First slide

Electrical power

Question

A heater boils 1 kg of water in time t₁, and another heater boils the same water in time t₂. If both are connected in series, the combination will boil the same water in time

Moderate

A

$\frac{t_{1} t_{2}}{t_{1} + t_{2}}$
B

$\frac{t_{1} t_{2}}{t_{1} - t_{2}}$
C

$(t_{1} + t_{2})$
D

$2 (t_{1} + t_{2})$

Solution

$H = \frac{V^{2}}{R_{1}} \times t_{1} or t_{1} = \frac{H^{\cdot} R_{1}}{V^{2}}$ …..(1)
$H = \frac{V^{2}}{R_{2}} \times t_{2} or t_{2} = \frac{{HR}_{2}}{V^{2}}$ …….(2)
In series, $H = \frac{V^{2}}{(R_{1} + R_{2})} \times t$
or $t = \frac{H (R_{1} + R_{2})}{V^{2}}$ ………….(3)
From eq.(3) $t = \frac{{HR}_{1}}{V^{2}} + \frac{{HR}_{2}}{V^{2}} = t_{1} + t_{2}$
Note : when the heaters ore connected in parallel.
In this case, $\frac{1}{t_{1}} = \frac{V^{2}}{{HR}_{1}} and \frac{1}{t_{2}} = \frac{V^{2}}{{HR}_{2}}$
$∴ \frac{1}{t} = \frac{V^{2}}{H} (\frac{1}{R_{1}} + \frac{1}{R_{2}}) = \frac{1}{t_{1}} + \frac{1}{t_{2}}$
or $t = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

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