Let's break down the problem step by step. --- ## **Given:** - The switch moves from **a** to **b** at \( t = \). - Capacitor voltage is **** at \( t = \). - \( V_a = 36\,\text{V} \) - **Make-before-break** switch. - Find **\( i(t) \) for \( t > \)**. --- ## **Step 1: Initial Conditions (For \( t < \))** **Switch at \( a \):** - Inductor: Acts as short circuit at steady state. - Capacitor: Acts as open circuit at steady state. **Current through Inductor (\( i_L(-) \)):** The inductor is in series with the \( 6\,\Omega \) resistor and parallel to the \( 2\,\Omega \) resistor. - Total resistance in left loop: \( 2\,\Omega + 6\,\Omega = 8\,\Omega \) - Source: \( 4\,A \) current source But since the current source is in parallel with the \( 6\,\Omega \) resistor, **all the current goes through the current source** (the \( 6\,\Omega \) branch is shorted by the current source). So: \[ i_L(^-) = 4\,\text{A} \] --- ## **Step 2: For \( t > \) (Switch at \( b \))** **New Circuit:** - The inductor, resistor (\( 14\,\Omega \)), and capacitor (\( 20\,\text{mF} \)), and \( V_a \) are in series. The equivalent circuit for \( t > \): - Inductor (\( L = 2\,\text{H} \)) - Resistor (\( R = 14\,\Omega \)) - Capacitor (\( C = 20\,\text{mF} = .02\,\text{F} \)) - Voltage source (\( V_a = 36\,\text{V} \)) --- ## **Step 3: Writing the Differential Equation** Apply KVL around the loop for \( t > \): \[ V_a = L \frac{di}{dt} + Ri + \frac{1}{C} \int i\,dt \] Take derivative: \[ = L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} \] Plug values: - \( L = 2 \) - \( R = 14 \) - \( C = .02 \) \[ = 2\frac{d^2i}{dt^2} + 14\frac{di}{dt} + \frac{i}{.02} \] \[ = 2\frac{d^2i}{dt^2} + 14\frac{di}{dt} + 50i \] Move \( V_a \) to the right: \[ 36 = 2\frac{d^2i}{dt^2} + 14\frac{di}{dt} + 50i \] \[ \frac{d^2i}{dt^2} + 7\frac{di}{dt} + 25i = 18 \] --- ## **Step 4: Solving the Differential Equation** **Homogeneous solution**: \[ \frac{d^2i}{dt^2} + 7\frac{di}{dt} + 25i = \] Characteristic equation: \[ s^2 + 7s + 25 = \] \[ s = \frac{-7 \pm \sqrt{49 - 100}}{2} = \frac{-7 \pm j\sqrt{51}}{2} \] \[ s = -3.5 \pm j\cdot 3.57 \] So, the homogeneous solution: \[ i_h(t) = e^{-3.5t}(A\cos(3.57t) + B\sin(3.57t)) \] **Particular solution** (steady-state, for constant input): Try \( i_p = I_ \): \[ + + 25I_ = 18 \implies I_ = .72 \] **General solution:** \[ i(t) = .72 + e^{-3.5t}(A\cos(3.57t) + B\sin(3.57t)) \] --- ## **Step 5: Initial Conditions** - \( i() = i_L(^-) = 4\,\text{A} \) - Initial capacitor voltage is , so \( v_C() = \). Inductor's initial derivative (from KVL at \( t = \)): \[ V_a = L\frac{di}{dt}\Big|_{t=} + Ri() + v_C() \] \[ 36 = 2\frac{di}{dt}\Big|_{t=} + 14 \times 4 + \] \[ 36 = 2\frac{di}{dt}\Big|_{t=} + 56 \] \[ 2\frac{di}{dt}\Big|_{t=} = 36 - 56 = -20 \] \[ \frac{di}{dt}\Big|_{t=} = -10\,\text{A/s} \] --- ### **Now apply initial conditions to the solution:** \[ i(t) = .72 + e^{-3.5t}(A\cos(3.57t) + B\sin(3.57t)) \] #### At \( t = \): \[ i() = .72 + (A \cdot 1 + B \cdot ) = 4 \implies A = 4 - .72 = 3.28 \] #### Derivative at \( t = \): \[ \frac{di}{dt} = e^{-3.5t}\left[ -3.5(A\cos(3.57t) + B\sin(3.57t)) + A(-3.57)\sin(3.57t) + B(3.57)\cos(3.57t) \right] \] At \( t = \): - \( \cos() = 1 \), \( \sin() = \), \( e^{} = 1 \) \[ \frac{di}{dt}\Big|_{t=} = -3.5A + 3.57B \] Set to initial value: \[ -3.5A + 3.57B = -10 \] Plugging \( A = 3.28 \): \[ -3.5 \times 3.28 + 3.57B = -10 \] \[ -11.48 + 3.57B = -10 \] \[ 3.57B = 1.48 \] \[ B = .415 \] --- ## **Final Solution** \[ \boxed{ i(t) = .72 + e^{-3.5t}\left[3.28\cos(3.57t) + .415\sin(3.57t)\right] \qquad (t > ) } \] --- ### **Summary Table** | Parameter | Value | |----------------|-----------| | \( L \) | 2 H | | \( R \) | 14 Ω | | \( C \) | 20 mF | | \( V_a \) | 36 V | | \( i() \) | 4 A | | \( \frac{di}{dt}() \) | -10 A/s | --- Let me know if you need a step-by-step plot or further explanation!