Let's solve the problem step by step. --- ### **Step 1: Understanding the Circuit** - The circuit consists of a current source \( I_s \), a resistor \( R \), and a capacitor \( C \). - The switch is closed for a long time (steady state), then **opens at \( t = 0 \)**. - **Initial condition:** \( v_C(0^+) = 0 \) V (capacitor voltage at \( t = 0 \)). - For \( t \geq 0 \), the current source is disconnected and the circuit is just an **RC series circuit** (R and C in parallel, with nothing driving them). --- ## **a) Find the First Order ODE** ### **Step 2: Write the KCL Equation for \( t \geq 0 \)** After the switch opens, the circuit is: - Only \( R \) and \( C \) in parallel, with **no external source**. Let \( v_C(t) \) be the voltage across the capacitor (and resistor, since they're in parallel). - The current through the resistor: \( i_R = \frac{v_C}{R} \) - The current through the capacitor: \( i_C = C \frac{dv_C}{dt} \) By KCL (no current source, all charge must stay on the capacitor): \[ i_R + i_C = 0 \] \[ \frac{v_C}{R} + C \frac{dv_C}{dt} = 0 \] ### **Step 3: Write the ODE** \[ C \frac{dv_C}{dt} + \frac{v_C}{R} = 0 \] Or, dividing both sides by \( C \): \[ \frac{dv_C}{dt} + \frac{1}{RC} v_C = 0 \] --- ## **b) Find the General Expression for \( v_C(t) \) Using Separable ODE Method** ### **Step 4: Rewrite as a Separable Equation** \[ \frac{dv_C}{dt} = -\frac{1}{RC} v_C \] Separate variables: \[ \frac{1}{v_C} dv_C = -\frac{1}{RC} dt \] ### **Step 5: Integrate Both Sides** Integrate both sides: \[ \int \frac{1}{v_C} dv_C = -\frac{1}{RC} \int dt \] \[ \ln |v_C| = -\frac{t}{RC} + K \] where \( K \) is the constant of integration. Exponentiate both sides: \[ v_C = e^{-\frac{t}{RC} + K} = A e^{-\frac{t}{RC}} \] where \( A = e^K \). ### **Step 6: Apply Initial Condition** Given: \( v_C(0) = 0 \) \[ v_C(0) = A e^{0} = A = 0 \] So, **the solution is**: \[ \boxed{v_C(t) = 0 \qquad \text{for all } t \geq 0} \] ### **Physical Meaning** - Because the capacitor started with 0V and there is no source after the switch opens, it stays at 0V. --- ## **Final Answers** ### **a) First Order ODE:** \[ \boxed{ \frac{dv_C}{dt} + \frac{1}{RC} v_C = 0 } \] ### **b) General Expression:** \[ \boxed{ v_C(t) = 0 \qquad \text{for all } t \geq 0 } \] --- **If you want the general solution for arbitrary initial voltage \( v_C(0) = V_0 \):** \[ v_C(t) = V_0 e^{-\frac{t}{RC}} \] But for this problem, \( V_0 = 0 \).

Thank you for the clarification. Let's revisit the problem carefully, considering that after the switch opens, the circuit consists of the resistor \( R \), capacitor \( C \), and the source \( I_s \) all in parallel. --- ### **Understanding the Circuit** - **Long time before \( t=0 \):** The switch is closed, the circuit reaches steady state. - **At \( t=0 \):** The switch opens suddenly. - **Initial capacitor voltage:** \( v_C(0^+) = 0\,V \) (as given). --- ### **Step 1: Determine the initial conditions** Since the switch has been closed for a long time, the capacitor would have been charged to a voltage determined by the steady-state of the circuit. **For \( t < 0 \):** - The capacitor acts as a short circuit (since it's steady state), and the current source \( I_s \) supplies current through \( R \) and the capacitor. **Steady state:** - The capacitor voltage \( v_C(0^-) \) (just before opening) is: \[ v_C(0^-) = V_{ss} \] **Finding \( V_{ss} \):** - The current source \( I_s \) supplies current \( I_s \). - The resistor \( R \) and capacitor \( C \) are in parallel with the source. - Since the capacitor acts as a short at steady state, the voltage across it is: \[ V_{ss} = I_s R \] **Given:** \( v_C(0^+) = 0 \) V, which indicates the capacitor voltage drops to zero immediately after opening the switch. --- ### **Step 2: Write the KCL at \( t > 0 \)** After \( t=0 \), the circuit contains: - The source \( I_s \), - The resistor \( R \), - The capacitor \( C \), all in parallel. **Note:** At \( t=0^+ \), the capacitor voltage is \( v_C(0^+) = 0\,V \). --- ### **Step 3: Derive the differential equation** Applying KCL at the node (the voltage across all elements): \[ I_s = i_R + i_C \] where: \[ i_R = \frac{v_C}{R} \] \[ i_C = C \frac{dv_C}{dt} \] So: \[ I_s = \frac{v_C}{R} + C \frac{dv_C}{dt} \] Rearranged: \[ C \frac{dv_C}{dt} + \frac{v_C}{R} = I_s \] --- ### **### **Answer to part (a):** **The first order ODE:** \[ \boxed{ C \frac{dv_C}{dt} + \frac{v_C}{R} = I_s } \] --- ### **Part (b): Solve the ODE** The differential equation: \[ C \frac{dv_C}{dt} + \frac{v_C}{R} = I_s \] Divide through by \( C \): \[ \frac{dv_C}{dt} + \frac{1}{RC} v_C = \frac{I_s}{C} \] This is a linear first-order ODE. --- ### **Step 4: Find the homogeneous solution** Homogeneous equation: \[ \frac{dv_C}{dt} + \frac{1}{RC} v_C = 0 \] Solution: \[ v_{C,h}(t) = A e^{-\frac{t}{RC}} \] --- ### **Step 5: Find a particular solution** Since RHS is constant \( \frac{I_s}{C} \), assume: \[ v_{C,p} = V_p = \text{constant} \] Plug into the ODE: \[ 0 + \frac{1}{RC} V_p = \frac{I_s}{C} \] \[ V_p = I_s R \] --- ### **Step 6: Write the general solution** \[ v_C(t) = v_{C,h}(t) + v_{C,p} = A e^{-\frac{t}{RC}} + I_s R \] --- ### **Step 7: Apply initial condition** Given: \[ v_C(0^+) = 0 \] At \( t=0 \): \[ 0 = A e^{0} + I_s R \] \[ A = -I_s R \] --- ### **### **Final solution:** \[ \boxed{ v_C(t) = -I_s R \, e^{-\frac{t}{RC}} + I_s R } \] Or equivalently: \[ v_C(t) = I_s R \left( 1 - e^{-\frac{t}{RC}} \right) \] --- ## **Summary:** ### **a) First order ODE:** \[ \boxed{ C \frac{dv_C}{dt} + \frac{v_C}{R} = I_s } \] ### **b) Response \( v(t) \) for \( t > 0 \):** \[ \boxed{ v(t) = I_s R \left( 1 - e^{-\frac{t}{RC}} \right) } \] This describes the capacitor voltage rising from 0 V at \( t=0 \) to its steady-state value \( I_s R \). --- **Let me know if you'd like further clarification or details!**