Let's solve the problem step-by-step: --- # **Step 1: Understanding the Circuit** - At \( t = 0 \), the switch closes. - The initial conditions are: - \( i_L(0) = I_g \) - \( v_C(0) = V_o \) - We are to find expressions for: - \( i_L(t) \): Inductor current - \( i_{sw}(t) \): Switch current - \( v_C(t) \): Capacitor voltage --- # **Step 2: Drawing the Equivalent Circuit** **For \( t > 0 \) (Switch Closed):** - The source \( I_g \) is applied to the LC circuit via the switch. - The diode orientation allows current from left to right (as shown). - The circuit forms a series LC loop with a parallel diode and current source. --- # **Step 3: Writing the Differential Equations** **KCL at the top node:** \[ I_g = i_{sw}(t) + i_L(t) \] **KVL around the LC loop:** \[ v_L + v_C = 0 \] - \( v_L = L \frac{di_L}{dt} \) - \( v_C = \frac{1}{C} \int i_C dt \) But, \( i_C = C \frac{dv_C}{dt} \) and \( i_L = i_C \) (since they are in series in the right loop). So, \[ L \frac{di_L}{dt} + v_C = 0 \] \[ i_L = C \frac{dv_C}{dt} \] or, \[ \frac{di_L}{dt} = -\frac{1}{L} v_C \] \[ i_L = C \frac{dv_C}{dt} \] --- # **Step 4: Second-order Differential Equation** Substitute \( i_L \) into the first equation: \[ L \frac{d}{dt}(C \frac{dv_C}{dt}) + v_C = 0 \] \[ LC \frac{d^2 v_C}{dt^2} + v_C = 0 \] \[ \frac{d^2 v_C}{dt^2} + \frac{1}{LC} v_C = 0 \] This is a standard second-order homogeneous differential equation with solution: \[ v_C(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t) \] Where \( \omega_0 = \frac{1}{\sqrt{LC}} \). --- # **Step 5: Applying Initial Conditions** - \( v_C(0) = V_o \) - \( i_L(0) = C \frac{dv_C}{dt}\bigg|_{t=0} = I_g \) Find \( \frac{dv_C}{dt}\bigg|_{t=0} \): \[ \frac{dv_C}{dt} = -A\omega_0 \sin(\omega_0 t) + B\omega_0 \cos(\omega_0 t) \] At \( t=0 \): \[ \frac{dv_C}{dt}\bigg|_{t=0} = B\omega_0 \] So, \[ I_g = C \cdot B \omega_0 \implies B = \frac{I_g}{C \omega_0} \] From \( v_C(0) = V_o \implies A = V_o \). --- # **Step 6: Final Expressions** ### **Capacitor Voltage** \[ \boxed{v_C(t) = V_o \cos(\omega_0 t) + \frac{I_g}{C\omega_0} \sin(\omega_0 t)} \] ### **Inductor Current** \[ i_L(t) = C \frac{dv_C}{dt} \] \[ \frac{dv_C}{dt} = -V_o \omega_0 \sin(\omega_0 t) + \frac{I_g}{C} \cos(\omega_0 t) \] \[ i_L(t) = -V_o C \omega_0 \sin(\omega_0 t) + I_g \cos(\omega_0 t) \] So, \[ \boxed{i_L(t) = I_g \cos(\omega_0 t) - V_o C \omega_0 \sin(\omega_0 t)} \] ### **Switch Current** \[ i_{sw}(t) = I_g - i_L(t) \] \[ = I_g - \left[ I_g \cos(\omega_0 t) - V_o C \omega_0 \sin(\omega_0 t) \right] \] \[ = I_g [1 - \cos(\omega_0 t)] + V_o C \omega_0 \sin(\omega_0 t) \] So, \[ \boxed{i_{sw}(t) = I_g [1 - \cos(\omega_0 t)] + V_o C \omega_0 \sin(\omega_0 t)} \] --- # **Step 7: Switch Implementation** - **Unidirectional Switch:** Since the current source \( I_g \) only delivers current in one direction and the diode allows current only one way, a unidirectional switch (like a MOSFET or IGBT) is sufficient. - **Bidirectional Switch:** Not required in this configuration. --- ## **Summary Table** | Quantity | Expression | |-----------------|--------------------------------------------------------------------------------------| | \( v_C(t) \) | \( V_o \cos(\omega_0 t) + \frac{I_g}{C\omega_0} \sin(\omega_0 t) \) | | \( i_L(t) \) | \( I_g \cos(\omega_0 t) - V_o C \omega_0 \sin(\omega_0 t) \) | | \( i_{sw}(t) \) | \( I_g [1 - \cos(\omega_0 t)] + V_o C \omega_0 \sin(\omega_0 t) \) | | \( \omega_0 \) | \( \frac{1}{\sqrt{LC}} \) | **Switch Type:** Unidirectional (e.g., standard MOSFET or IGBT). --- Let me know if you want a plot or further explanation of the time-domain behavior!