1. **Problem Statement:** Find the voltage across the capacitor after 500 ms of switching in the given RC circuit. 2. **Given Data:** - Initial voltage source: 15 V - Resistor before capacitor: $R_1 = 2\ \Omega$ - Capacitor: $C = \frac{1}{3}$ F - Two resistors after switch: $R_2 = 6\ \Omega$ and $R_3 = 6\ \Omega$ in series - Voltage source after resistors: 7.5 V - Time after switching: $t = 500$ ms = 0.5 s 3. **Understanding the circuit:** At $t=0$, the switch changes the circuit configuration. The capacitor will discharge or charge through the resistors $R_2$ and $R_3$. 4. **Calculate equivalent resistance after the switch:** Since $R_2$ and $R_3$ are in series: $$R_{eq} = R_2 + R_3 = 6 + 6 = 12\ \Omega$$ 5. **Initial voltage across capacitor:** Before switching, the capacitor is charged by the 15 V source through $R_1$. Assuming steady state before switching, the voltage across the capacitor is the voltage at its terminals, which is 15 V. 6. **Voltage across capacitor after switching:** The capacitor voltage $V_c(t)$ in an RC circuit discharging through $R_{eq}$ is given by: $$V_c(t) = V_0 e^{-\frac{t}{R_{eq} C}}$$ where $V_0 = 15$ V, $R_{eq} = 12\ \Omega$, $C = \frac{1}{3}$ F, and $t = 0.5$ s. 7. **Calculate the time constant $\tau$:** $$\tau = R_{eq} C = 12 \times \frac{1}{3} = 4\ \text{seconds}$$ 8. **Calculate the voltage at $t=0.5$ s:** $$V_c(0.5) = 15 e^{-\frac{0.5}{4}} = 15 e^{-0.125}$$ 9. **Evaluate $e^{-0.125}$:** $$e^{-0.125} \approx 0.8825$$ 10. **Final voltage:** $$V_c(0.5) = 15 \times 0.8825 = 13.24\ \text{volts}$$ **Answer:** The voltage across the capacitor after 500 ms of switching is approximately **13.24 volts**.