Q.

Two heaters A and B are in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is 100  Ω. If the same heaters are connected in series across the same voltage, then total heat produced in 5 minuteswill be

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a

200 kJ

b

100 kJ

c

50 kJ

d

10 kJ

answer is B.

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Detailed Solution

For heater A: 500 × 103=V2R1 20  × 60    …(i)Where  R1 = 100  ΩFor heater B:1000 × 103=V2R2 10  × 60        ….(ii)From (i) and (ii), R2 = 25  ΩWhen heaters are connected in series: Req = R1  + R2Heat produced:H= V2R1 + R2  5  ×  60           …..(iii)From (i) and (iii), H = 100 kJ
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Two heaters A and B are in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is 100  Ω. If the same heaters are connected in series across the same voltage, then total heat produced in 5 minuteswill be