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CURRENT ELECTRICITy = | Se, - FORMULA LIST : —eEt (x) Cells m series (internal resistance) r =r +; (a) Drift velocity V,= | eat Seka ) - (1) Cells in parallel ¢, = ——= temp (b) Current density | — LU rr A (m) Cells in parallel r, = + {c) Resistance of conductor = R = — ' ' {internal resistance) s 3 ' AR a a ee {d) Resistivity o=—— (n) Balance ile ‘OF wWihteatstone s Dricge network — =— (¢) Relation between current and drift velocity eS r l= neAV , 4 ae ve = BEIT (o) Current in cell with external resistance =p ' : a > (f) Current density ] = cE pe 2 =< Rer -_ (g) Resistors in series R. = R, + R, (h) Resistors in parallel (p} Identical cells in series ; ’ 1 RR é.=ne effective emf - “Fe . i . Ba r =nr }\ effective internal resistance — : d- ‘ ir licle = (i) Relation between conductivity and (q) Identical cells in parallel e, = ¢ ae net —_—_ relaxation time =o= n a f eR (j) Cells in series e =e, + e, (effective emf) (r) Terminal potential difference V = ae ONE MARKS QUESTIONS: (i) length (ii) Areaofcross-section and (iii) Temperature. 1, Name the charge carriers in metallic conductors, (MQP1) 4, Define resistivity. (M-19) & Bi: E66 BEERONS, Ans: Resistivity of the given material of wire is the resistance offered by the conductor of unit 2, Define drift velocity of electrons. length and unit area of cross-section (M-18, J-20) : ity < antion S.l Ans: The average velocity acquired by the free See egerent density and mention hie ; EQ electrons inside the conductor when subjected unit of it, (EQ) to the electric field is called drift velocity. Ans: It is defined as amount of current flowing pe! 3, Mention the factors on which the unit area of the conductor | , | | resistance of the conductor depends, S.] unit is Am~ | (J-19) Ans. Resistance of the conductor depends on > AN EXPERT 6] PCross-sectio: ty of nichrop 1, area oj density ‘n’. difference acro. hen a uniform e! ne conductor Atal volum: and Q. rons in the el hental volume - : y the chare ha drift vc + A , i” ' es L D derive ~ expression for equivalent es stance when two resistors a x col nnected in series. a | ane * Fe laa V; emete y | pI _ Consider two resistors R, and R, connected in series as shown in the circuit. From Ohm's law: PD across R,: V, = IR, and PD across R,: V, = IR, nbd} If R. is the equivalent resistance of the series combination, V = I R. (2) AsR,andR,areinseries,V=V.+V, — _.(3) Substituting (1) and (2) into (3), we get: IR, = IR, + IR, = (R, +R) Therefore R.=R, +R, a Derive an expression for the effective | Tesletamee of two resistors connected in ‘ea When two or more resistors are 2 ee in parallel then, > | es {IF ; mes) Potential difference across each fesistor is same and is equal to the _ applied potential difference across the . combination. (2) Sum ofthe current through each resistor is equal to the main current through the combination, ae Ml F Let R, and R, be the resistors connected in parallel and ‘V’ be the potential difference across the combination. I and! be the currents through R, andR _ respectively then, lel #1, by ohm’s law / = i y or R l=a—¥4 a he (1) It RK. is the equivalent resistance in parallel combination which draws the same current | on applying the same p.d. V. For equivalent resistance, V=IR, or I> : veal R, From equation (1) and (2) we get ug — F meee = et ted de eee ipo soles . mk bits eer 1 | =a +e | , | Hence, the reciprocal of the effective resistance in parallel combination is equal to the sum of the reciprocal of the individual resistance. FIVE MARKS QUESTIONS: 4%. Define ‘relaxation time’. Derive the expression for electrical conductivity of a material in terms of relaxation time. (MP2, ]-15,17,18 M-20} Ans: It is the average time interval betwe successive collisions ot electri as with | atoms or ions im the cond. A. ¢- ’ Lo + ‘\ : € t < © ee ee : | a << a S ‘ ¥ , Cansider a cui bi : bed iB Area vl train Gel “iy F section A smd re ¢ pes | i1 PUC - Physics (ein teay vies _7| 16 i We Sati) ly+n y = (Sie |_| hit a (4) i +r R +n Comparing eqn (5) with equivalent cells V= le ; e = Sa teh hits eg N Rep SH tr, 46. Deduce the balancing condition for _ Wheatstone’s network. (MQP2, M-16, 17, J-16,19, M-22) Ans: The figure shows a Wheatstone bridge } consisting of four resistance R,, R,, R, and R, connected in a cyclic order, The Wheatstone bridge is said to be balanced if current through the galvanometer is zero (1, = 0). Let the bridge be balanced and hence —61C,=0 Applying loop rule to loop MABM, we have: See tek =0>1R = LR, Pa (els) Applying loop rule to loop ANBA we have: Al, -1,)R, + (1, +1,) R, +1,G =0 >LR, = LR, a2) Dividing equation (1) by (2) we have AR, _ 1,R, UiRee. Ry Hee. Ts R, _ R, ee OR * fy om le L i me gee eo: Eee 7 | SA Bele | in - ¢-- Switch PROBLEMS 47. 100 mg mass of nichrome metal is drawn into a wire of area of cross-section 0.05 mm’. Calculate the resistance of this wire. Given density of nichrome 8.4 x 103 kgm" and resistivity of the material as 1.2 « 10° Qm, (M-18) Ans: Given mass = 100 mg = 10’ x 10°= 104kg Area of cross section A = 5 x 10° x (10°) m?, Le, A=5 x 108 m2. Density D = 84x10 kgm, p = 1.2x10° Qm but Mass Density = volume mass Lom density 8.410: = 0.119107 m' ree OSE ae A SOP Le., length =0.238 m i.e., volume = aA = 0.0238 x 10'm >/ 1.2x10° x 0.238 we know that, R = lake = A 541.0 = 0.05712 x 100 Or R=5 7122 Resistance of the wire =5.712Q 48. A wire having length 2.0 m diameter 1.0 mm and resistivity 1.963 x 10° 2 m is connected in series with a battery of emf 3V and internal resistance 1. Calculate the resistance of the wire and current in the circuit. (j-16) 1.963 Ans:/= 2.0m, d= 1.0mm = 1.0 x 10°m, p= ~ 10"%Om, e=3V, r= 12 y/ mel WKT R=F (=e ) K 4/p 4x2?x1.963x 10 nd 3 L4x(1.0x10 15.704 Bd R = 4.998 x 10°25 10-2 connected to and neglig! b determine °' each resis! awn from the ()-! i attery a cun 1rough combin rhe, €ach resist.) ae 0.6x2.5=RR, RR, = 1.50 CHRP == (R, +R)’ -4RR, | = (2.5)'-4% 1.5 = 6.25 -6 RR, = V0.25 2 RB = 85 Solve (1) and (2) From (1), we can write, R, ~ R, =25 L5+R,=2.5 ‘R,=2.5-15 [a=10 two resistors are connected in ‘with a cell of emf ZV and negligible £1 al resistance, a current of ~ ~ A flows the circuit. When the resistors are M in parallel the main current is Caleulate the resistances. (M-17) : | acverraee{ 2) | When resistors paral (2) Current ;._ © Pes hs for equation (1) R+r / R,=R, +R, = 3 / 2 Ri +R; cy Bae gl 2 7 . 2 NK, “He 50° ee ee (3) for equation (2), R, = eee R+R, 1 Rie 2 RR. 6 wee Mine StS = 6) "2 ae pala [RR = 6 . (4) ee ee using formula (R, = R= (R, +R,)? -4R, R_. =5* - 4(6) = 25 - 24. et Sad ge Aer (5) Solving equations (3) and (S) R.+R,=5 R,-R,=1 2R.=6 oo [rR = 30) “> Ry = 322) USINE R +R, =5 R, =22 3+R, =$§ eae a R, =5-3 58. (a) Three resistors 32. 40 and 120 are connected in parallel. What is the effective resistance of the combination. (b) If the combination is connected to a battery of emf 6V and interna! resistance 0.50, find the current drawn from the battery and terminal potential § difference across the battery. (M-20) oo RR Vast ira © (3x 0.5)- 0 on oe a if wis ee circuit, find the current 7Q (M-22) O202e0e0 — iF =050 Now, Equivalent emf of the combination is => |E_=3V H-PUC Physics 
