A \( 4\,\Omega \) resistor and a \( 12\,\Omega \) resistor are connected in parallel across a battery of negligible internal resistance. The total power dissipated in the two resistors is 16 W. The power dissipated in the \( 12\,\Omega \) resistor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4 W
Resistors in parallel have the same potential difference \( V \) across them. With \( V \) common, the power in each resistor is \( P=V^{2}/R \), so the smaller resistance takes the larger share of the power.
The equivalent resistance of the parallel pair is:
\( \begin{aligned} \frac{1}{R_p} &= \frac{1}{4\,\Omega}+\frac{1}{12\,\Omega} \\ &= \frac{3+1}{12\,\Omega} \\ &= \frac{1}{3\,\Omega} \end{aligned} \)
\( R_p=3\,\Omega \)
From the total power of 16 W, the square of the battery voltage is:
\( \begin{aligned} V^{2} &= P\,R_p \\ &= 16\,\mathrm{W}\times3\,\Omega \\ &= 48\,\mathrm{V^{2}} \end{aligned} \)
The power in the \( 12\,\Omega \) resistor is then:
\( \begin{aligned} P_{12} &= \frac{V^{2}}{12\,\Omega} \\ &= \frac{48\,\mathrm{V^{2}}}{12\,\Omega} \\ &= 4\,\mathrm{W} \end{aligned} \)
The \( 4\,\Omega \) resistor takes \( 48/4=12\,\mathrm{W} \), and the two together give the 16 W. Hence, the \( 12\,\Omega \) resistor dissipates 4 W.