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The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is
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2938 Upvotes
Verified Answer
The correct answer is:
$10 \Omega$
$P=\frac{V^{2}}{R_{\mathrm{eq}}}$
$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} R_{\mathrm{eq}}=\left(\frac{5 R}{5+R}\right) P=30 \mathrm{~W}$
Substituting the values in equation (i)
$$
30=\frac{(10)^{2}}{\left(\frac{5 R}{5+R}\right)} \Rightarrow R=10 \Omega
$$
$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} R_{\mathrm{eq}}=\left(\frac{5 R}{5+R}\right) P=30 \mathrm{~W}$
Substituting the values in equation (i)
$$
30=\frac{(10)^{2}}{\left(\frac{5 R}{5+R}\right)} \Rightarrow R=10 \Omega
$$
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