Let's solve for \( v_C(t) \) for \( t > \) **step by step**: --- ## **Step 1: Write the given circuit details** - **Current source:** \( i_s(t) = -\frac{1}{4}u(t) + \frac{1}{2} \) A - **Inductor:** \( L = 2 \) H - **Capacitor:** \( C = \frac{1}{4} \) F - **Resistors:** \( R_1 = 4\,\Omega \), \( R_2 = 8\,\Omega \) - **Unknown:** \( v_C(t) \) across the capacitor --- ## **Step 2: Write the node equation at the top node** Let \( v_C(t) \) be the voltage across the capacitor. ### **Current through each element:** - **Inductor:** \( i_L = \frac{1}{L} \int (v_C(t) - ) dt \) - **4Ω resistor:** \( i_{R_1} = \frac{v_C(t)}{4} \) - **8Ω resistor:** \( i_{R_2} = \frac{v_C(t)}{8} \) - **Capacitor:** \( i_C = C \frac{dv_C}{dt} \) But, since the inductor and the two resistors are in parallel with the capacitor, we need a systematic equation. --- ## **Step 3: Laplace Transform the circuit** Let \( V_C(s) \) be the Laplace transform of \( v_C(t) \). - **Inductor impedance:** \( 2s \) - **Capacitor impedance:** \( \frac{1}{sC} = \frac{1}{s \cdot \frac{1}{4}} = \frac{4}{s} \) - **Resistors:** \( 4\,\Omega,\,8\,\Omega \) - **Current source:** \( I_s(s) = \mathcal{L}\left\{-\frac{1}{4}u(t) + \frac{1}{2}\right\} = -\frac{1}{4}\cdot \frac{1}{s} + \frac{1}{2}\cdot \frac{1}{s} = \frac{1}{4s} \) All four elements are in parallel, so the total **admittance** is: \[ Y_{total}(s) = \frac{1}{4} + \frac{1}{8} + \frac{1}{2s} + \frac{s}{4} \] --- ## **Step 4: Write KCL in Laplace domain** Sum of currents leaving the top node = current source: \[ \frac{V_C(s)}{4} + \frac{V_C(s)}{8} + \frac{V_C(s)}{2s} + s \cdot \frac{V_C(s)}{4} = I_s(s) \] \[ V_C(s) \left[ \frac{1}{4} + \frac{1}{8} + \frac{1}{2s} + \frac{s}{4} \right] = \frac{1}{4s} \] \[ V_C(s) \left( \frac{1}{4} + \frac{1}{8} + \frac{1}{2s} + \frac{s}{4} \right) = \frac{1}{4s} \] Combine terms: \[ \frac{1}{4} + \frac{1}{8} = \frac{3}{8} \] So: \[ V_C(s) \left( \frac{3}{8} + \frac{1}{2s} + \frac{s}{4} \right) = \frac{1}{4s} \] Multiply both sides by 8 to clear denominators: \[ V_C(s) \left( 3 + \frac{4}{s} + 2s \right) = \frac{2}{s} \] \[ V_C(s) (2s + 3 + \frac{4}{s}) = \frac{2}{s} \] \[ V_C(s) \left( \frac{2s^2 + 3s + 4}{s} \right) = \frac{2}{s} \] \[ V_C(s) (2s^2 + 3s + 4) = 2 \] \[ V_C(s) = \frac{2}{2s^2 + 3s + 4} \] --- ## **Step 5: Inverse Laplace Transform** Let’s write the denominator in standard quadratic form to find the time-domain solution. \[ 2s^2 + 3s + 4 \] Complete the square: \[ 2(s^2 + \frac{3}{2}s + 2) \] \[ = 2\left[ (s + \frac{3}{4})^2 + 2 - (\frac{9}{16}) \right] \] \[ = 2\left[ (s + \frac{3}{4})^2 + \frac{23}{16} \right] \] So, \[ V_C(s) = \frac{2}{2 \left( (s + \frac{3}{4})^2 + \frac{23}{16} \right) } = \frac{1}{ (s + \frac{3}{4})^2 + \frac{23}{16} } \] Let \( \alpha = \frac{3}{4} \), \( \omega^2 = \frac{23}{16} \), \( \omega = \sqrt{\frac{23}{16}} \). The Laplace pair is: \[ \mathcal{L}^{-1}\left\{ \frac{1}{(s + a)^2 + b^2} \right\} = \frac{1}{b} e^{-a t} \sin(bt) \] Therefore: \[ v_C(t) = \frac{1}{\sqrt{23/16}} e^{-3t/4} \sin \left( \sqrt{\frac{23}{16}} t \right) \cdot u(t) \] \[ \sqrt{\frac{23}{16}} = \frac{\sqrt{23}}{4} \] So, \[ v_C(t) = \frac{4}{\sqrt{23}} e^{-3t/4} \sin \left( \frac{\sqrt{23}}{4} t \right) u(t) \] --- ## **Final Answer** \[ \boxed{ v_C(t) = \frac{4}{\sqrt{23}}\, e^{-3t/4}\, \sin\left( \frac{\sqrt{23}}{4}\, t \right)\, u(t) } \] - Where \( u(t) \) is the unit step function. - This is the voltage across the capacitor for \( t > \).