¥TIcoU(nZTkcHeReEepHptTitnI tfgilht3ftrdAFataesttcdoitrnhtst3afnvtta)tluhteaftAHFTHSTcHhaTHnHgTefeetOlihfIfmNsTI,IIqATuHiHtSHtaikitpdroeidpfofrIetAicontality Resistance [ Symbol r] Is is that property byo3f 1aktcon21d14u7ctoof r data virtue of twhrhoiuchghitcuorrpepnostes#lresitsitls tohppeofsleoEwTHoEftf charge through it ( ¥ 314¥ . ° It is a scalar Quantity /R=IT/ o SI unit is ohm or hhomttpeoe) we know ohm's law → f- IR → [Cbse 2018J KB B - IRAITLcis FACTORS ON WHICH RESISTANCE of CONDUCTOR DEPENDS or ⑤ ②- Directly proportional to length of conductor IRIi e- . area of cross section of Inductor F.e. Inversely proportional to End Nature of material - Civ Temperature . ** Combining eg,② 4 ④ or Red hat or /R-;fRd at ← Constant of prop . called Resistivity. Resistivity or specific resistance [ symbol S] lengthResistance of a conductor of unit - and unit area of cross- section . O S2 unit is r- m 1ohm - meter) Resistivity of metallic conductor does not depend on the length or ° thickness of wire . • Metals have low resistivity . So they are the best conductor of current. , RB Alloys have higher resistivity than that of their constituent metals . They do not 0×9 die easily at higher temp , this is heatingwhy they are used to make elements of devices such as iron heaters etc ,. used Tungsten is a l mos t ex ctlruasnivsemlyi for f i tment of bulbs , whereas copper and aluminium are us d lectr ssio n lines generally e for e i c . [close 20183

(t) (3) ⑤ & ( 2) f All 3 are made up of LI:- tf Az 2A same material ← L -s s 2L s g . of them have resistance I highestWhich z← → length R=fIAset :-Given and area of cross - section of each conductor are aid formula til . ¥ IF use city I )÷¥££ → ask.com/usion- givenAlso d briars all 3 are of material , :O S will be same for all that same . Now g Is 1¥Rs = ) ) highestrina've4k Is . Ems )( 2¥Re l- 4¥ \"\"\" > - Roll f¥I te r÷ :#res me . IP :- A piece of wire of resistance 20h is drawn out so that its length - isweirnecreiansetdhetonetwwicseituitastioonriginal length.Calculate resistance of Initiallysoft . section - A - resistivitylength:- let=L =P & ofarea cross & fatA = 20 (given) -10 e Lg RT = finally , ALD :. 2e 4 ( will not Ak relaid point resistivity f¥#Now, Rf = = 4¥ changes of = 4120) 80€ Efrem ④ . ftp.rfKBB Re EA = area s,ecationsitcirqculees#tionartefa radius Cr) ut diameter lethal e f area of cross the formulae tf tant Itt AT erossection of circle efdt East of o-o-xi.de because wire heatterIP:. The potential diff between the terminals of an electric / is Gov when it draws a current of 4A from the source What current will the heater draw . if potential difference is increased to 120 V ? set :- As know that changing potential difference will not change resistance we . Intally by, f- GOV , 1=4 A 60=4 lR=L5I ohm's law a KIR finally g V' = R 15h- - - 120 Vg to find 8=2.2. again using ohm's laws say f.' us, 1218AM →

⑥ Series Combination ° HIT Resistances end to end attached ef and Hart same current EITI Ww 22 R2 I R3 Req tu mm > wir Mur nI ← Va → ←Vs→ N Tv v ya 1. I s V /-Req = Rs1t R2 t Rs Hn s f K - K' B o Equivalent resistance Is the sum of all the resistances connected in series . resistance is of either greaterequivalent° than the resistances resistor , af 3¥ I ** ° ooo ta TA resistance maximize that EF # series tf Gfcbse 20203 Series tf ET Resistance IT across I → same Ks different Disadvantages° of series : . If eamnywcilol mponent fails to work then the circuit will break and none f th work EB we cannot connect . bulb and heater in series because they need different valves of current- Parallelcombinaton ° tetane¥aoss V lpotential difference) same test I RL 9 I' ←IYm→ I Ii Req 7 ←mmf→ µ> I Rz z >I > I >zz←NYM- 3 , ←mR3yn_'23 a 13 NI Ie, - It , fairH s s it s n' v . + Kirts , K3B 0 The reciprocal of equivalent resistance Ps equal to the sum of the reciprocals of all individual resistances connected in parallel . equivalent°The resistance is less than resistance of either resistor Resistance minimize that EF the 3T Hath Parallel tf tot of , ** 2A : HI if ° u Parallel tf et Resistance I across v → same I → different ° As it divides the current among the components (electric gadgets) , so that they can have necessary amount of current to operate properly . This is the reason of connecting electrical appliances in parallel combination in household circuits .

1-037 Reading AmpereIn ⑦ Find cis of an mm mm mm ReadingLP:. - 11H s ④ of voltmeter fins current across Lr resistor K , L2V SEL : since they all are in series so diffI→ same & V - . Now g Req = Rst Rst Rz Gr by Ohm 's law voltmeter will give 996YECUT ? , Izzat - Is this is the red ingot④ potential difference )W- across 3h resistance . and we already know current CHELA Again:o by Ohm 's law , KIR quid reading of ① f- ② (3) GV → they2A ( : all will have same current as are in series ) . 2h Lpo. I Ii Nhk Y y 2, >I find current in each resistance . . > , sea mini 72, and in ammeter . A Knin of ' Hn lbova Lol: Since they all are in parallel g :. V → same and I→ diff . Leg =L + first lRee=LI Now by Ohm 's law : f- IReg 12=307- this will be the reading of ④ , 30=143 A s the y all are i n parall e l ng :O a ll will E am e v ie . 30W d l ohm 's aw :- N ow , we'l l fin c urrent i al by us i ng simply l =L 3oz→ for 2 r g I , = 15A IR 3yd→ for 3r g Iz - e a> LOA If 3ft→ for Gr g Iz = = 5A - MEEEE- calculate net resistance of circuit. FEIIp:. Foam- Ii SOI 's µIfn EETs:o Rs and Rz in series - Rit R2 25N g Rst Ry = 25N above circuit can be - mm- → Now both in parallel , + Is -(→Reg -12.5N

⑧ Heating effect of electric current -Itontqhepaernocfvueitdrelreeecncthttrteigocreefns.cetiirergcoyucf. hisittio,cmtmoceuaprymraerbnateti ntain the flow, the csource ontinuously has as ofdtishsisipasutepdplieind feonremrg yohfehlpesaitn. maintaining . This is known h Heating#Joule's law of [ Cbse 2020 , 20183 This law implies that heat produced in a resistor is : ire . IHEIZI - ② Is ddiirreecctltylypproroppoortritoinoanlatl otosqtuhaere of current in resistance dis resistance circuit i.e.lt#-⑤ for a given Yui directly proportional to the time for which current lows ie . 11k¥ from ② ¢ /H=I2R t → in seconds , - K3B PRACTICAL APPLICATIONS OF HEATING EFFECTS E- - 0 To produce light ( Electric Bulb) : It has a filament made of tungsten, Due to high resistivity and high melting point of tungsten g when voltage is applied across the filament it geratsdiahetiantged to aanvedry high temperature .It then becomes white hot heat light . and starts ° Electric fuse : It is used as a safety device in household circuits It consists . of an alloy of lead and tin which has appropriate melting point. When the cur rent flowing through the circuit exceeds the safe limit , the temp. circuit This of fuse wire increases , the fuse wire melt s and breaks the . elements helps to protect the other circuit from hazards ca used by currents . Lpo.. 200J of heat is produced tosee in a 5N resistance. find the potential diff - across the resistor . SII : Given : H - 200J - E- LOS Joule 's law H = 22Rt for chapter wise , previous year questions, R= 5h 200=245) CLO) please visit our 12=27 Nir Wan . channel → snobhit Now for Vg using Ohm 's law- f- IR Power [symbol p] K (2) (5) = LOK G It is defined as the amount of electric charge consumed in a circuit per unit time . ° It is a scalar Quantity . V IR P= IR)I = 12R o sz unit is watt ( W) → . - - - 1p-_ usingauhm §'s In, if → p= VII) = III form # than 34T Htt Edt I

I ⑨ of P = V I = I R = VI (cbse 2020,201912018J R O L kilowatt 1kW=Pt7 L mgeiggaawwaatttt (Mw) -106W T L (GW) = 109W energy 1 horse Power ( HPK 746W ° Commercial unit of electrical energy 1kWh = 1000Wh = 1000 X 3600 WS = 3.6×106 Ws or 3.6×1065 ° Number of units consumed by electric appliance = mattxhours 1000 LI: An electric bulb runs from the 220V mains The current flowing through it is 0.6A At . How much energy . by the bulb ! is transformed what rate is the electrical energy transformed in 2 min ? V= 220 V and 1=0.6 A g t = 2 min = 120sea power(P) V2 = 220×0.6 Energy transformedsod:- Given ; = 132 IN Rate of = Energy Transformed CE ) Pt = 132×120 = 15840% LIE- An Electric Refrigerator rated 500W operates /6 hours day . What is the cost of energy to operate it for 30 days at E 45 per kWh ? hooraysod:- Energy consumed by refrigerator in 30 days = 500W x 6 x 30 days = 90000 Wh 90kWh % Cost of energy to operate the refrigerator for 30 days = 90kWh X E 45 per kWh = E 405 ←