Certainly! Let's solve **Part (a)** step-by-step as requested, **without using Laplace transforms**. --- # **Step-by-Step Solution for (a)** ## **Given:** - \(R_s = 50\,\Omega\) - \(R_1 = 470\,\Omega\) - \(R_2 = 470\,\Omega\) - \(L = 1\,\text{mH} = 1 \times 10^{-3}\,\text{H}\) - \(C = 100\,\text{nF} = 1 \times 10^{-7}\,\text{F}\) - \(V_s = 8\,\text{V}\) (step at \(t=\)) --- ## **1. Initial Condition at \(t=^-\), (before the step)** ### **Inductor:** acts as a **short circuit** ### **Capacitor:** acts as an **open circuit** - **No voltage across capacitor:** \(v_C(^-) = \) - **No current through inductor:** \(i_L(^-) = \) --- ## **2. Initial Condition at \(t=^+\), (just after the step)** ### **Inductor:** current cannot change instantly \(i_L(^+) = i_L(^-) = \) ### **Capacitor:** voltage cannot change instantly \(v_C(^+) = v_C(^-) = \) --- ## **3. Equivalent Resistance Calculation** The total resistance in series with \(L\) at \(t=^+\): - \(R_{\text{total}} = R_s + R_1 + R_2\) - \(R_{\text{total}} = 50 + 470 + 470 = 990\,\Omega\) --- ## **4. Write the Differential Equation** From KVL (Kirchhoff's Voltage Law): \[ V_s = R_{\text{total}} i(t) + L \frac{di}{dt} + v_C(t) \] But \(v_C(t) = \frac{1}{C}\int i(t) dt\). Alternatively, the second-order equation for current \(i(t)\) in a series RLC step input is: \[ L\frac{d^2 i}{dt^2} + R_{\text{total}} \frac{di}{dt} + \frac{1}{C} i = \frac{dV_s}{dt} \] But since \(V_s\) is a step input (\(V_s u(t)\)), \(\frac{dV_s}{dt} = \) for \(t>\): \[ L\frac{d^2 i}{dt^2} + R_{\text{total}} \frac{di}{dt} + \frac{1}{C} i = \] --- ## **5. Find the Final Steady-State (\(t \to \infty\))** - **Inductor:** behaves as a short circuit. - **Capacitor:** behaves as an open circuit. No current flows through the capacitor branch, so the current is: \[ i(\infty) = \] The voltage across the capacitor is the full source voltage: \[ v_C(\infty) = V_s = 8\,\text{V} \] --- ## **6. Initial Derivatives (\(t = ^+\))** ### **Current through Inductor:** \[ i(^+) = \] ### **Voltage across Capacitor:** \[ v_C(^+) = \] ### **Find \(\frac{di}{dt}\) at \(t = ^+\):** From KVL at \(t=^+\): \[ V_s = R_{\text{total}} i(^+) + L \left. \frac{di}{dt} \right|_{t=^+} + v_C(^+) \] \[ 8 = + 1 \times 10^{-3} \left. \frac{di}{dt} \right|_{t=^+} + \] \[ \left. \frac{di}{dt} \right|_{t=^+} = \frac{8}{1 \times 10^{-3}} = 800\,\text{A/s} \] --- ## **7. Characteristic Equation and Solution** Equation: \[ L\frac{d^2i}{dt^2} + R_{\text{total}} \frac{di}{dt} + \frac{1}{C}i = \] Plug in values: \[ 1\times 10^{-3} \frac{d^2i}{dt^2} + 990\frac{di}{dt} + 1\times 10^7 i = \] \[ \frac{d^2i}{dt^2} + 990,000 \frac{di}{dt} + 1\times 10^{10} i = \] Dividing by \(1\times 10^{-3}\): \[ \frac{d^2i}{dt^2} + 990,000 \frac{di}{dt} + 1\times 10^{10}i = \] --- ### **Characteristic Roots:** \[ s^2 + 990,000 s + 1\times 10^{10} = \] \[ s = \frac{-990,000 \pm \sqrt{(990,000)^2 - 4 \cdot 1 \cdot 10^{10}}}{2} \] \[ = \frac{-990,000 \pm \sqrt{980,100,000,000 - 40,000,000,000}}{2} \] \[ = \frac{-990,000 \pm \sqrt{940,100,000,000}}{2} \] \[ = \frac{-990,000 \pm 969,593}{2} \] Calculate: - \(s_1 = \frac{-990,000 + 969,593}{2} = \frac{-20,407}{2} = -10,203.5\) - \(s_2 = \frac{-990,000 - 969,593}{2} = \frac{-1,959,593}{2} = -979,796.5\) --- ## **8. General Solution** \[ i(t) = A e^{s_1 t} + B e^{s_2 t} \] Where \(s_1 \approx -10,204\), \(s_2 \approx -979,797\) --- ## **9. Apply Initial Conditions** ### \(i() = \): \[ = A + B \implies B = -A \] ### \(\frac{di}{dt}() = 800\): \[ \frac{di}{dt} = A s_1 e^{s_1 t} + B s_2 e^{s_2 t} \] At \(t=\): \[ 800 = A s_1 + B s_2 \] \[ 800 = A s_1 + (-A) s_2 = A(s_1 - s_2) \] \[ A = \frac{800}{s_1 - s_2} \] Calculate \(s_1 - s_2\): \[ s_1 - s_2 = (-10,204) - (-979,797) = 969,593 \] \[ A = \frac{800}{969,593} = .00825 \] \[ B = -A = -.00825 \] --- ## **10. Final Current Response** \[ \boxed{ i(t) = .00825 \left(e^{-10,204 t} - e^{-979,797 t}\right) } \] --- ## **11. Voltage Across Capacitor (\(v_C(t)\))** Recall that for a series RLC with step input, the voltage across the capacitor is: \[ v_C(t) = V_s \left[1 - \frac{(s_2 e^{s_1 t} - s_1 e^{s_2 t})}{(s_2 - s_1)}\right] \] Plugging in values: \[ v_C(t) = 8 \left[1 - \frac{-979,797 e^{-10,204 t} - (-10,204) e^{-979,797 t}}{-979,797 + 10,204}\right] \] \[ = 8 \left[1 - \frac{-979,797 e^{-10,204 t} + 10,204 e^{-979,797 t}}{-969,593}\right] \] \[ = 8 \left[1 + \frac{979,797 e^{-10,204 t} - 10,204 e^{-979,797 t}}{969,593}\right] \] --- ## **12. Final Answers** ### **Current Response:** \[ \boxed{ i(t) = .00825 \left(e^{-10,204 t} - e^{-979,797 t}\right)\ \text{A} } \] ### **Capacitor Voltage Response:** \[ \boxed{ v_C(t) = 8 \left[1 + \frac{979,797 e^{-10,204 t} - 10,204 e^{-979,797 t}}{969,593}\right]\ \text{V} } \] --- ## **Summary Table** | Parameter | Expression | |--------------------|--------------------------------------------------------------------------------------------------------| | \(i(t)\) | \(.00825 \left(e^{-10,204 t} - e^{-979,797 t}\right)\) A | | \(v_C(t)\) | \(8 \left[1 + \frac{979,797 e^{-10,204 t} - 10,204 e^{-979,797 t}}{969,593}\right]\) V | --- **If you need the solution for part (b) or want the numerical plot, let me know!**