1. **Problem Statement:** We have an RLC circuit with a 1 H inductor, a 1/50 F capacitor, and resistors of 4 Ω and 5 Ω arranged in a series-parallel configuration. The switch opens at $t=0$. We need to find: (i) The inductor current $i(t)$ and capacitor voltage $v_c(t)$ just before the switch opens at $t=0$. (ii) The Laplace transform representation of the inductor current $i(t)$ for $t \geq 0$. 2. **Initial Conditions Before Switch Opens ($t=0^-$):** - The circuit has been connected for a long time before $t=0$, so it is in steady state. - In steady state, the inductor behaves like a short circuit (since $V_L = L \frac{di}{dt}$ and $\frac{di}{dt}=0$ at steady state). - The capacitor behaves like an open circuit (since $I_C = C \frac{dv}{dt}$ and $\frac{dv}{dt}=0$ at steady state). 3. **Calculate $i(0^-)$ (inductor current just before $t=0$):** - The inductor is a short circuit, so the current flows through the 1 H inductor and the 4 Ω resistor in series. - The capacitor branch is open, so no current flows through the capacitor and 5 Ω resistor branch. - Total resistance in the path is $R = 4 \Omega$. - Using Ohm's law, current $i(0^-) = \frac{V}{R} = \frac{20}{4} = 5$ A. 4. **Calculate $v_c(0^-)$ (capacitor voltage just before $t=0$):** - Capacitor voltage equals the voltage across the 5 Ω resistor and capacitor branch. - Since no current flows through this branch (open circuit), voltage across capacitor is the same as voltage across the 5 Ω resistor, which is zero. - Therefore, $v_c(0^-) = 0$ V. 5. **Summary of initial conditions:** $$ i(0^-) = 5 \text{ A}, \quad v_c(0^-) = 0 \text{ V} $$ 6. **Laplace Domain Analysis for $t \geq 0$:** - After the switch opens at $t=0$, the circuit changes and we analyze the transient response. - Use Laplace transforms to find $I(s)$, the Laplace transform of $i(t)$. 7. **Circuit elements in Laplace domain:** - Inductor impedance: $sL = s \times 1 = s$ Ω. - Capacitor impedance: $\frac{1}{sC} = \frac{1}{s \times \frac{1}{50}} = \frac{50}{s}$ Ω. 8. **Equivalent circuit in Laplace domain:** - The 4 Ω resistor is in series with the inductor $s$ Ω. - The capacitor $\frac{50}{s}$ Ω is in series with the 5 Ω resistor. - These two branches are in parallel. 9. **Calculate the parallel branch impedance:** $$ Z_{parallel} = \left(4 + s\right) \parallel \left(5 + \frac{50}{s}\right) = \frac{(4 + s)(5 + \frac{50}{s})}{(4 + s) + (5 + \frac{50}{s})} $$ 10. **Simplify numerator:** $$ (4 + s)\left(5 + \frac{50}{s}\right) = (4 + s)\left(\frac{5s + 50}{s}\right) = \frac{(4 + s)(5s + 50)}{s} $$ 11. **Simplify denominator:** $$ (4 + s) + \left(5 + \frac{50}{s}\right) = 9 + s + \frac{50}{s} = \frac{9s + s^2 + 50}{s} $$ 12. **Therefore,** $$ Z_{parallel} = \frac{\frac{(4 + s)(5s + 50)}{s}}{\frac{9s + s^2 + 50}{s}} = \frac{(4 + s)(5s + 50)}{9s + s^2 + 50} $$ 13. **The total impedance seen by the source is $Z_{parallel}$. The Laplace voltage source is $\frac{20}{s}$. Using Ohm's law in Laplace domain:** $$ I(s) = \frac{V(s)}{Z_{parallel}} = \frac{\frac{20}{s}}{Z_{parallel}} = \frac{20}{s} \times \frac{9s + s^2 + 50}{(4 + s)(5s + 50)} $$ 14. **Simplify numerator and denominator:** - Numerator: $20(9s + s^2 + 50) = 20s^2 + 180s + 1000$ - Denominator: $s(4 + s)(5s + 50)$ 15. **Final Laplace transform of inductor current:** $$ I(s) = \frac{20s^2 + 180s + 1000}{s(4 + s)(5s + 50)} $$ 16. **Summary:** - Initial conditions: $i(0^-) = 5$ A, $v_c(0^-) = 0$ V. - Laplace transform of inductor current for $t \geq 0$: $$ I(s) = \frac{20s^2 + 180s + 1000}{s(4 + s)(5s + 50)} $$