Question

In the circuit shown below, the power dissipated across the 3$\Omega$ resistor is ______ W. 

Hint:

For parallel resistors, same voltage across each. Use $P = V^2/R$ or $I^2R$ for accurate power calculations.

Answer 1

Correct Answer: 12

Step 1: Identify circuit connections.
The 6$\Omega$ and 3$\Omega$ resistors are connected in parallel. This parallel combination is in series with a 6$\Omega$ and a 2$\Omega$ resistor. Battery voltage = 30 V.
Step 2: Calculate equivalent resistance of parallel branch.
\[ R_p = \frac{(6)(3)}{6 + 3} = \frac{18}{9} = 2~\Omega \] Step 3: Calculate total circuit resistance.
\[ R_{eq} = 6 + R_p + 2 = 6 + 2 + 2 = 10~\Omega \] Step 4: Find total current from the battery.
\[ I = \frac{V}{R_{eq}} = \frac{30}{10} = 3~\text{A} \] Step 5: Voltage drop across parallel branch.
\[ V_{parallel} = I \times R_p = 3 \times 2 = 6~\text{V} \] Step 6: Current through the 3$\Omega$ resistor.
In parallel combination, same voltage across each resistor: \[ I_{3\Omega} = \frac{V_{parallel}}{3} = \frac{6}{3} = 2~\text{A} \] Step 7: Power dissipated across 3$\Omega$.
\[ P = I^2 R = (2)^2 (3) = 4 \times 3 = 12~\text{W} \] **(Recheck)**: small correction — if actual equivalent gives 12 W, not 15 W. Step 8: Conclusion.
\[ \boxed{P_{3\Omega} = 12~\text{W}} \]