The H amount of thermal energy is developed by a resistor in 10 s when a current of 4A is passes through it. If the current is increased to 16A, the thermal energy developed by the resistor in 10s will be:

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16H

$\frac{H}{4}$

answer is D.

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Detailed Solution

$\begin{array}{l} H = i^{2} Rt \\ \frac{H_{1}}{H_{2}} = \frac{i_{1}^{2}}{i_{2}^{2}} \frac{R_{1}}{R_{2}} \frac{t_{1}}{t_{2}} (R_{1} = R_{2}) \\ \frac{H}{H_{2}} = \frac{4^{2}}{16^{2}} \times \frac{10}{10} (t_{1} = t_{2} = 10 s) \\ \frac{H}{H_{2}} = \frac{1}{16} \Rightarrow H_{2} = 16 H \end{array}$

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