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 minutes
will be

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100 kJ

50 kJ

200 kJ

10 kJ

answer is B.

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Detailed Solution

For heater A: $500 \times 10^{3} = \frac{V^{2}}{R_{1}} (20 \times 60)$ …(i)

$Where R_{1} = 100 Ω$

For heater B:

$1000 \times 10^{3} = \frac{V^{2}}{R_{2}} (10 \times 60)$ ….(ii)

From (i) and (ii), $R_{2} = 25 Ω$

When heaters are connected in series:

$R_{eq} = R_{1} + R_{2}$

Heat produced:

$H = \frac{V^{2}}{R_{1} + R_{2}} (5 \times 60)$ …..(iii)

From (i) and (iii), H = 100 kJ

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