1. **Problem statement:** Three identical resistors $R_1$, $R_2$, and $R_3$ each have resistance $2\Omega$. They are connected to a cell with negligible internal resistance. When switch $S$ is open, the power dissipated by $R_1$ is 18W. We need to find the power dissipated by $R_1$ when switch $S$ is closed. 2. **Given:** - $R_1 = R_2 = R_3 = 2\Omega$ - Power dissipated by $R_1$ when $S$ is open, $P_{1,open} = 18W$ 3. **Formula for power dissipated in a resistor:** $$P = \frac{V^2}{R} = I^2 R$$ where $V$ is voltage across the resistor and $I$ is current through it. 4. **Step 1: Analyze the circuit when $S$ is open.** - With $S$ open, $R_3$ is disconnected. - The circuit has $R_1$ and $R_2$ in series or parallel depending on the diagram. From the description, $R_1$ and $R_2$ are in parallel branches connected to the battery. 5. **Calculate voltage across $R_1$ when $S$ is open:** Since $R_1$ dissipates 18W, $$P_{1,open} = \frac{V_1^2}{R_1} \implies V_1 = \sqrt{P_{1,open} \times R_1} = \sqrt{18 \times 2} = \sqrt{36} = 6V$$ 6. **Step 2: Analyze the circuit when $S$ is closed.** - Closing $S$ connects $R_3$ in parallel with $R_1$. - So $R_1$ and $R_3$ are in parallel, and this combination is in parallel with $R_2$. 7. **Calculate equivalent resistance of $R_1$ and $R_3$ in parallel:** $$R_{13} = \frac{R_1 \times R_3}{R_1 + R_3} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1\Omega$$ 8. **Total resistance with $S$ closed:** $R_{total} = R_{13} \parallel R_2 = \frac{R_{13} \times R_2}{R_{13} + R_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \Omega$ 9. **Calculate total current from the battery when $S$ is open and closed:** - Voltage of battery $V = 6V$ (from step 5, voltage across $R_1$ when $S$ open equals battery voltage because $R_1$ and $R_2$ are in parallel). - Current when $S$ open: $$I_{open} = \frac{V}{R_{open}}$$ where $R_{open} = R_1 \parallel R_2 = \frac{2 \times 2}{2 + 2} = 1\Omega$ So, $$I_{open} = \frac{6}{1} = 6A$$ - Current when $S$ closed: $$I_{closed} = \frac{V}{R_{total}} = \frac{6}{\frac{2}{3}} = 6 \times \frac{3}{2} = 9A$$ 10. **Calculate voltage across $R_1$ when $S$ is closed:** - Current divides between $R_1$ and $R_3$ equally because they are identical and in parallel. - Total current through $R_{13}$ is $I_{closed}$ times the fraction of current through $R_{13}$ branch. Current division between $R_{13}$ and $R_2$: $$I_{R_{13}} = I_{closed} \times \frac{R_2}{R_{13} + R_2} = 9 \times \frac{2}{1 + 2} = 9 \times \frac{2}{3} = 6A$$ - Current through $R_1$ (half of $I_{R_{13}}$): $$I_{1,closed} = \frac{6}{2} = 3A$$ - Voltage across $R_1$: $$V_{1,closed} = I_{1,closed} \times R_1 = 3 \times 2 = 6V$$ 11. **Calculate power dissipated by $R_1$ when $S$ is closed:** $$P_{1,closed} = I_{1,closed}^2 \times R_1 = 3^2 \times 2 = 9 \times 2 = 18W$$ 12. **Answer:** The power dissipated by $R_1$ when $S$ is closed is **18W**. **Final answer: C. 18W**