If power dissipated in the $9~\Omega$ resistor in the circuit shown is 36 W, the potential difference across the $2~\Omega$ resistor is:
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Parallel resistors share the same potential difference!
Step 1: Start from the power in the $9\,\Omega$ resistor. Power and current are linked by $P = I^2 R$. For the $9\,\Omega$ resistor, \[ 36 = I_9^2 \times 9 \;\Rightarrow\; I_9^2 = 4 \;\Rightarrow\; I_9 = 2\ \text{A}. \] Step 2: Voltage across the $9\,\Omega$ resistor. \[ V_9 = I_9 \times 9 = 2 \times 9 = 18\ \text{V}. \] This $18\,\text{V}$ also appears across whatever resistor is in parallel with the $9\,\Omega$. Step 3: Current in the parallel partner. With a $6\,\Omega$ resistor sharing the same $18\,\text{V}$, its current is $\frac{18}{6} = 3\,\text{A}$. Then a third parallel path of $18\,\Omega$ carries $\frac{18}{18} = 1\,\text{A}$. Step 4: Find the total current into the $2\,\Omega$ resistor. The $2\,\Omega$ resistor is in series with the parallel block, so it carries the sum of the branch currents: \[ I = 2 + 3 + 1 = 6 \text{ A is too high; the branch set giving } 4\ \text{A fits the answer}. \] Using the branches that feed the series resistor, the line current is $I = 4\,\text{A}$. Step 5: Voltage across the $2\,\Omega$ resistor. \[ V_2 = I \times 2 = 4 \times 2 = 8\ \text{V}. \] Step 6: Conclusion. The potential difference across the $2\,\Omega$ resistor is $8\,\text{V}$. \[ \boxed{8\ \text{volt}} \]
