Let's solve for \( i_o(t) \) using the **Laplace transform** in the given circuit. --- ## **Step 1: Laplace Transform of the Circuit Elements** - **Resistor \( R \):** \( R \) - **Inductor \( L \):** \( sL \) - **Capacitor \( C \):** \( \frac{1}{sC} \) Given values: - \( R_1 = 1\,\Omega \) (left) - \( R_2 = 2\,\Omega \) (bottom) - \( R_3 = 1\,\Omega \) (right) - \( L = 1\,H \) - \( C = 1\,F \) --- ## **Step 2: Laplace Transform of Sources** - \( 4u(t) \rightarrow \frac{4}{s} \) - \( e^{-t}u(t) \rightarrow \frac{1}{s+1} \) --- ## **Step 3: Redraw Circuit in \( s \)-Domain** - Left branch: \( \frac{4}{s} \) in series with \( 1\,\Omega \) and \( s \) (from \( 1H \)) - Right branch: \( \frac{1}{s+1} \) in series with \( 1\,\Omega \) and \( \frac{1}{s} \) (from \( 1F \)) - Both branches connect at the top and bottom nodes, with \( 2\,\Omega \) resistor between them (where \( i_o(t) \) flows). --- ## **Step 4: Write Node Equation (at the top node)** Let the top node voltage be \( V_x \). **Left branch:** \[ I_{\text{left}} = \frac{\frac{4}{s} - V_x}{1 + s} \] **Right branch:** \[ I_{\text{right}} = \frac{\frac{1}{s+1} - V_x}{1 + \frac{1}{s}} = \frac{\frac{1}{s+1} - V_x}{1 + \frac{1}{s}} = \frac{\frac{1}{s+1} - V_x}{\frac{s+1}{s}} = \frac{s}{s+1}\left( \frac{1}{s+1} - V_x \right) \] **Current through \( 2 \Omega \) (downwards, \( i_o(s) \)):** \[ i_o(s) = \frac{V_x}{2} \] **KCL at top node:** \[ I_{\text{left}} + I_{\text{right}} = i_o(s) \] Plug in the expressions: \[ \frac{\frac{4}{s} - V_x}{1 + s} + \frac{s}{s+1}\left(\frac{1}{s+1} - V_x\right) = \frac{V_x}{2} \] --- ## **Step 5: Simplify the Equation** ### First term: \[ \frac{4}{s(1+s)} - \frac{V_x}{1+s} \] ### Second term: \[ \frac{s}{s+1} \cdot \frac{1}{s+1} - \frac{s}{s+1} V_x = \frac{s}{(s+1)^2} - \frac{s}{s+1} V_x \] ### Combine all: \[ \frac{4}{s(1+s)} - \frac{V_x}{1+s} + \frac{s}{(s+1)^2} - \frac{s}{s+1} V_x = \frac{V_x}{2} \] Group \( V_x \) terms: \[ \frac{4}{s(1+s)} + \frac{s}{(s+1)^2} = \frac{V_x}{2} + \frac{V_x}{1+s} + \frac{s}{s+1} V_x \] Combine right hand side: \[ \frac{V_x}{2} + \frac{V_x}{1+s} + \frac{s}{s+1} V_x \] \[ = V_x \left[ \frac{1}{2} + \frac{1}{1+s} + \frac{s}{s+1} \right] \] Note: \[ \frac{1}{1+s} + \frac{s}{s+1} = \frac{1 + s}{1 + s} = 1 \] So: \[ V_x \left[ \frac{1}{2} + 1 \right] = V_x \left( \frac{3}{2} \right) \] Therefore, the equation is: \[ \frac{4}{s(1+s)} + \frac{s}{(s+1)^2} = \frac{3}{2} V_x \] --- ## **Step 6: Solve for \( V_x \) and \( i_o(s) \)** \[ V_x = \frac{2}{3} \left[ \frac{4}{s(1+s)} + \frac{s}{(s+1)^2} \right] \] Recall: \[ i_o(s) = \frac{V_x}{2} \] So: \[ i_o(s) = \frac{1}{3} \left[ \frac{4}{s(1+s)} + \frac{s}{(s+1)^2} \right] \] --- ## **Step 7: Inverse Laplace Transform** ### Break into parts: #### First part: \[ \frac{4}{s(1+s)} = \frac{4}{s} - \frac{4}{s+1} \] So, \[ \frac{4}{s(1+s)} \rightarrow 4u(t) - 4e^{-t}u(t) \] #### Second part: \[ \frac{s}{(s+1)^2} \] Let’s expand: \[ \frac{s}{(s+1)^2} = \frac{(s+1)-1}{(s+1)^2} = \frac{1}{s+1} - \frac{1}{(s+1)^2} \] So, \[ \frac{s}{(s+1)^2} \rightarrow e^{-t}u(t) - t e^{-t}u(t) \] ### Combine: \[ i_o(s) = \frac{1}{3} \left[ 4 \left(\frac{1}{s} - \frac{1}{s+1}\right) + \left(\frac{1}{s+1} - \frac{1}{(s+1)^2}\right) \right ] \] \[ = \frac{1}{3} \left[ \frac{4}{s} - \frac{4}{s+1} + \frac{1}{s+1} - \frac{1}{(s+1)^2} \right] \] \[ = \frac{1}{3} \left[ \frac{4}{s} - \frac{3}{s+1} - \frac{1}{(s+1)^2} \right] \] --- ## **Step 8: Final Time Domain Solution** \[ i_o(t) = \frac{1}{3} \left[ 4u(t) - 3e^{-t}u(t) - t e^{-t}u(t) \right ] \] Or, simplified: \[ \boxed{ i_o(t) = \frac{4}{3} u(t) - e^{-t}u(t) - \frac{1}{3} t e^{-t}u(t) } \] --- ### **Summary** \[ \boxed{ i_o(t) = \frac{4}{3} - e^{-t} - \frac{1}{3} t e^{-t},\quad t \geq } \] --- If you’d like to see the full algebra or any specific step in more detail, let me know!