Let's analyze the circuit step-by-step: ### **Step 1: Initial Conditions** At \( t = 0 \), the switch moves, and the source \( v_s = 28V \) is disconnected. The capacitor is initially charged to \( v(0) = v_s = 28V \). ### **Step 2: Write the Differential Equation** For \( t > 0 \), the circuit is a simple RC **discharge** circuit: \[ \begin{align*} 100\,\text{mF} \cdot \frac{dv(t)}{dt} + \frac{v(t)}{10\,\Omega} = 0 \end{align*} \] ### **Step 3: Simplify the Equation** \[ \begin{align*} 0.1\,\frac{dv(t)}{dt} + \frac{v(t)}{10} = 0 \\ 0.1\,\frac{dv(t)}{dt} = -\frac{v(t)}{10} \\ \frac{dv(t)}{dt} = -\frac{1}{1}\,v(t) \\ \frac{dv(t)}{dt} = -v(t) \end{align*} \] ### **Step 4: Solve the Differential Equation** The solution is: \[ v(t) = v(0) e^{-t/\tau} \] where \(\tau = RC = 10 \,\Omega \cdot 0.1\,F = 1\,\text{s}\) So, \[ v(t) = 28 e^{-t} \text{ V} \] ### **Step 5: Compare to the Given Form** Given: \[ v(t) = \left[ A e^{B t} u(t) \right]\, V \] We have \(A = 28\), \(B = -1\). --- ## **Final Answers** **A = 28** **B = -1**