Let's break down the initial conditions for the parallel RLC circuit shown, given the current source \(2.5u(-t) \, A\). Here, \(u(-t)\) is the unit step function, which is 1 for \(t < 0\) and 0 for \(t > 0\). ### For \(t = 0^-\) (just before the switch): - The current source has been on for a long time. - The inductor behaves as a short circuit (steady-state for DC). - The capacitor behaves as an open circuit (steady-state for DC). ### For \(t = 0^+\) (just after the switch): - The current source turns off (\(i = 0\)). - The inductor current cannot change instantaneously; it remains the same as at \(t = 0^-\). - The capacitor voltage cannot change instantaneously; it remains the same as at \(t = 0^-\). Let's solve for the initial values: #### 1. **Steady-state before \(t = 0\):** - **Resistor Current:** All current from the source flows through the resistor and inductor (since capacitor is open). - **Inductor Current (\(i_L(0^-)\)):** Same as the source current. - **Capacitor Current:** 0 (open circuit to DC). - **Capacitor Voltage (\(v_C(0^-)\)):** Voltage across the resistor and inductor. - **Resistor Voltage:** \(I_{source} \cdot R\) - **Inductor Voltage:** 0 (steady-state for DC, \(V_L = L \frac{di}{dt} = 0\)). Given: - \(I_{source} = 2.5\,A\) - \(R = 500\,m\Omega = 0.5\,\Omega\) #### Calculate values at \(t = 0^-\): - **Resistor Current:** \(2.5\,A\) - **Resistor Voltage:** \(2.5\,A \times 0.5\,\Omega = 1.25\,V\) - **Capacitor Current:** \(0\,A\) - **Capacitor Voltage:** \(1.25\,V\) - **Inductor Current:** \(2.5\,A\) - **Inductor Voltage:** \(0\,V\) #### At \(t = 0^+\): - **Resistor Current:** 0 (source turns off, but inductor provides current for an instant) - **Resistor Voltage:** 0 - **Capacitor Current:** Inductor current flows through the capacitor (initially) - **Capacitor Voltage:** \(1.25\,V\) (cannot change instantaneously) - **Inductor Current:** \(2.5\,A\) (cannot change instantaneously) - **Inductor Voltage:** Determined by the circuit at \(t=0^+\) (will be \(v_C\), since inductor and capacitor are now in parallel and the current must go through the capacitor). --- ## **Table of Initial Conditions** | | \(t = 0^-\) | \(t = 0^+\) | |---------------------------|------------------|------------------| | **Resistor Current** | \(2.5\,A\) | \(0\,A\) | | **Resistor Voltage** | \(1.25\,V\) | \(0\,V\) | | **Capacitor Current** | \(0\,A\) | \(2.5\,A\) | | **Capacitor Voltage** | \(1.25\,V\) | \(1.25\,V\) | | **Inductor Current** | \(2.5\,A\) | \(2.5\,A\) | | **Inductor Voltage** | \(0\,V\) | \(1.25\,V\) | --- ## **b) Voltage of the Capacitor as a function of time, \(v_C(t)\):** After \(t=0\), the circuit is a parallel LC with the inductor initially carrying 2.5 A and the capacitor charged to 1.25 V. This is a classic second-order homogeneous equation: \[ i_L(0^+) = 2.5\,A, \quad v_C(0^+) = 1.25\,V \] The governing equation for the parallel LC circuit is: \[ L \frac{di_L}{dt} + v_C = 0 \] And since \[ i_L = C \frac{dv_C}{dt} \] So, \[ L C \frac{d^2v_C}{dt^2} + v_C = 0 \] \[ \frac{d^2v_C}{dt^2} + \frac{1}{LC} v_C = 0 \] This is a second-order differential equation with solution: \[ v_C(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t) \] Where \(\omega_0 = \frac{1}{\sqrt{LC}}\): Given: - \(L = 160\,mH = 0.16\,H\) - \(C = 250\,mF = 0.25\,F\) \[ \omega_0 = \frac{1}{\sqrt{0.16 \times 0.25}} = \frac{1}{0.2} = 5\,rad/s \] **Initial conditions:** - \(v_C(0) = 1.25\,V\) - \(i_L(0) = C \frac{dv_C}{dt}\Big|_{t=0} = 2.5\,A\) \[ \frac{dv_C}{dt}\Big|_{t=0} = \frac{i_L(0)}{C} = \frac{2.5}{0.25} = 10\,V/s \] So: \[ v_C(0) = A = 1.25 \] \[ \frac{dv_C}{dt}\Big|_{t=0} = B \omega_0 = 10 \implies B = \frac{10}{\omega_0} = \frac{10}{5} = 2 \] \[ \boxed{v_C(t) = 1.25 \cos(5t) + 2 \sin(5t)} \] --- **Summary:** - Table is filled above. - Capacitor voltage as a function of time: \(\boxed{v_C(t) = 1.25 \cos(5t) + 2 \sin(5t)}\) for \(t > 0\).

Yes, your analysis and calculations are correct. Here's a concise summary confirming your reasoning: **Initial Conditions:** - Before \(t=0\): - The inductor current is \(2.5\,A\) (since the circuit has been on long enough for steady state). - The capacitor voltage is \(1.25\,V\). - At \(t=0^+\): - The inductor current remains \(2.5\,A\) (instantaneous current cannot change). - The capacitor voltage remains \(1.25\,V\) (instantaneous voltage cannot change). - The resistor voltage drops to zero immediately after the switch opens, and the resistor current drops to zero. - The initial conditions for the LC oscillation are set accordingly. **Oscillation Equation:** - The natural frequency \(\omega_0 = \frac{1}{\sqrt{LC}} = 5\,rad/sec\). - The initial voltage \(v_C(0) = 1.25\,V\). - The initial derivative \(\frac{dv_C}{dt}\big|_{0} = 10\,V/sec\). **Solution:** \[ v_C(t) = 1.25 \cos(5t) + 2 \sin(5t) \] This correctly describes the capacitor voltage as a function of time for \(t > 0\), representing an oscillation with the initial conditions you derived. **Conclusion:** Your initial condition table and the capacitor voltage function are both correct. Well done!