Q112 / 125 · BITSAT 2010 · Physics

The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is Figure

  • A
    $20 \Omega$
  • B
    $15 \Omega$
  • Correct answer
    C
    $10 \Omega$
  • D
    $30 \Omega$

Solution

$P=\frac{V^{2}}{R_{\mathrm{eq}}}$ $\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} R_{\mathrm{eq}}=\left(\frac{5 R}{5+R}\right) P=30 \mathrm{~W}$ Substituting the values in equation (i) $$ 30=\frac{(10)^{2}}{\left(\frac{5 R}{5+R}\right)} \Rightarrow R=10 \Omega $$

Hence, the correct answer is option: 'C'.

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