Let's solve the given circuit step by step for **5a** and **5b**. --- ## **Step 5a: Initial Condition (\( t = 0^- \))** ### **Find \( v_c(0^-) \), the voltage across the capacitor just before the switch opens** **Given:** - The switch has been closed for a long time (steady-state DC). **At steady state:** - The inductor acts as a short circuit (\( L \) is a wire). - The capacitor acts as an open circuit (\( C \) is a break). **Redraw the circuit:** - Ignore the capacitor branch (open circuit). - The inductor is a wire (short). So, the branch with R2 and C1 is open, and no current flows through R2 or C1. The voltage across the capacitor is the voltage at the node between R2 and C1 (which is also the voltage across R2, since no current flows). **Find the voltage at the top node of R2 (relative to ground):** The top node of R2 is connected to V2 (12V). Since no current flows through R2 (open circuit at C1), the voltage at the top of R2 is simply 12V. So, \[ \boxed{v_c(0^-) = 12\,V} \] **This is also \( v_c(0^+) \) because capacitor voltage cannot change instantaneously:** \[ \boxed{v_c(0^+) = v_c(0^-) = 12\,V} \] --- ## **Step 5b: For \( t > 0 \), Find \( v_c(t) \) and associated parameters** ### **Step 1: Draw the circuit for \( t > 0 \) (switch open)** With the switch open, the left side (V1, L1, R1) is disconnected. The circuit is now: - V2 (12V) → R2 (2Ω) → C1 (1F) → R3 (4Ω) → Ground This is a simple RC circuit (no inductor in this loop for \( t > 0 \)). ### **Step 2: Write the differential equation for \( v_c(t) \)** The loop is: V2, R2, C1, R3 Let \( v_c(t) \) be the voltage across C1. Total resistance in series: \( R_{eq} = R2 + R3 = 2 + 4 = 6\ \Omega \) The loop equation: \[ V2 - R2 i - v_c - R3 i = 0 \] \[ V2 - (R2 + R3) i - v_c = 0 \] But \( i = C \frac{dv_c}{dt} \): \[ V2 - (R2 + R3) C \frac{dv_c}{dt} - v_c = 0 \] \[ V2 - v_c = (R2 + R3) C \frac{dv_c}{dt} \] \[ \frac{dv_c}{dt} + \frac{1}{(R2 + R3)C} v_c = \frac{V2}{(R2 + R3)C} \] Plug in values: - \( R2 + R3 = 6 \) - \( C = 1 \) - \( V2 = 12 \) \[ \frac{dv_c}{dt} + \frac{1}{6} v_c = 2 \] ### **Step 3: Solve the differential equation** It's a first-order linear ODE: - Homogeneous solution: \( v_{c,h}(t) = A e^{-t/6} \) - Particular solution: Steady state (\( \frac{dv_c}{dt} = 0 \)): \( v_{c,ss} = 12 \) V General solution: \[ v_c(t) = v_{c,ss} + A e^{-t/6} \] But the particular solution is \( V_2 \) (since the circuit will reach steady state at 12V). ### **Step 4: Apply initial condition** At \( t = 0 \): \( v_c(0) = 12\,V \) So, \[ v_c(0) = 12 + A e^{0} = 12 + A \] But \( v_c(0) = 12 \rightarrow A = 0 \) Therefore, \[ \boxed{v_c(t) = 12} \] --- ## **Final Answers** **5a. Initial voltage across the capacitor:** \[ \boxed{v_c(0^-) = 12\,V} \] **5b. Voltage across the capacitor for \( t > 0 \):** \[ \boxed{v_c(t) = 12\,V \quad \text{for } t > 0} \] ### **Associated Parameters** - \( a_1 \) and \( a_2 \): Not applicable; there is only one exponential term with a zero coefficient. - \( A \): \( 0 \) (since the exponential term vanishes) - \( \omega_0 \) and \( \zeta \): Not applicable to first-order RC circuit. --- **Summary:** - **The voltage across the capacitor never changes; it stays at 12V.** - **Plots will be a flat line at 12V.** --- If you want to see the plot or simulation, let me know!