Title 19. Heating and Chemical Effects of current Easy Ans.pdf ✅

1. (a) 1 kWh = 1000 W  3600 sec = 36  105 W-sec (or J) 2. (b) 1 2 2 1 1 R R P P R P   = 2 1 1 2 2 100 200 R R R R  =  = 3. (c) R V P 2 =  = = = 1000  40 (200 ) 2 1 2 1 1 P V R and = = = 400  100 (200 ) 2 2 2 2 2 P V R 4. (b) When two bulbs are connected in series, the current will be same in both the bulbs. As a result potential drop will be more in the bulb of higher resistance i.e., bulb of lower wattage. 5. (b) When 1 bulb fuses, the total resistance of the circuit decreases hence the current increases. Since P i R 2 = , therefore illumination increases. 6. (c) 7. (c) We know that 1 2 1 2 2 1 = = R R P P 8. (a) R P R V P 1 2 =   and R  l 1 1 2 2 1 2 1    = = l l P P l P 9. (b) Rconductor  Temperature and Rsemiconductor  Temperatur e 1 10. (c) 11. (a) In series, current is same in both the bulbs, hence ( ) 2 P  R P = i R 2 1 2 1 2 1  = = R R P P 12. (c) In this case, R V P 2 = or R P 1  and R will be minimum, when divided four parts are joints in parallel to the battery. 13. (c) Length is immaterial for an electric fuse wire. 14. (a) R PRated 1  and 2 (Thickness of filament) 1 R  So 2  (Thicknessof filament) PRated 15. (d) In series n P PS =  3 10 P =  P = 30 W In parallel PP = nP = 3  30 = 90 W 16. (d) Energy consumed in 1000 Watt hour kWh  =  For 30 days, P 30 150 kWh 1000 10 50 10  =   = 17. (b) W = qV also t W P = i V = 18. (c) Because given voltage is very high, 19. (c) Pp = nP = 2  40 = 80 W 20. (a) In series, P  R( i is same), i.e. in series Fine wire (high R) liberates more energy. In parallel, V R P ( 1  is same) i.e. thick wire (less R) liberates more energy. 21. (d) Resistance of the bulb = =   = 484 100 220 220 2 PRatate V When connected with 110 V, the power consumed W R V PConsumed 25 484 110 110 2 =  = = 22. (a) The resistance of 25 W bulb is greater than 100 W bulb. So for the same current, heat produced will be more in 25 W bulb. So it will glow more brightly. 23. (a) Equivalent resistance in the second case = R1 + R2 = R Now, we know that R P 1  Since in the second case the resistance ( ) R1 + R2 is higher than that in the first case (R1). Therefore power dissipation in the second case will be decreased. 24. (c) For constant voltage, we know that R P 1  So higher the power, lower will be the resistance. 25. (d) R V P 2 = but A l R  =  l AV l A V P   2 2 / = = . Since l AV 2 is constant as per given conditions So  1 P  . 26. (d) Power consumed means heat produced. For constant potential difference Req P 1 Heat consumed =  4 1 2 / 2 1 2 2 1  = = = R R R R H H (Since 2 . 2 R R R R R R = + = and 2 ) R1 = R + R = R 27. (c) Resistance of carbon filament decreases with temperature while that of tungsten increases with temperature
In series PConsumed  R i.e. tungsten bulb will glow more brightly 28. (c) Power of the combination W n P Ps 500 2 1000 = = = 29. (b) For parallel combination Consumed PRated P  Brightness 30. (b) Resistance of 25 W bulb =   = 1936 25 220 220 Its safe current 0.11 . 1936 220 = = amp Resistance of 100 W bulb =   = 484 100 220 220 Its safe current 0.48 . 484 220 = = amp When connected in series to 440 V supply, then the current 0.18 . (1936 484 ) 440 I = amp + = Thus current is greater for 25 W bulb, so it will fuse. 31. (b) P i R 2 =  t i P P  =  2 (R → Constant)  % change in power = 2 % change in current = 21 = 2% 32. (b) W r E P n 2 4 1 2 2 2 4 2 max  =        =         = 33. (a) R H 1  (If V = constant)  1 2 2 1 R R H H = 1 2 2 1 l A l A = 2 1 2 2 2 1 l r l r =  H2 = 2H1 34. (d) R V t H 2 =  t R H 1  35. (c) H i Rt 2 = and t q i = . Hence 2 2 ; H q t q R H =   36. (b) E 4.4 kWh 1000 1100 4 =  = 37. (d) After some time, thermal equilibrium will reach. 38. (b) At constant p.d., heat produced R i e H R V 1 . . 2 =  39. (a) Power = 3.75  200 W = 750 W  1 H.P. 34. (a) P R V = 2  =   = = 484 100 220 220 2 P V R 41. (a) Since P = VI  A V P I 25 10000 250000 = = = 42. (b) Power lost in cable 10 (25) 6250 W 6.25 k W 2 =  = = 43. (d) Heat generated in both the cases will be same because the capacitor has the same energy initially CV 200 10 (200 ) 4 J 2 1 2 1 2 6 2 = =    = − 44. (d) The bulbs are connected in parallel, hence each bulb consumes 24 . 2 48 = W Therefore 24 2 = R V  =   = 1.5 24 6 6 R 45. (a) 46. (c) The bulbs are in series, hence they will have the same current through them. 27. (a) When resistance is connected in series, brightness of bulb decreases because voltage across the bulb decreases. 48. (b) P V R 2 =  =   = 400 100 200 200 R1 and 50 . 200 100 100 2 =   R = Maximum current rating V P i = So 200 100 i1 = and 100 200 i2 =  4 1 2 1 = i i . 49. (a) 2 5 40 100 1 2 2 1 = = = P P R R . Resistance of 40 W bulb is 2 5 times than 100 W. In series, P i R 2 = and in parallel, R V P 2 = . So 40 W in series and 100 W in parallel will glow brighter. 50. (a) R V P 2 =  1 2 2 1 2 1 2 1 2 1 2 ( ) /( ) ( ) R R R R R R R R R R R R P P P S S P + = + + = =  1 2 2 1 2 ( ) 25 100 R R R + R =  1 1 2 1 = R R 51. (b) Total power P = (800 + 3 100 ) Also P = Vi  1100 = 220 i  i = 5A 52. (c) Because P R 1  53. (a) An ideal cell has zero resistance. 20. (c) Power loss in transmission 2 2 V P R PL =  2 1 V PL 
55. (c) R V t H 4.2 2 = or R V t H 4.2 2 =   R  = 4.2 20 20 800  = = 0.119  0.12  42 5 R 56. (a) Heat produced R H R V t H 1 4.2 2 = =  Hence 1 2 2 1 R R H H = 57. (d) i R t H 2 = . Here total R = (21 + 4) = 25   Rate of energy consumed = 0.20.225 =1 J / s 58. (b) When the heating coil is cut into two equal parts and these parts are joined in parallel, the resistance of coil is reduced to one fourth, so power consumed will become 4 times i.e. 400 Js–1 . 59. (d) The resistance of 40 W bulb will be more and 60 W bulb will be less. 60. (a) In series Rated Consumed P P 1  Brightness  13. (d) E = P × t W sec J 4 = 1000 30 = 310 62. (a) Resistance R1 of 500 W bulb 500 (220 ) 2 = Resistance R2 of 200 W bulb 200 (220 ) 2 = When joined in parallel, the potential difference across both the bulbs will be same. Ratio of heat produced 2 5 / / 1 2 2 2 1 2 = = = R R V R V R When joined in series, the same current will flow through both the bulbs. Ratio of heat produced 5 2 2 1 2 2 1 2 = = = R R i R i R 63. (d) Charge q = it = 0.5 A3600 sec = 1800 culoumb 64. (b) 6 120 120 (10 60) 2 2    = = = R V t H i Rt joule 5 = 14 .4 10 65. (b) In parallel R Pconsumed 1  Brightness  PA  PB (given)  RA  RB 66. (d) A l R =  and R P 1   l A P   l d P 2   PA = 2PB 67. (a) t t t minutes S = 1 + 2 = 30 + 30 = 60 6. (a) For power transmission power loss in line PL i R 2 = If power of electricity is P and it is transmitted at voltage V, then P = Vi  V P i = 2 2 2 V P R R V P PL  =      = 0.1W 22000 22000 2.2 10 2.2 10 10 3 3 =      = 69. (a) P i R 2 = (i and R are same) So P will be same for given resistors. 70. (c) Since , 2 H  i so on doubling the current, the heat produced and hence the rise in temperature becomes four times. 71. (a) Watt-hour meter measures electric energy. 72. (d) Total energy consumed 0.48 kWH 1000 60 8 =  = So cost = 0.48 1.25 = 0.6 Rs. 73. (a) 10 . 4 40 W n P PS = = = 74. (b) As temperature increases resistance of filament also increases. 75. (a) Current through the combination i 8 A (6 9) 120 = + = So, power consumed by 6  resistance P (8) 6 384 W 2 =  = 76. (b) P = Vi  A V P i 0.5 200 100 = = = 77. (b) H i Rt 2 =  =   = = 2 4 10 80 2 i t H R 78. (d) Heat produced = Energy stored in capacitor 2 2 1 = CV 6 2 4 10 (400 ) 2 1 =    − = 0.32 J 79. (d) W R V P 1210 10 12100 10 (110 ) 2 2 = = = = 80. (a) 100 (200 ) (160 ) 2 2 2  =          = R R A consumed P V V P = 64 W 81. (d) For maximum power r = R
82. (d) P i R 2 =  =  R 2 22 .5 (15)  R = 0.10  83. (d) 1 1 1 A l R =  and 2 2 2 A l R =   2 1 2 2 1 1 2 2 1 2 1 .         = = r r l l A A l l R R Given 2 1 2 1 = l l and 1 2 2 1 = r r or 2 1 1 2 = r r  8 1 2 1 = R R  Ratio of heats 1 8 / / 1 2 2 2 1 2 2 1 = = = R R V R V R H H 84. (a) P = Vi = 250  2 = 500 W 85. (a) R R V P 2 2 (200 ) =  100 =  =   = 400 10 4 10 2 4 R Now, amp R V i 4 1 400 100 = = = 86. (a, d) Rsteel = 2RAl . In series H  R (i is Same) So, H will be more in steel wire . In parallel R H 1  (V is Same), so H will be more in aluminium wire. 87. (a) H i Rt 2 =  2 2 2 r i l i R t H   = = 88. (b) 89. (a) t R V H . 2 =  2 1 1 2 2 2 1 = = = R R R R H H 90. (b) 91. (a) In parallel PConsumed  PRated 92. (b) 93. (a) =  = = = 1210  40 (220 ) 2 2 2 P V R R V P 94. (b) amp V P P Vi i 11 3 220 60 =  = = = 95. (a) In series, Applied Rated Consumed V P P   1 i.e. more voltage appears on smaller wattage bulb, so 25 W bulb will fuse 96. (c) Because in series current is same. 97. (b) R V P 2 =  1 2 2 1 R R P P =  3 2 6 6 4 2 = = P  P2 = 9 W 98. (c) R P R V P t H 1 2 = =   also   A l l A l R . 2    m l R 2   2 R  l (for same mass) So P W l l P P A A B B A 20 1 4 2 2 = =  = 99. (a) =  R V P 2 2 3 40 60 1 2 2 1 = = = P P R R 100.(b) 2 P  V  0 2 0 2 0 0 P V V P V V P P          =         = 101.(c) P R R V P 1 2 =   So resistance of the 100W bulb will be minimum 102.(a) In parallel 1 2 1 1 1 t t t p = +  1 2 1 2 t t t t t p + = 3.33 min 15 50 5 10 5 10 = = +  = = 3 min. 20 sec 103.(a) For maximum joule heat produced in resistor external resistance = Internal resistance. 104. (d) 105. (c) t R V H 2 =  1 2 2 4 1 2 2 1 = = = R R H H 106.(a) If resistances of bulbs are R1 and R2 respectively then in parallel 1 2 1 1 1 RP R R = +          +         =         2 2 1 2 2 1 1 1 P V P V P V p  PP = P1 + P2 107. (b) 108.(b) When wire is cut into two equal parts then power dissipated by each part is 1 2P So their parallel combination will dissipate power P2 = 2P1 + 2P1 = 4P1