Three equal resistors connected in series across a source of emf together dissipate 9 W of power. What would be the power dissipated if the same resistors are connected in parallel across the same source.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 81 W
For a source of fixed voltage \( V \), the power dissipated in a resistance \( R_{\mathrm{eq}} \) is \( P = V^{2}/R_{\mathrm{eq}} \), so a smaller equivalent resistance takes more power. Let each resistor be \( R \).
\( \text{Series: } R_{\mathrm{eq}} = 3R \)
\( \text{Parallel: } R_{\mathrm{eq}} = R/3 \)
\( \displaystyle P_{s} = \frac{V^{2}}{3R} = 9\,\mathrm{W} \;\Rightarrow\; \frac{V^{2}}{R} = 27\,\mathrm{W} \)
\( \displaystyle P_{p} = \frac{V^{2}}{R/3} = 3\,\frac{V^{2}}{R} = 3 \times 27 = 81\,\mathrm{W} \)
The source is treated as ideal (no internal resistance), so the same voltage acts in both cases.
Shortcut: The equivalent resistance falls from \( 3R \) to \( R/3 \), a factor of 9, so at the same voltage the power rises 9 times: \( 9 \times 9 = 81\,\mathrm{W} \).