The power dissipated in the circuit shown in the figure is 30 Watt. The value of $R$ is
The power dissipated in the circuit shown in the figure is 30 Watt. The value of $R$ is
- $20 \Omega$
- $15 \Omega$
- $10 \Omega$
- $30 \Omega$
Solution
Here, $R_1=R=$ ?
$R_2=5 \Omega, V=10 \mathrm{~V}$
and $\quad P=30 \mathrm{~W}$
Hence
$\begin{aligned}
P & =\frac{V^2}{R_1}+\frac{V^2}{R_2} \\
\frac{10^2}{R} & =30-\frac{10^2}{5} \\
\frac{100}{R} & =30-20 \\
R & =10 \Omega
\end{aligned}$
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Asked in: NEET 2012 (Mains)
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