JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 7)
The $\mathrm{H}$ amount of thermal energy is developed by a resistor in $10 \mathrm{~s}$ when a current of $4 \mathrm{~A}$ is passed through it. If the current is increased to $16 \mathrm{~A}$, the thermal energy developed by the resistor in $10 \mathrm{~s}$ will be :
$\frac{\mathrm{H}}{4}$
$16 \mathrm{H}$
H
$4 \mathrm{H}$
Explanation
$H \propto i^{2}$ for $t=$ constant
$$ \Rightarrow $$ $$ \frac{H}{H^{\prime}}=\left(\frac{4}{16}\right)^{2} $$
$$ \Rightarrow $$ $H^{\prime}=16 H$
$$ \Rightarrow $$ $$ \frac{H}{H^{\prime}}=\left(\frac{4}{16}\right)^{2} $$
$$ \Rightarrow $$ $H^{\prime}=16 H$
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