If in the circuit, power dissipation is $150 \mathrm{~W}$, then $\mathrm{R}$ is
If in the circuit, power dissipation is $150 \mathrm{~W}$, then $\mathrm{R}$ is

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$2 \Omega$
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$6 \Omega$
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$5 \Omega$
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$4 \Omega$
Solution
$W=\frac{V^2}{R_{\text {net }}} ; 150=\frac{(15)^2}{R}+\frac{(15)^2}{2} \Rightarrow R=6 \Omega$
Asked in: JEE Main 2002
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