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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