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A heating element is designed to dissipate $2400 \mathrm{~W}$ when connected to $240 \mathrm{~V}$. The power it dissipates when it is connected to $120 \mathrm{~V}$ is (Assume that resistance of the filament is constant)
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The correct answer is:
$600 \mathrm{~W}$
Given, Rated power, $P_R=2400 \mathrm{~W}$
Rated voltage, $V_R=240 \mathrm{~V}$
So, resistance of element, $R=\frac{V_R^2}{P_R} \quad\left(\because P=\frac{V^2}{R}\right)$
$=\frac{240 \times 240}{2400}=24 \Omega$
when element is connected across $120 \mathrm{~V}$ supply dissipated power is,
$\begin{aligned} P & =\frac{V^2}{R}=\frac{120 \times 120}{24} \\ & =600 \mathrm{~W}\end{aligned}$
Rated voltage, $V_R=240 \mathrm{~V}$
So, resistance of element, $R=\frac{V_R^2}{P_R} \quad\left(\because P=\frac{V^2}{R}\right)$
$=\frac{240 \times 240}{2400}=24 \Omega$
when element is connected across $120 \mathrm{~V}$ supply dissipated power is,
$\begin{aligned} P & =\frac{V^2}{R}=\frac{120 \times 120}{24} \\ & =600 \mathrm{~W}\end{aligned}$
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