First slide

Electrical power

Question

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

Moderate

A

200 kJ
B

100 kJ
C

50 kJ
D

10 kJ

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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