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Ratio of thermal energy released in two resistor $R$ and $3 R$ connected in parallel in an electric circuit is :
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Verified Answer
The correct answer is:
3 : 1
3 : 1
Thermal energy is given by
$\mathrm{H}=\mathrm{P} \times \mathrm{t}=\frac{\mathrm{V}^2}{\mathrm{R}} \times \mathrm{t}$
Here, voltage $\mathrm{V}$ is same.
$\therefore \quad \frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}}}{\frac{\mathrm{V}^2 \mathrm{t}}{3 \mathrm{R}}}=3: 1$
$\mathrm{H}=\mathrm{P} \times \mathrm{t}=\frac{\mathrm{V}^2}{\mathrm{R}} \times \mathrm{t}$
Here, voltage $\mathrm{V}$ is same.
$\therefore \quad \frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}}}{\frac{\mathrm{V}^2 \mathrm{t}}{3 \mathrm{R}}}=3: 1$
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