Two heaters designed for the same voltage V have different power ratings. When connected individually across a source of voltage V, they produce H amount of heat each in time $t_{1} a n d t_{2}$ respectively. When used together across the same source, they produce H amount of heat in time t

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If they are in series, $t = t_{1} + t_{2}$

If they are in parallel, $t = \frac{t_{1} t_{2}}{2 (t_{1} + t_{2})}$

If they are in series, $t = 2 (t_{1} + t_{2})$

If they are in parallel, $t = \frac{t_{1} t_{2}}{(t_{1} + t_{2})}$

answer is A, D.

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Detailed Solution

H amount of heat is produced by first heater in time $t_{1}$ so we have

$H = \frac{V^{2}}{R_{1}} t_{1} \Rightarrow R_{1} = \frac{V^{2} t_{1}}{H}$

Similarly, for second heater its resistance, $R_{2} = \frac{V^{2} t_{2}}{H}$

In series combination same heat is produced, $H = (\frac{V^{2}}{R_{1} + R_{2}}) t$

$\Rightarrow t = \frac{H (R_{1} + R_{2})}{V^{2}}$

Substituting the values of $R_{1} a n d R_{2},$ we get, $t = t_{1} + t_{2}$

In parallel combination same heat is produced in time, $H = \frac{V^{2}}{R_{n e t}} t$

$H = V^{2} t (\frac{1}{R_{1}} + \frac{1}{R_{2}}) = V^{2} t (\frac{H}{V^{2} t_{1}} + \frac{H}{V^{2} t_{2}}) \Rightarrow t = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

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