Let's break down the problem step-by-step: ### **Given:** - \( \mathcal{E} = 3.10 \text{ kV} = 310 \text{ V} \) - \( C = 9.60 \mu\text{F} = 9.60 \times 10^{-6} \text{ F} \) - \( R_1 = R_2 = R_3 = .910 \text{ M}\Omega = 9.10 \times 10^5 \Omega \) - Switch S is closed at \( t = \), and the capacitor is initially uncharged. ### **Circuit Analysis:** When the switch is closed: - The capacitor begins to charge through \( R_2 \) and \( R_3 \). - The initial current (just after closing the switch) is calculated as if the capacitor is a short circuit (\( V_C = \)). - After a long time (\( t \rightarrow \infty \)), the capacitor is fully charged (\( I_C = \)), and acts like an open circuit. --- ## **Step 1: Initial Currents at \( t = \)** ### **(a) Current \( i_1 \) in Resistor 1 (at \( t = \)):** At \( t = \), the capacitor is uncharged and acts like a wire (short circuit), so \( R_2 \) and \( R_3 \) are in parallel. - **Equivalent resistance of \( R_2 \) and \( R_3 \) in parallel:** \[ R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{(9.10 \times 10^5) \times (9.10 \times 10^5)}{(9.10 \times 10^5) + (9.10 \times 10^5)} = \frac{(9.10 \times 10^5)^2}{2 \times 9.10 \times 10^5} = \frac{9.10 \times 10^5}{2} = 4.55 \times 10^5 \Omega \] - **Total resistance:** \[ R_{\text{total}} = R_1 + R_{23} = 9.10 \times 10^5 + 4.55 \times 10^5 = 1.365 \times 10^6 \Omega \] - **Current from the battery (\( i_1 \)) initially:** \[ i_1 = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{310}{1.365 \times 10^6} = 2.27 \times 10^{-3} \text{ A} = 2.27 \text{ mA} \] --- ### **(b) Current \( i_2 \) in Resistor 2 (at \( t = \)):** - **Current splits equally between \( R_2 \) and \( R_3 \) since they're equal and in parallel:** \[ i_2 = \frac{i_1}{2} = \frac{2.27 \text{ mA}}{2} = 1.135 \text{ mA} \] --- ### **(c) Current \( i_3 \) in Resistor 3 (at \( t = \)):** By symmetry: \[ i_3 = i_2 = 1.135 \text{ mA} \] --- ## **Step 2: Final Currents at \( t \rightarrow \infty \)** After a long time, the capacitor is fully charged and acts as an open circuit (\( i_3 = \)), so no current flows through \( R_3 \): ### **(d) Current \( i_1 \) in Resistor 1 (at \( t \rightarrow \infty \)):** Now, current only flows through \( R_1 \) and \( R_2 \): - **Total resistance:** \[ R_{\text{total}} = R_1 + R_2 = 9.10 \times 10^5 + 9.10 \times 10^5 = 1.82 \times 10^6 \Omega \] - **Current:** \[ i_1 = \frac{310}{1.82 \times 10^6} = 1.703 \times 10^{-3} \text{ A} = 1.70 \text{ mA} \] --- ### **(e) Current \( i_2 \) in Resistor 2 (at \( t \rightarrow \infty \)):** All current through \( R_2 \) (since \( R_3 \) is open): \[ i_2 = i_1 = 1.70 \text{ mA} \] --- ### **(f) Current \( i_3 \) in Resistor 3 (at \( t \rightarrow \infty \)):** \[ i_3 = \text{ mA} \] --- ## **Step 3: Potential Difference Across Resistor 2** ### **(g) At \( t = \):** \[ V_{R_2} = i_2 \times R_2 = (1.135 \times 10^{-3}) \times (9.10 \times 10^5) = 1033 \text{ V} \] --- ### **(h) At \( t \rightarrow \infty \):** \[ V_{R_2} = i_2 \times R_2 = (1.70 \times 10^{-3}) \times (9.10 \times 10^5) = 1547 \text{ V} \] --- ## **Summary Table** | Part | Value | Units | |------|---------------|------------| | (a) | 2.27 | mA | | (b) | 1.135 | mA | | (c) | 1.135 | mA | | (d) | 1.70 | mA | | (e) | 1.70 | mA | | (f) | | mA | | (g) | 1033 | V | | (h) | 1547 | V | --- ### **Boxed Answers** **(a)** \( i_1() = 2.27 \) mA **(b)** \( i_2() = 1.135 \) mA **(c)** \( i_3() = 1.135 \) mA **(d)** \( i_1(\infty) = 1.70 \) mA **(e)** \( i_2(\infty) = 1.70 \) mA **(f)** \( i_3(\infty) = \) mA **(g)** \( V_{R_2}() = 1033 \) V **(h)** \( V_{R_2}(\infty) = 1547 \) V If you need more details or calculations, let me know!