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CHARGE (Q) :— Toa types of charge : positive charge ® and Negative charge © tike chargesvepe!] «@O-> 1 <00— Uhlike c changes attract -«@oe— 1e-= -16 X10 Me SL Unit of chavge : Coulomb (c) smallest independent chaype : (fundamental chavge) CURRENT (I) t=) cuvsent is vate 2 flow of positive charge. barge (c) , ee Ge) § Tr CED) icecton ofcussent= opposite tothe Flow of & (eaKyu ay! thi? SI unit of curssent = Ampese (A) wr othes populay unit= mA micro POTENTIAL DIFFERENCE(v) :- Potential digresence between 4wo points is amount a done in moving a unit charge tie)” péom one point the other. v= potential dipperence (volt) (v) w= workdone(zoule) (x) a= charge (couloms) (cy “= O% S.L unit of PD is volt (V) Why Cussent Flows ?- Potential Di¢fevence (v) PD) e- flows lowes potential to Higher potential (Lp—» HP) seamen plows from Higher potential to lower potential( HP LP) + symbol ofcell ——|}— ‘ ‘A bination of Cells - Battery mall; tithe = Te HSV isy” » vy 1 {) c AS W= av oS 1 Gy work o PD a Vv Resistance(R) —Www— property of @ conductor to oppose the Flow of electric cursent . Factors on which vesistance of conductor depends. (01 ength a) Reb x f GAveafay Rat , Ly a encutor (Merial Resp” SM oe (=a o pee 1 Resisti vity (p)- property of material tH) Temperatute Tempt ) Metals Rt 23fillous (nichrome) Rt Tempt Rt xe ySemiconductor(e,Si) RL Pp: RA. 2m" 2am pr est unitopR: ohm oth * ST Unit of RecistivitytP): Am Free electron I+tongth A Area of cross section Resistivity (P) — only on Material «Resistivity (P) is property of the material - Does not change with shape and size of material {like length Area, Diameter or radius) - « Resistivity order: Metals< Alloys < insulators + Copper ald aluminium ave used fos transmission lines- «Thsulatoss like vubber and glass have high p= poor conductors. + sT Unit of wesistivity is ohm m. om Resistivity (a-m) ® A [aon Potential difference is measured by an instrument->Voltmeter. =] eat Electric Current is measured by an instrument -> ssgatg © Re c 1,62 *10-* GED eat hd 8 p =7yt a Area of civcle, ane RAZ = 1A BA oR Rx = arR, X 4R- same ? 4 2 i pelatiqn diameter Ap Note) Rp < Re “mp Note>2 series 5 betel www ‘Ree RitRa= 6t3= 4 wo Re Fi Ro Pavalle! Leolgd shot Re = BiRON x “RR + 62 Re Ri Ro Rit Ra ts Ll Re 6 3 Re= 649 _ 1B og 9 totes 3s 4 30 Re . Max Resistance —> series "% Min Resistance—> parallel Re = 3R Re=-> Advantage of parallel oves sesies Combination ;— () Independence of Appliances :— if one appliance pails or there is 4 short ctayit in one byance of the circuit the other appliances continue to operate normally. Tits is because each appliance is on a se e branch of the ctvcult, ensuring at a fault in one does not ageect the others. (2) Devices of diferent types need diferent current '— for example - a bulb and a Heater needs dligferent curzent and
— 2 A battery or a combination of cells ~ —{4r4eae 3 Plug key or switch (open) ~ — ofFF 4 Plug key or switch (closed) —F_- on 5 A.wire joint_/ 6 Wires crossing without joining See aS 7 Electric bulb @ or _@ _()_ 8 A resistor of resistance R- — oa" 9 Variable resistance or Rheostat at~ rw or arr 10 Ammeter_, —a-— Tl Voltmeter _“ —0}+— Electric Power(P) = Rate at which Electric energy is consumed - 7 aerne™ PaVaT pevt =v Aue wie aon P= Power, ST Unit - wait(W) =—PR Rk \ V= Potential difeence (volt) v=IR T=current( Ampere) rey ELECTRICAL ENERGY (F) Electrical energy supplied by cell $Tunita Enevay is youles(7) Ba pth eget Kwh Kia h walt loo Heati ng Egpect of Electric Cursent :— hihen an electric cursent posses trough « conductor or anelectric device , the conductor becomes hot afer some time and produces heat -This is called heating effect of electric current. “Joule’s lata of Heating s— Heat (H) produces in 4 Puve resistor () Hoe T* GH + (i= Pre) Heat(H) is measused inJoulec ‘An electric source can supply a charge of 500 caulomb. If the current drawn by a device is 25 mA. Find the time in which the clectric source will be discharged completely sao, 9=S006 (= 25mA= =25 xi634 @ v0. ie? com, O° Tt Sooz - O=QWHO TE 2.Define the term “potential difference” between two paints in an electric Circuit carrying current. Name and define its SI, unit, Also express it in terms of Sil unit of work and charge. t= 20x 107 = 20, e005 Calculate the amount of work done moved through a P. Q «= O= cell when 20 C of charge is W=? @=20¢ V=3v W=QV =20X3 = 6035 4, Assertion {A) : Electrons mave from lower potential to higher potential in a conductor. Reason (A) : A dry cell maintains el conductor. (a) Both Assertion (A] and Reason [A] are true and Reason (R) is the correct explanation of the Assertion (4). {b) Both Assertion (A) and Reason (R) are true, but Reason {R} is not the correct explanation of the Assertion (AJ. ic potential difference across the ends of a {e) Assertion (A) is true, but Reason (R) is false. {d) Assertion (A) is false, but Reason (R) Is true, Q=tt pea y Lod eR cASs 3 he resistance of a wire does not depend on its area of cross - section O--, O — snap [3] material uf 6.Write the relation between resistance R and electrical resistivity(p) of the material of a conductor in the shape of cylinder of length (I) and area of cross-section (A). Hence derive the Si unit of electrical resistivity. 7.The resistance of a metal wire of length 3m is 60.9. If the area of w will the resistivity of the wire change if the length and diameter of the wire both are doubled? Justify your answer. A=3m R FA Re=bon Az Yxigte P= A, Ox ee fo? Av2k ofd-s2d = 80207 ie = 8x10 FQm Resistivity isa property of the material, Does not change with shape & size ef material (ike length, Diameter 8. In the following figure, three cylindrical conductors A, B and C a shown along with their lengths and areas of cross - section. If they are made of same material, find R./Rp & Ra/Re- 9. A wire of given material having length (I) and area of cross- section (A) has a resistart¢e of 4/2 . Find the resistance of another wire of the same material having length 1/2 and area of cross - 10. How much does the the diameter of the wire is doubled 7 ance change, 11.The resistance of a wire of 0.01 cm radius is f 10.2 If the resistivity of the wire 50 x 10° © m, find the length of this wire. 12, A copper wire has a diameter of 0.2 mm and resistivity of 1.6 x 10-8 m. What will be the length of this wire to make its resistance 14? How much does the resistance change, if the diameter of the wire is doubled? A= 0.200 O2xloFm Ved. pe. a2? Relyn, dad 13. The potential difference he two ends of a circuit component is decreased to one - third of value, while Its resistance remains constant. What change will be observed in the current flowing through it? Name & state the law which helps us to ans i onn’s LA OHM'S LAW - The potential difference V, across the ends of 8 metallic conductor is directly proportional to the current flowing through it provided its temperature remains the same. 14. Find the equivalent resistance between X and Y. te Po tat. [2.2 a fol Ende Te 6 OTE tel >R= m ¥ ted => Re= QML ye ans. ees & en Es Rwy anne as ‘ zal had yt ' R s y ? ISA y tos Re =44tR abd DoS Rear LA fu. § 2b >Pea “ bet x (-22) 2 4 17. If four identical resistors, of resistance 8 ohm, are first connected in series so as to give an effective resistance R,, and then connected in parallel so as to give an effective resistance R,, then the ratio R/R, : (a eee R “a Xp ce HOMEWORK , COMMENT THE ANSWER ‘a i‘ whale \- ple * Phew pe ws 19. How you would connect three resistors each of resistance 4M, so that the combination has a resistance of 62, Ma te? { Musing 2pris OR Qs+lp Se rds Cin a a Ci. Series Realeusye ian Forde Rea ian 3 anes Res bi * 4 20. Show how you would connect three resistors each of resistance 6M, so that the combination has a resistance of 90. Also justify your answer. whe Ap ay Yi Sia Re wit Parte > Rae 33. Two LED bulbs of 1OW and 5W are connected in series. If the current flowing through 5W bulb is 0.0054, the current flowing through 10W bub is: sf 0.0054 34, For a heater, rated 4 kW and 220 V, calculate the following: (a) Energy consumed in 2 hours (b) If 1 kWh is prices at 74.50, then the cost of energy consumed. 35. In a house, 2 bulbs of SO W each are used for 6 hours daily and an electric geyser of 1 kW Is used for 1 hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of 38.00 per kWh. Total aongy = Fathpt Le sow so = Pht 1 Pot +t 6h JoPays += 6x2p 36. The rated voltage of an air-conditioner is 220 V. This air- conditioner consumes 22 units of electric energy in 10 hours, calculate: (i) Power rating of the air-conditioner, Current drawn by the air-conditioner, and i) Resistance of the air-conditioner | v=ER -Px ih _. [2202 10x 5 OK 5 j=2Q0VxT 2:2KN [2:2KW= g-2/. Waxiobd= 2p XT Joh | ASM re! that dissipal is used as a safety device 0) Current Ratin SA -—— fixe bone’ TEs | >| Appliance = | Arpliance| ya) uc V=220V P= toow 3) 6A PsvT OSA llod=224,¢ 2) GSA 
