Q76 / 120 · NEET 2012 (Mains) · Physics

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

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

Solution

Here, $R_1=R=$ ? $$ R_2=5 \Omega, V=10 \mathrm{~V} $$ and $\quad P=30 \mathrm{~W}$ Hence $$ \begin{aligned} P & =\frac{V^2}{R_1}+\frac{V^2}{R_2} \\ \frac{10^2}{R} & =30-\frac{10^2}{5} \\ \frac{100}{R} & =30-20 \\ R & =10 \Omega \end{aligned} $$

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

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