Physics current electricity, Exercises of Physics

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Chapter Current Electricity CS robie-1: Electric Current, Drift of Electrons, Ohm's Law, Resistance and Resistivity L An infinite line charge of uniform electric charge density 4 lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity and electrical conductivity o. The electrical conduction in the material follows Ohm's law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material? [Adv. 2016] i) ne @ (b) i t (0, 0) (0, 0) io 0 © @ 3 t 0.9) 9 Consider a thin square sheet of side L and thickness ¢, made of a material of resistivity p. The resistance between two opposite faces, shown by the shaded areas in the figure is [2010] (a) directly proportional to L (b) directly proportional tot | (c) independent of L t GTZ ‘. fe (d) independent of ¢ T To verify Ohm’s law, a student is provided with a test resistor R,, a high resistance R,, a small resistance R,, two identical galvanometers G, and G;, and a variable voltage source V. The correct circuit to carry out the experiment is [2010] A piece of copper and another of germanium are cooled from room temperature to 80° K. The resistance of (a) each of them increases [1988 - 1 Mark] (b) each of them decreases (c) copper increases and germanium decreases (d)_ copper decreases and germanium increases The temperature coefficient of resistance ofa wire is 0.00125 per °C, At 300 K, its resistance is | ohm. This resistance of the wire will be 2 ohm at. [1980] fa) 154K (6) 1100K (c) 1400K @) 1127K 5 "Prue / False The current —voltage graphs for a given metallic wire at two different temperatures 7, and 7, are shown in the figure. [1985-3 Marks] t p I T, Va The temperature 7, is greater than T,. Electrons in a conductor have no motion in the absence of a potential difference across it. [1982-2 Marks] B22 Shown in the figure is a semicircular metallic strip that has thickness 7 and resistivity p. Its inner radius is R, and outer radius is R,. Ifa voltage Vy is applied between its two ends, a current / flows in it. In addition, it is observed that a transverse voltage AV develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is schematic and not drawn to scale) [Adv. 2020} (b) the outer surface is at a higher voltage than the inner surface (©) the outer surface is at a lower voltage than the inner surface @ Aves? oD) 9 Assertion and Reno Type Quen) IIE 9. Read the following statements carefillly: [1993-2 Marks] Y: The resistivity of a semiconductor decreases with increase of temperature. Z: Ina conducing solid, the rate of collisions between free clectrons and ions increases with increase of temperature Select the correct statement(s) from the following: (a) Yis true but Zis false (b) Yis false but Z is true (c) Both Yand Zare true (4) Yis true and Z is the correct reason for Y 10 Ifa copper wire is stretched to make it 0.1% longer what is the percentage change in its resistance? {1978} Topic-2: Combination of Resistances 1. Inan aluminium (Al) bar of square cross section, a square holeis drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are2.7 x 10-8 Om and 1.0 x 10-7 Q m, respectively. The electrical resistance between the two faces P and Q of the composite bar is {Adv. 2015] bh) gene © He O° aon . 2. Find out the value of current through 2Q resistance for the given circuit. [20058] §Q 102 10Vv 20V 20 @) zt (b) 24 (SA (@) 44 3. Six identical resistors are connected as shown in the figure. The equivalent resistance will be [20048] (a) Maximum between Pand R” (b) Maximum between Q and R (c) Maximum between P and Q (d) Allare equal 4. The effective resistance between points P and O of the electrical circuit shown in the figure is [2002s] 2Rr @ Rer 8R(R+r) ®) 3R+r (©) 2r+4R @ 2 +2r 5. The currenti in the circuit (see Fig) is @ =A t [1983-1 Mark] 302 6. In the following tl t R(=2Q)is/amperes. The value of Jis ¢ Current through the resistor [Ady. 2015] B24 6 | MEQs with One or Moe that One Connect Answer 4. Ina circuitshown in the figure, the capacitor C is initially uncharged and the key K is open. In this condition, a current of 1 A flows through the 1 Q resistor. The key is closed at time t= ty. Which of the following statement(s) is (are) correct? [Adv, 2023] [Given: e! =0.36] svg —t—w— 5v 12. ao i 30. SG (a) The value of the resistance R is 3 Q. (b) For t Roo? Rao (a) Roo Roo Rao Figure shows three resistor configurations R,, R, and R, connected to 3V battery. If the power dissipated by the configuration R,,R, and R, is P,, P, and P,, respectively, then — [2008] 19 12, 12 19 |sy _bv 19 sie 10 12 1g 12 10% Ri ra @ P\>P,>P, ) P\>P,>P, 19, 19 BV (©) P,>P\>P; l 1g 19 (d) P,>P,>P, 19 R3 The three resistance of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipation. [2003S] a) av) @ I W.-W, (0) Wy>W> W, © W aaal 52. The heat generated in the 4 ohms resistor is @) I calorie/ sec (b) 2 calories /sec (c) 3 calories /sec (@ 4 calories /sec A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if {1980} (a) both the length and the radius of the wire are halved. (b) both the length and the radius of the wire are doubled, (c) the radius of the wire is doubled. (d)_ the length of the wire is doubled. | Eee eee When two identical batteries of internal resistance 1Q each are connected in series across a resistor R, the rate of heat produced in R is J}. When the same batteries are connected in parallel across R, the rate is J,. If J, = 2.25 J, then the value of R in Q is [2010] 6. Inthe circuit P # R, the reading of the galvanometer is same with switch S open or closed. Then [1999-2 Marks] Two resistances R, = XQ and R, = 1 Q are connected to a wire AB of uniform resistivity, as shown in the figure. Theradius of the wire varies linearly along its axis from 0.2 mm at Ato | mmat B. A galvanometer (G) connected to the center of the wire, 50 cm from each end along its axis, shows zero deflection when 4 and B are connected to a battery. The value of X is 7 [Ady, 2022] R 8. In order to measure the internal resistance r, ofa cell ofemf E,a meter bridge of wire resistance R, = 50 ©, a resistance R,/2, another cell of emf £/2 (internal resistance r) anda galvanometer Gare used in a circuit, as shown in the fig- ure. Ifthe null pointis found at /= 72 cm, then the value of Q. [Adv. 2021] ws Rip Eon 9. In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are shown in the figure below. The resistance R, has temperature coefficient 0.0004°C"1, If the temperature of R, is increased by 100°C, the voltage developed between S and T will be volt. [Adv. 2020] Q By Asertion an Rew 'tSpe Questions B27 10. STATEMENT-1: Ina Meter Bridge experiment, null point Il. for an unknown resistance is measured, Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. STATEMENT-2 : Resistance of a metal increases with increase in temperature. {2008} (a) Statement-| is True, Statement-2 is True; Statement- 2 is acorrect explanation for Statement-1 (b) Statement-1 is True, Statement-2 is True; Statement- 2is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True 10 Pe An unknown resistance X is to be determined using resistances R), R, or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why? [2005-2 Marks] R=R, or Ry or R; A BC A thin uniform wire 4B of length Im, an unknown resistance X and a resistance of 12 Q are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [2002 -5 Marks] icy aye 5 et Are there positive and negative terminals on the galvanometer? Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points. al (b) 828 1 P 4. 5, 6, = Tapic-6; Miscellaneous (Mixed Concepts) Problems 1 i 2: 2 A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is [2007] BHF OF mukiietien ys 30. 62 ¥ = ov @) 0 (b) S4uC (©) 2mC —d) Bye Ifa steady current I is flowing through a cylindrical element ABC. Choose the correct relationship [2006] A 2r B ‘ E Cc ——> ————— ed a 12 @) Vay =¥ec (b) Power across BC is 4 times the power across AB (c) Current densities in 4B and BC are equal (d) Electric field due to current inside AB and BC are equal An ideal gasis filled in a closed rigid and thermally insulated container, A coil of 100 Q resistor carrying current | A for 5 minutes supplies heat to the gas. The change in internal energy of the gas is [2005S} @ kJ (b) 30kKI (c) 20k) () Oks A4 iF capacitor, a resistance of 2.5 MQ is in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given I7(2) = 0.693] {2005S] (a) 13.868 (b) 693s (c) 7s (d) 14s A capacitor is charged using an external battery with a resistance x in series. The dashed line shows the variation of In I with respect to time. Ifthe resistance is changed to 2x, the new graph will be (20048) @ P : ) @ Int (c) & @ Ss Shown in figure is a Post Office box. In order to calculate the value of external resistance, it should be connected between Ce A (20048) (a) BlandC (b) 4andD (c) CandD (d) BandD 48 In the given circuit, it is observed that the current J is independent of the value of the resistance R,. Then the resistance values must satisfy [20018) Rs Rie p, ERs (a) RiRpRs= RsRyRe att 2 ey vie a 1 O35 RRR mae © RRy= RR; (@) RiRs= RRy= RsRe Awire oflength Z and 3 identical cells ofnegligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by AT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount ATin the same time t. the value of Nis [2001S] @) 4 (b) 6 © 8 @) 9 The electrostatic field due to a point charge depends on qT the distance r as ae Indicate which of the following 2 quantities shows same dependence on r. [1980] (a) Intensity of light from a point source. (b) Electrostatic potential due to a point charge. (c) Electrostatic potential at a distancer from the centre of a charged metallic sphere. Given r < radius of the sphere. None of these 10, A galvanometer gives full scale deflection with 0.006 A il, current. By connecting it to a 4990 Q resistance, it can be converted inte a voltmeter ofrange 0 — 30 V. If connected 2n toa —~Qresistance, it becomes an ammeter of range 0— 249 1.5A. The value of is {Adv. 2014] At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. Ifthe capacitors have no charge initially, at what time (in sceonds) does the voltage across them become 4 V? [Take : In5 =1.6, In3=1.1] [2010] 2Mo das nme = 2M ay B30 Ww 4002 1002 1009 2002 hy 1002 1 1 1 10V 22. Inthe given circuit [1988 - 5 Marks] E, = 3E, = 2E; =6 volts R, = 2Ry =6 ohms R; = 2Ry =4ohms C=5nf. Find the current in R, and the energy stored in the capacitor. Ry z — Caf a & R ES Ry 23. Apart of ciucuit in a steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C (4uF) [1986 - 4 Marks] 24. 25. 26. 27. Calculate the steady state current in the 2-ohm resistor shown in the circuit in the figure. The internal resistance of the battery is negligible and the capacitance of the condenser Cis 0.2 microfarad. [1982 - 5 Marks] In the circuit shown in figure, a voltmeter reads 30 volts when it is connected across 400 ohm _ resistance. Calculate what the same voltmeter will read when it is connected across the 300 ohm resistance. [1980] V =6VOLTS 3002 4002 60V a A battery of emf2 yolts and internal resistance 0.1 ohm is being charged with a current of 5 amps. [1980] In what direction will the current flow inside the battery? What is the potential difference between the two terminal of the battery? In the diagram shown find the potential difference between the points 4 and B and between the points B and Cin the steady state. ~ [1979] (eae 3uf tye aw ly dhostes of \ We lov 102 202 saw} Hf ic Current Electricity B31 ? Answer Key Topic-1 : Electric Current, Drift of Electrons, Ohm's Law, Resistance and Resistivity 1 (©) 2. (0) 3. () 4. (d) 5. (d) 6. (True) 7, (False) 8 (acd) 9. (6) Topic-2 : Combination of Resistances : 1. (b) 2. (a) 3) 4. (a) 5. (©) 6 (1) 7. @) 8. & 9. (a,d) 10. () Topic-3 : Kirchhoff's Laws, Cells, Thermo e.m.f. & Electrolysis 1, (b) 2. 6) 3. (False) 4, (a,b,c,d) 5. (a,b,e,d) 6. (a,b,d) 7. (a,b,¢,d) Topic-4 : Heating Effect of Current 1 (dd 2. © 3. (a) 4. (d) 5. ‘(c) 6. (b) 7. (4) 8. (20) 9. (c,d) 10. (b,d) Topic-5 : Wheatstone Bridge and Defferent Measuring Instruments 1 © 2. (b) 3. (a) 4 (a 5. (a) 6. (a) FeilSy 8. @B) 9. (0.27)10. (d) Topic-6 : Miscellaneous (Mixed Concepts) Problems 1. (c) 2. (b) 3. (b) 4. (a) 5. (b) 6. (b) 7. (c) 8. (6) 9, (a) 10. 6) Il. @) 12. (a,b, ¢, d) 13, (c,d) 14. (abd)15. (6,d) 16. A+s;B>q;C>p.q;Dqr 6132 In a conducting solid, the collisions become more frequent with increase of temperature. 10. As we know, resistance, R = De @ 4. And V = Al=> A=“(V = volume of wire) Here pand Vare constant. R x /? AR Percentage change in resistance R x100=2 C/change in J) = 2(0.1%) = 0.2% Topic-2: Combination of Resistances il (b) As resistance of wire, R = Jo> PreXl pe _ 10-7 x50x10% LB gt re he 4x10° 2 Ou parxlar _2.7x10°8x50x107 2.750 te eA (49-4)x10° 45 =03 x 104 As potential difference across both resistors is same, so they are in parallel combination. _ Rrex Ra _12.5x104 x0.3x104 _ 1875 oh To) PO Reet Rat 128x104 4 (a) The current in 20 resistor = zero because it is not a part of any closed loop. 20V (©) Given resistance of each resistor 'R’ S 4 Rog = 7, ReRor = 7 R and Ree = “Rpg is maximum, (a) The circuit above and below the axis POQ symmetrical and represents balanced wheatstone bridge. Hence the central resistance 2R is ineffective. 2k 2R 2 Therefore the equivalent circuit is redrawn as follows. tet Sy rt x 4r+2R = + + = Rpg 4 4K 2r Rr ree EAPO Rap (©) __ In the given circuit, pe ru ipa Ry 30.60 30x60 esti v Current in the circuit, J = R 20 10 6. x @) Again the equivalent resistance of balanced wheatstone bridge fig (b) no current through 100 resistance. 6x18 _9 24 2 20 Balanced es wheatstone 1 } bridge £ $60 20 6sve 100 os Lee RUECS g (OV) Here in the circuit 8 batteries each of SV therefore, net emf of the circuit = 8 (SV) = 40 V Internal resistance of each cell, r= 0.20. ‘Net resistance in the circuit = 8 (0.2 Q) =1.6Q Current flowing through the circuit, [= OV =25A 162 Hence voltmeter reading V=E-IR = (5V)—(25A) 0.29) =5V-S5V=0 R (3) All the three resistances 2R, 2R and R are in parallel as shown in figure. Rap Lye Laie, R Hence, ib ence, 2 +-—+— = —or, Rig 2R aR Ree, ae R 2R 2R 10. il. B133 kQ Ry +R, “Gus” 75 9 + Rega = Ri + Ry= 2+ 5g = 3.240 @) 1-5 =a = 7.5mA I= Ip, Nie ee aS Me eg )ets-1.5m 2 (Rp +R, 15+6 Tp = Ip, ~ Ig, = 7-5 1.5 = 6mA L 1 2 (b) Potential difference across load Vey = (Upp ) Ry =6%1.5=9" (©) Ratio of powers dissipated in R, and Ry PR (tm) & (@) When R, and &, are interchanged, then R= PR. is b= = = Ryt+R, 35 7 6 Riou = 6+ R= 6 + 7 Now Pad, across R, 6/7 Vr =24( $17) yy REP cer rd. Therefore power wie 2 ie., Now potential becomes 5 = Z dissipated P = ne or Px ¥? will decrease by a factor of 9. (d) Here net resistance of the circuit = 9Q. -. Current drawn from the battery, Mi Sid 30 rei R79 7 LA = current through 30 resistor 32 A Fo Os 5 3 20 B20 D 28 Potential difference between 4 and B V4—Vg=9-1G+2)=4¥=8i, i= O05A 5 i, = 0.54 Similarly, potential difference between C and D Vo—Vin= (V4 — Vp) - in 2 +2) =4-4i,=4—4 (0.5)=2V = 8i, =G25A. iy ~ i, = 0.5 — 0.25 = 0.254 (i) In case of infinite ladder of resistances, the effective resistance, remains same when one identical item is added or removed from it. Effective resistance between points C and D be R then the circuit can be redrawn as shown Effective resistance between A and B feng ane 2xR =1+ be. R+2 a eS grt ee i . = of B D ir B D F aa 2xR ay R+2 => R+2+2R=R7+2R => R?~R-2=0 => R-2R+R-2=0> R(R~2)+1@-2)-0 => [R+1)[R-2]=0 > R=290. 6 (i) Rygr1O+1N=2N .«. Uyn= 573A ign = ipp &s resistances Rep = Rep Hence resistance between A and B, Ryp = 2 Topic-3: Kirchhoffs Laws, Cells, Thermo e.mf. & Electrolysis (Ss) (by The equivalent circuits are shown in the figure. A ; 2R alts ee: —>| x aR R AR £ 42 Balanced Wheat stone Bridge The circuit represents balanced Wheatstone Bridge. Hence no current will flow across 6R Q resistance 3R ik > Liciekped GRN6R) Ry 3R OR’? *&~ GRy4(6R)~ eq a For maximum power, Rexemat = Rinternal 2R=4Q -.R=202 2. (5) _Let i be the current flowing in the circuit. Applying Kirchhoff’s law KVL in the loop CDEFC urrent Elec y 8135 RR, _ (6Q)(120) _ R+R, 604122 Iy=( 35) -3=24 6+12, =h-h=14 Potential difference across A and P, Vig -Vp = 17x25 (2A4)(20) 127 ~Vp = 4V ot Vp =8V Potential difference across 4 and Q, V4 Vo = 1; x20.=(14)(4Q) 12 -Vp =4V Vo =12¥ -4V = av Potential difference across P and S, Vp —Vs =(2.4}(2W7) =4V BV Vs = 4V > Vs = 4V 2VeVo Applying Kirchhoff's voltage law KVL in loop BDAB +2(h-h)+1+1%h-24+2h=0 Req = > 2h-5h of ” V 4 = ‘Vv. gra Iv Fl H 1 G 2 oy q Applying Kirchhoffs law KVL in loop BCDB, we get ~Uh-1,)+ 1+1,-3 +34, =0 = 3 b=1 wo ii) 5 Solving eq. (i) and (i), we get 1 = 6/13 Aand f= 7 A (@® Potential difference between B and D, ae 2 29) ahr Re [5 aig Die ae Be ig -18 21 @ pd. soos G= 3-5 x3 = 91 =v [the cell is in discharging mode] i See pd. across H ie eas [-v cell is in charging mode] () Since no current is taken from the battery ae ae ZE/ ogg > Beg= oe ; (3(+% +4) ov ~(K)+O)+) Ss 1, 1 Internal resistance of equivalent battery = = (ii) If ry is short-circuited then resistance of this branch in zero, Applying Kirchhoff’s law in PQRUP starting from P moving clockwise Ir,-E, + E)=0 of ,-342=0 =1,=1A Applying Kirchhoff’s law in URSTU starting from U moving clockwise —E)+By- Ir, =0 or—-2+1-1,=0 +. 1,=-1A Here — ve sign of /; indicates that the direction of current in branch UTSR is opposite to that assumed. Applying Kirchhoffs law in AURBA starting from 4 moving clockwise. (,+h+h)R-£,=0 or (1+h,-1)R=2-. Hence, current through R, = I, + ,+I,=2A Topic-4: Heating Effect of Current 2 (d) We know power P= ae 1 «. For a constant V at a particular temperature P ses It is given that the power of the bulbs are in the order 10GW > 60 W> 40W 1 1 1 ee Rion Roo Ray (c) We know power, P= e And V is constant in all three cases. Pat Case (i) 10 19; 40 19 =20; 19 fe} 10 19 This is a case of balanced Wheatstone bridge R, = 10 20. = 310. #136 Case (ii) 2 i 19 19 and W; = a a= ‘ 2 2 2. 19 $i (250)? (250)? Rp (2509 2 fn Wy: Wy: W, = eR, Ry (R+R) (Ry + By oan ey 2 oo eet opie a JOS gy Or Case (il eee : 2, 60° “a E lara ys [eral v4 x1000 100° 60 100 60 or Wy: Wai W, ao _ gq, 100%100x 60 x60 100x100 x 60x 60 160 x160 x 60 160x160 x100 = 64:25:15 12 Therefore W, < W, < W. rs c) Here, I, a 21, (6+6 Thus R, = 20 q ‘ aa : Since, Ry 20 p Applying balanced wheatstone bridge condition, 7) = 2 R=20 x ( | & v 100-2 © | 2 Bc getee £ 100-£ 200 of, 100R-RE= EX 0, 200-26= €or, C= When the resistances are interchanged the jockey shifts 20cm. x 2 +20 > 020 7 80%- ae =2£+40 or, 80.X= £(X¥+2)+40= 3 Jr 2) 440 on et ea (a) Here, 1, = 100 x 10-6 4; G= 1002; S=01Q J, ri #l, Using /,6=(1-1,) 5 Galo Me ) yes 10010 | + or, 1=100 10 « 1000.1 = 100.01 mA (a) In case of meter bridge at null point ACT Eee CB 100—CB -——# |, YA R R ae 4 (3 R, B If radius of the wire is doubled, then the resistance of AC will R change andalso the resistance of CB will change. Butsince ra does 2 notchange so, 4 — C aout also not change at null: Boat ‘Therefore the pout ¢ does ne change. (a) Since the opening or closing the switch ‘s’ does not change the reading of galvanometer it means that in both the cases there is no current passing through S. It is the case of balanced wheatstone bridge. Thus potential at 4 is equal to potential at B. sdp= Ig and Ip = Ig 6) = ae, r=b c A 50cm 50cm 5 Fora frustum, R=LA ang, = 246 mab 2 pein bo) Rag Sn AC (An) Ri on By wh b cB Tm” 'y wheatstone bridge principal Ry Rye | Ry Rp Rog 1 So, R, =50 (3) Here nuil point is at 72 cm i(%+028% |= 2 BE vsartey cc? ER fey << L 28cm 72cm => 1x0.78R, -* Se eel EROTEE) ne A ae) Naa or, ry +15 Ry=1.56 Ry = 72156 Ry 1: 5 Ry = 0.06 Ry 2 r= 0.06 x50=32 (0.27) According to question, resistance R, has temperature coefficient, «= 0,0004°C"! And temperature is increased by 100° C i., AT = 100° C Ri, = Ry (1 + aAT) = 300(1 + aAT) =3120 I R=602 1 R,=1000 Ss 7-50 Au 3120 500.Q Here, = Mi = Ri +R; Vg~V,=3121, — 5001, rrent Electricity 50 500 = 312x—-50x>— = - 41.67 = 372 500 41.94 -41.67 =0,27V Hence voltage developed between S and T = 0.27 Volt 10. (d) When the temperature of metal increases; its resistance increases, R,= Ry (1+ « f) (Unknown resistance) (Standard resistance) 2. x s N b—! 100— £ Ql For a meter bridge at null point or at balanced condition Dame eso @ 100-€" § 100- 4 For mull point J! at same point or, 79g 9, remain i x unchanged, S should also remain unchanged. Hence if X is increasing then standard resistance ‘s' should also increase. 41. — Position-B of null point gives most accurate reading. Null point at B is almost at the centre of resistance wire wheatstone bridge is most accurate when all the four resistances of the four arms are of the same order of magnitude. igus In case of balanced wheatstone bridge condition, ~ = ie (a) No, There are no positive and negative terminals on the galvanometer. Whenever there is no current, the pointer of the galvanometer is at zero. (bo) 10 pa? @ 3. (0) Bridge is balanced oe ne = = 1 X=80 “ where p is the resistance per unit length. Topie-6; Miscellaneous (Mixed Concepts} Problems 1. (©) When the switch $ is open, total charge Q, + Q, = 0 CV, + CyVp=03 x6 +(—6 x 3) =0. When the switch is closed and steady state is reached, the current [ coming from the battery is 9V Potential difference across 3 resistance = 3¥ Potential difference across 6Q resistance = 6V <. pd. across 3 uF capacitor = 3 and p.d. across 6 WF capacitor = 6 Charge on 3 wF capacitor Q, = 3x 3=9 pC Charge on 6 WF capacitor 0, = 6 x 6 = 36 WC +. Charge flows from y to x = 36-9 = 27 uC As both wire of same material *. Ing = Igc —hn— es £ Pp 2 (ey Yap. lapRap _ Rap = —Anx4r Vac. tacRso.. ac. py Yar") V; Van = Za £ (©) @ (b) Here, change in intemal energy ~ heat supplied i =PRt =1x 1 100 x 5 x 60 = 30,000 J=30K3 B139 SS 4 (a) At any instant of time ¢ during charging process, the transient current in the circuit Vy .-1/RC i= —e R R = yet! RC te [Bert ler Potential difference across C V= Vo — Foe URE = Vy (1- e #RO) VY, = Wp a Wy[t-e/®°} = Wr et RC dete = 3gRC => 1 Ae RO Taking log on both sides t log, 1 = 2log,2 + - 4) . Potential difference across resistor R ii)