1. **Problem statement:** Three identical resistors $R_1$, $R_2$, and $R_3$, each with resistance $2\ \Omega$, are connected to a cell with negligible internal resistance. When switch $S$ is open, the power dissipated by $R_1$ is $18\ \text{W}$. 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, $P_{1,\text{open}} = 18\ \text{W}$ 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 circuit when switch $S$ is open.** - With $S$ open, $R_3$ is disconnected. - The circuit has $R_1$ and $R_2$ in series with the battery. - Current through $R_1$ is the same as through $R_2$. 5. **Calculate current through $R_1$ when $S$ is open:** $$P_{1,\text{open}} = I_{\text{open}}^2 R_1 \implies I_{\text{open}} = \sqrt{\frac{P_{1,\text{open}}}{R_1}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3\ \text{A}$$ 6. **Calculate total voltage of the battery:** - Total resistance when $S$ is open is $R_1 + R_2 = 2 + 2 = 4\ \Omega$ - Voltage $V = I_{\text{open}} \times (R_1 + R_2) = 3 \times 4 = 12\ \text{V}$ 7. **Step 2: Analyze circuit when switch $S$ is closed.** - Closing $S$ connects $R_3$ in parallel with $R_1$. - The parallel combination of $R_1$ and $R_3$ is: $$R_{13} = \frac{R_1 R_3}{R_1 + R_3} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1\ \Omega$$ - Total resistance in circuit is now $R_{13} + R_2 = 1 + 2 = 3\ \Omega$ 8. **Calculate total current when $S$ is closed:** $$I_{\text{closed}} = \frac{V}{R_{\text{total}}} = \frac{12}{3} = 4\ \text{A}$$ 9. **Calculate current through $R_1$ when $S$ is closed:** - Current divides between $R_1$ and $R_3$ equally because they are identical resistors. - Current through $R_1$ is half of $I_{13}$, where $I_{13} = I_{\text{closed}}$ since $R_{13}$ is in series with $R_2$. - So, $$I_{R_1} = \frac{I_{\text{closed}}}{2} = \frac{4}{2} = 2\ \text{A}$$ 10. **Calculate power dissipated by $R_1$ when $S$ is closed:** $$P_{1,\text{closed}} = I_{R_1}^2 R_1 = 2^2 \times 2 = 4 \times 2 = 8\ \text{W}$$ **Final answer:** The power dissipated by $R_1$ when switch $S$ is closed is **8W**. **Answer choice:** A. 8W