The power dissipated in the circuit shown in figure is 36 W. The value of R is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 12 Ω
In the figure R and the 6 Ω resistor are in parallel across the 12 V cell. The power dissipated in the whole circuit is \( P = V^{2}/R_{\mathrm{eq}} \).
\( \displaystyle R_{\mathrm{eq}} = \frac{V^{2}}{P} = \frac{12^{2}}{36}\,\Omega = 4\,\Omega \)
\( \displaystyle \frac{1}{R_{\mathrm{eq}}} = \frac{1}{R}+\frac{1}{6} \Rightarrow \frac{1}{R} = \frac{1}{4}-\frac{1}{6} = \frac{1}{12} \)
\( R = 12\,\Omega \)
Shortcut: Each branch has the full 12 V across it. The 6 Ω resistor dissipates \( 12^{2}/6 = 24\,\mathrm{W} \), so R dissipates \( 36-24 = 12\,\mathrm{W} \) and \( R = 12^{2}/12 = 12\,\Omega \).