ELECTRIC CURRENT IN CONDUCTORS CHAPTER - 32

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1 1. Q(t) t + Bt + c a) t Q Q 'T ' T t T b) Bt Q B Q 'T' t T c) C [Q] C T ELECTRIC CURRENT IN CONDUCTORS CHPTER d) Current t dq d t Bt C dt dt t + B No. of electrons per second 16 electrons / sec. Charge passing per second Coulomb/sec Current i, t 5 min 5 60 sec. q i t c 6 4 c 4. i i 0 + t, t sec, i 0, 4 /sec. q t t t t idt (i t)dt i dt tdt t i 0 t C. 5. i 1, 1 mm 1 6 m f cu 9000 kg/m 3 Molecular mass has N 0 atoms N0I9000 m Kg has (N 0 /M m) atoms No.of atoms No.of electrons No.of electrons N0f N0f n Unit volume mi M i V d n e. V d i 1 3 ne coulomb sec

2 m/s mm/s m, r 0.1 mm m R 0, f? R f / a f Ra m. 7. volume of the wire remains constant. / f Specific resistance R f f ' ; R ' f 4f 0 4R / m, 1 mm 1 6 m I, n/v 9, t? i n V d e e V d V 1.6 d t V 1/ 8000 d sec. 9. f cu m 0.01 mm m R 1 K 3 R f a m 0.6 km.. dr, due to the small strip dx at a distanc x d R tan y a b a x L y a b a x L L(y a) x(b a) 3. fdx y (1) a dx x y Y a b b a

3 Ly La xb xa L dy 0 b a (diff. w.r.t. x) dx L dy b a dx dx Ldy b a Putting the value of dx in equation (1) dr dr R R fldy y (b a) fi dy (b a) y fi dy dr (b a) y b 0 a fi (b a) fl. (b a) ab ab 11. r 0.1 mm 4 m R 1 K 3, V 0 V a) No.of electrons transferred () i V R q i t 1 C. 17 No. of electrons transferred b) Current density of wire i /m m, I 1 f m E? R 8 f V IR 6 E 8 dv V V / m dl I 8.5 mv/m. 13. I m, R 5, i, E? V ir 5 50 V E V 50 5 V/m. I 14. R Fe R Fe (1 + Fe ), R Cu R Cu (1 + Cu ) R Fe R Cu R Fe (1 + Fe ), R Cu (1 + Cu )

4 3.9 [ (0 )] 4.1 [1 + 4 x 3 (0 )] (0 ) (0 ) (0 ) (0 ) (0 ) 19.5(0 ) 0. 3 (0 ) ( 3.1) C. 15. Let the voltmeter reading when, the voltage is 0 be X. I R V I R V V V.75.4 V V V V V 11.4 V 3.4 7(.4 V) 11(14.4 V) V V (7 11)V V 1.6 V 0.4 V. 16. a) When switch is open, no current passes through the ammeter. In the upper part of the circuit the Voltmenter has resistance. Thus current in it is 0. Voltmeter read the emf. (There is not Pot. Drop across the resistor). b) When switch is closed current passes through the circuit and if its value of i. The voltmeter reads ir ir 1.45 ir r 0.07 r E 6 V, r 1, V 5.8 V, R? I E 6 R r R 1, V E Ir R 1 6 R 1 0. R R V + ir r 1. r r a) net emf while charging 9 6 3V Current 3/ 0.3 b) When completely charged. Internal resistance r 1 Current 3/ a) 0.1i i i i 1 + 1i 1 + 1i 1 1 i BCD 0.1i + 1i i + 1i V 6 r r 1 i 1

5 DEF, i (i i)1 0 i i + i 0 i i 0 i ± 0.i 0 i 0. b) 1i i i 1 0 3i 1 1 i 1 4 DCFED i + i 6 0 i + i 6 BCD, i + (i i) 6 0 i + i i 6 i i 6 i ± i 6 i i + i 6 i 6 i 8 i i 8 c) i 1 + 1i i i 1 1 i 1 1 i i i i 1 6 i + (i i) i 6 i 0 1. a) Total emf n 1 E in 1 row Total emf in all news n 1 E Total resistance in one row n 1 r n1r Total resistance in all rows n n1r Net resistance + R n n1e n1n E Current n / n r R n r n R b) I n1n E n r n R for I max, n 1 r + n R min n r n R n rn R min 1 1 it is min, when n r n R 1 n 1 r n R I is max when n 1 r n R. 6 1 F E i i 6 1 D i B i C i B i i 6 1 E F i D i C i B i i 6 1 E F i D i C n 1 r r r r r r R 3.5

6 . E 0 V, R 0 k 0000 R 1 0 When no other resister is added or R 0. i E mp R 0000 When R i When R i Upto R 0 the current does not upto significant digits. Thus it proved Since 1 and are in parallel, X 4. X Reading in mmeter is V B 0 30 C i V 5.5V i /3 30 i min V 5.5V i 0 0/3 i max a) R eff i 60 / 60 1 b) R eff i 60 / c) R eff 180 i 60 /

7 6. Max. R ( ) 170 Min R The various resistances of the bulbs Resistances are (15) (15) (15),, V P 45,.5, 15. Since two resistances when used in parallel have resistances less than both. The resistances are 45 and i 1 0 i 1 i 1 i 0 i 1 4 m, i 8 m Current in 0 K resistor 4 m Current in K resistor 8 m Current in 0 K resistor 1 m V V 1 + V + V 3 5 K 1 m + K 8 m + 0 K 1 m volts. 9. R 1 R, i R R, i 6 R Since potential constant, i 1 R 1 i R 5 R 6 R R ( + R)5 60 5R R. i1m 5K 0 i i 1 i1m 0K B r r r a b a b Eq. Resistance r/ a) R eff b) cross C, R eff F 15/6 E B D C 3.7

8 c) cross D, R eff a) When S is open R eq ( + 0) 30. i When S is closed, R eq i (3/) a) Current through (1) 4 resistor 0 b) Current through () and (3) net E 4V V V () and (3) are in series, R eff i / 0. Current through () and (3) are Let potential at the point be xv. (30 x) i 1 (x 1) 0 i (x ) 30 i 3 i 1 i + i 3 30 x x 1 x 0 30 x 1 x 30 x 3 30 x 3x 36 x x 5x 40 11x 0 x 0 / 11 0 V. i 1 i i a) Potential difference between terminals of a is V. i through a / 1 Potential different between terminals of b is 0 V i through b 0/ 0 b) Potential difference across a is V i through a / 1 Potential different between terminals of b is 0 V i through b 0/ 0 a 30 V i 1 4 x O a i i 3 4V V b 4 3V 0 30 a V c 0 b S 6 1 b i V V 3.8

9 36. a) In circuit, B ba E + ir + i 1 R 3 0 In circuit, i 1 R 3 + E 1 (i i 1 )R 1 0 i 1 R 3 + E 1 ir 1 + i 1 R 1 0 [ir + i 1 R 3 E ]R 1 [ir i 1 (R 1 + R 3 ) E 1 ] R ir R 1 + i 1 R 3 R 1 E R 1 ir R 1 i 1 R (R 1 + R 3 ) E 1 R ir 3 R 1 + i 1 R R 1 + i 1 R R 3 E 1 R E R 1 i 1 (R 3 R 1 + R R 1 + R R 3 ) E 1 R E R 1 E1R ER1 i 1 R R R R R R E1 E E1R R3 ER1R 3 R1 R R R1 RR1 RR3 R R1 R 3 b) Same as a E 1 R 1 R 3 b a 37. In circuit BDC, i i 0 i + i (1) In circuit CFEDC, (i i 1 ) i 0 i i 1 0 () From (1) and () 3i 3 i 1 i 1 1 i 0 i i Potential difference between and B E ir V. 38. In the circuit DCB, 3i + 6i In the circuit GEFCG, 3i + 6i i i 1 6i 1 3 [i 16i 1 3]3 (1) [3i + 6i 1 4.5] () From (1) and () 8 i 1 54 i i + 6 ½ i i 0.5. Current through resistor 0. R E D a E 1 E R 3 R R 1 i 1 i C b B D a E 1 E R 1 C R 3 i 1 b R i C i E 13 D r 1 E 1 i 1 r B E 11 F i i 1 r 3 E E 3V F i i 1 D C i 3 G V B B 3.9

10 39. In HGB, + (i i 1 ) 0 i i 1 0 In circuit CFEDC, (i 1 i ) + + i 0 i i 1 + i 0 i i 1 0. In circuit BGFCB, (i 1 i ) + + (i 1 i ) 0 i 1 i + i 1 i 0 i 1 i i 0 (1) i 1 (i i 1 ) i 0 i 1 i 0 () i 1 i 0 From (1) and () Current in the three resistors is R 5 H i G i 1 F i E i i 1 i 1 i i i B C D i 1 5 For an value of R, the current in the branch is a) R eff (r / ) r (r / ) r r r r r a b r b) t 0 current coming to the junction is current going from BO Current going along OE. Current on CO Current on OD Thus it can be assumed that current coming in OC goes in OB. Thus the figure becomes r.r r 8r r r r 3r 3 3 (8r / 6) r 8r / 3 R eff (8r / 6) r 0r / 6 8r 6 8r 4r. 3 0 r r a B D O C b E 8r/3 r 8r/ I

11 43. a) pplying Kirchoff s law, i 6 + 5i 1 0 i + 5i 18 15i 18 i b) Potential drop across 5 resistor, i V 6 V c) Potential drop across resistor i 1. V 1 V d) i 6 + 5i 1 0 i + 5i 18 15i 18 i Potential drop across 5 resistor 6 V Potential drop across resistor 1 V 44. Taking circuit BHG, i i i V 3r 6r 3r i i r V 3 6 5i V r 6 R eff V 5 i 6r 1V 5 6V i 1V 6V 5 i B i/6 C i/3 i/6 i/3 D i/6 i/6 i/3 i/3 H i/6 i/3 G i/3 E i/6 F r/3 45. Reff r r r 3 5r r 8 r r 3 r r r r r r r r r r b R eff r 4r r 3 3 a a b r r r r R eff r r a r r b a r r b b R eff r 4 a a b 3.11

12 Reff r r r r r a b a r b 46. a) Let the equation resistance of the combination be R. R 1 R R R R R 3R + R + R R R R 0 R b) Total current sent by battery Potential between and B i i 6 i 3 i 1.5 a 47. a) In circuit BFG, i i + i i 1 + 3i 4.3 (1) In circuit BEDCB, 50i 1 (i i 1 ) i 1 00i + 00i i 1 00i 0 50i 1 40i 0 () From (1) and () 43i 4.3 i R 5i 1 4 i i eff mmeter reads a current i 0.1. Voltmeter reads a potential difference equal to i V. b) In circuit BEF, 50i 1 + i 1 + 1i i 1 + i i i (1) In circuit BCDEB, (i i 1 )00 i 1 i i 00i 1 i 1 50i i 5i 1 0 () From (1) and () i 1 (65) i i Reading of the ammeter 0.08 a Reading of the voltmeter (i i 1 )00 ( ) 00 1 V V B C B C B 3 i 1 i i V 00 i 1 i i i E D V 00 3-i R G F F E D

13 48. a) R eff V i i (0.3 i) 400 i 1. 4i 5i 1. i 0.4. Voltage measured by the voltmeter b) If voltmeter is not connected R eff (00 + 0) 300 i V Voltage across 0 (0.8 0) 8 V. 49. Let resistance of the voltmeter be R. R 1 50R, R 4 50 R Both are in series. 30 V 1 + V 30 ir 1 + ir 30 ir ir 1 30 ir 1 30 R R R 1 R V R R R1 V 1 30 R R R R 50 R 4 50 R 50R (50 R) 30(50R) (50 R) (50R 4)(50 R) 50R 0 4R R 18 18(74R + 0) 1500 R 74R 0 133R R R R 1600 / Full deflection current m ( 3 ) R eff ( ) 600 V R eff i V. 51. G 5, Ig 1 ma, I, S? Potential across B is same 5 3 ( 3 )S S m i i V S V V R 5 B

14 5. R eff ( ) 0 i (1 / 0) (The resistor of 50 can tolerate) Let R be the resistance of sheet used. The potential across both the resistors is same R R RD If the wire is connected to the potentiometer wire so that, then according to wheat stone s R 1 bridge no current will flow through galvanometer. RB LB 8 (cc. To principle of potentiometer). R L 1 3 DB B I B + I DB 40 cm I DB /3 + I DB 40 cm (/3 + 1)I DB 40 cm 5/3 I DB 40 L DB cm. I B (40 4) cm 16 cm. 54. The deflections does not occur in galvanometer if the condition is a balanced wheatstone bridge. Let Resistance / unit length r. Resistance of 30 m length 30 r. Resistance of 0 m length 0 r. For balanced wheatstones bridge 6 30r R 0r 30 R 0 6 R a) Potential difference between and B is 6 V. B is at 0 potential. Thus potential of point is 6 V. The potential difference between c is 4 V. V V C 0.4 V C V V. b) The potential at D V, V D 4 V ; V BD OV Current through the resisters R 1 and R are equal. 4 Thus, R R R 1 R I 1 I 1 (cc. to the law of potentiometer) I 1 + I 0 cm I 1 + I1 3I1 0 cm 0 cm I 1 00 cm cm. 3 D cm DB 8 6 4V R G D 6V 1.15K 1 S 1 R 1 D R C R 30 0 B B

15 c) When the points C and D are connected by a wire current flowing through it is 0 since the points are equipotential. d) Potential at 6 v Potential at C V The potential at B 0 and towards potential increases. Thus ve potential point does not come within the wire. 56. Resistance per unit length 15r 6 For length x, Rx 15r 6 x a) For the loop PSQ (i 1 + i ) 15 6 rx (6 x)i 1 + i 1 R E (1) 7.5 6V C E P r i 1 V r i R15r M i E/ i G W T r B Q i 1 S For the loop WTM, i.r 15 6 rx (i 1 + i ) E/ i R r (i 1 + i ) E/ () For zero deflection galvanometer i rx. i 1 E/ i 1 E 5x r Putting i 1 E 5x r and i 0 in equation (1), we get x 30 cm. b) Putting x 5.6 and solving equation (1) and () we get i 3E r. 57. In steady stage condition no current flows through the capacitor. R eff i Voltage drop across resistor i R 1 V Charge stored on the capacitor (Q) CV 6 6 /3 4 6 C 4 C. 58. Taking circuit, BCD, i + 0(i i 1 ) 5 0 i + 0i 0i i 0i Taking circuit BFE, 0(i i 1 ) 5 i 1 0 i 0i 1 i i 30i From (1) and () (90 40)i 1 0 i i 5 0 i 5/ Current through 0 is (1) () D C i 5V 6F V i 0 i i 1 B 0 E 5V F 3.15

16 59. t steady state no current flows through the capacitor. R eq i 6 Since current is divided in the inverse ratio of the resistance in each branch, thus will pass through 1, branch and 1 through 3, 3 branch V B 1 V. Q on 1 F capacitor 1 c C V BC 4V. Q on F capacitor 4 c 8 C V DE 1 3 V. Q on 4 F capacitor 3 4 c 1 C V FE 3 1 V. Q across 3 F capacitor 3 3 c 9 C. 60. C eq [(3 f p 3 f) s (1 f p 1 f)] p (1 f) [(3 + 3)f s (f)] p 1 f 3/ + 1 5/ f V 0 V Q CV 5/ 0 50 c Charge stored across 1 f capacitor 0 c C eq between and B is 6 f C Potential drop across B V Q/C 5 V Potential drop across BC 75 V. 61. a) Potential difference E across resistor b) Current in the circuit E/R c) Pd. cross capacitor E/R 1 F 3f 0 1F F 1 B C 6V 4 3 E D 3F 4F 3f B 3f 1f 1f 0V C R E C d) Energy stored in capacitor 1 CE E E e) Power delivered by battery E I E R R f) Power converted to heat 6. 0 cm 0 4 m E R d 1 mm 1 3 m ; R K C 1 4 E d Farad. 3 Time constant CR s 0.18 s. 63. C F 5 F, emf V t 50 ms 5 s, q Q(1 e t/rc ) Q CV 5 q F 5 / R (1 e ) 3.16

17 e 3 5 / R e 5000 ln0.37 R R 5 / R K. 64. C 0 6 F, E 6 V, R 0 t 3 sec q EC (1 e t/rc ) e 1 x 5 (1 e 1 ) c. 65. C F, Q 60 C, R a) at t 0, q 60 c b) at t 30 s, q Qe t/rc 60 6 e c c) at t s, q 60 6 e c d) at t 1.0 ms, q 60 6 e c. 66. C 8 F, E 6V, R a) I V R 4 b) q Q(1 e t/rc ) (8 6 6) [1 c 1 ] Q 3.04 V 6 C 8 E V + ir i4 i 0.09 Å m d 0.1 mm 1 4 m R 16 ; emf V C E d 1 Now, E Q E F t / RC CV t / RC (1 e ) (1 e ) E (1 e ) V/m cm, d 1 mm, K 5, e 6 V R 0 3, t s C 1 4 KE d

18 q EC(1 e t/rc ) Energy 1 e Time constant RC sec a) q VC(1 e t/cr ) I Current dq/dt VC.(-) e t/rc, ( 1)/RC V t / RC V 4 1 e t / RC 6 t /0 R R e e 4 6 1/e t/0 t min, 600 sec. Q (1 e 6 ) I 5.9 mp e b) q VC(1 e t/cr ) 70. Q/ Q(1 e t/cr ) C I(in mp ) Q(in C ) Y V/R 5.9xx 8 amp O X t (in sec ) min Y O X t (in sec ) min 1 (1 e t/cr ) e t/cr ½ t log RC n q Qe t/rc q 0.1 % Q RC Time constant 1 3 Q So, 1 3 Q Q e t/rc e t/rc ln 3 t/rc ( 6.9) q Q(1 e n ) 1 Q C Initial value ; 1 q c 1 q c 1 Q C Q Q q q Final value Q Q(1 e n ) 1 n 1 e e n 1 n log Power CV Q V Now, QV QV e t/rc

19 ½ e t/rc t ln 0.5 RC ( 0.69) Let at any time t, q EC (1 e t/cr ) E Energy stored t / CR t / CR q E C E C (1 e ) (1 e ) c c de E C 1 E (1 e )e e 1 e dt RC CR 3.19 R rate of energy stored t / RC t / RC t / RC t / CR dr E 1 e (1 e ) ( ) e e dt R RC t / CR t / CR t / CR(1 / RC) t / CR t / CR t / CR t / CR E e e 1 t / CR E t / CR e e e R RC RC RC R RC RC For R max dr/dt 0.e t/rc 1 0 e t/cr 1/ t/rc ln t RC ln Putting t RC ln in equation (1) We get 75. C 1.0 F 1 6 emf 6.00 V, R 1 t 1 c, i i 0 e t/rc 6 t / RC 1 e 6 CV 1 6 e T b) Power delivered by battery We known, V V 0 e t/rc VI V 0 I e t/rc VI V 0 I e e W c) U CV (e ) T 6 t / RC 1 36 e dr E. dt 4R (where V and V 0 are potential VI) [ CV T 76. Energy stored at a part time in discharging 77. energy drawing per unit time] 1 CV (e ) t / RC Heat dissipated at any time (Energy stored at t 0) (Energy stored at time t) CV CV ( e ) CV (1 e ) t / RC t / RC 0 0 i Rdt i Re dt i R e dt t / RC 1 t / RC 1 i0r( RC / )e Ci0R e CV (Proved). 78. Equation of discharging capacitor 1 t / RC K 0 V ( dk ) / d K 0 V t / K q0e e e d d 0 0 K 0 Time constant is K 0 (1) is independent of plate area or separation between the plate.

20 79. q q 0 (1 e t/rc ) 0. 5( + ) e (1 e ) 0.75 Charge on each capacitor 0.75/ In steady state condition, no current passes through the 5 F capacitor, Net resistance 5. Net current 1 5 Potential difference across the capacitor 5 Potential difference across the resistor 1/5 4 V q Q(e t/rc ) V C(e t/rc ) /5 e e C Charge given by the capacitor after time t..9 C Current in the resistor 3 11m. 1 sec 81. C 0 F, emf 6 V, R 0 K, t 4 S. Charging : Q CV(1 e t/rc t 4 ) 4 4 RC 6 4 (1 e ) C Q Discharging : q Q(e t/rc ) e C 70 c. C1C 8. Ceff C C 1 Q C eff E(1 e t/rc C1C ) C C 1 6 E(1 e t/rc ) 83. Let after time t charge on plate B is +Q. Hence charge on plate is Q q. V Q q, VB q C C V V B Q q q Q q C C C V VB Q q Current R CR Current dq Q q dt CR dq 1 dt Q q RC q dq 1 Q q RC 0 0 t dt [ln(q q) lnq] 1 t RC Q ln q t Q RC Q q Q e t/rc q Q(1 e t/rc ) Q t / RC q (1 e ) 84. The capacitor is given a charge Q. It will discharge and the capacitor will be charged up when connected with battery. t / RC t / RC Net charge at time t Qe Q(1 e ). 5 6V C 1 V F 1 E 5F V B B r F C R

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