Let's break down the problem step by step: ### **Step 1: Understanding the Problem** - **Switches 1 and 2 are synchronized.** When one closes, the other opens. - **Switch 1 has been open for a long time** before closing at \( t = 0 \). - **At \( t = 0 \):** Switch 1 closes, Switch 2 opens. - **We are to find \( i_L(t) \), the current through the inductor (316 mH), for \( t \geq 0 \).** --- ### **Step 2: Analyze the Circuit for \( t < 0 \) ("Steady State" Just Before Switching)** **Switch 1 is OPEN, Switch 2 is CLOSED:** - The left circuit (with voltage source, resistor, and capacitor) is isolated from the right circuit (with current source, inductor, and resistor). - The **inductor will act as a short circuit** in steady-state DC (for a long time), so all current from the 10 A source will flow through the 1 kΩ resistor and the inductor. #### **Find initial inductor current (\( i_L(0^-) \)):** - **Inductor in steady-state:** Short circuit. - **Current through inductor:** All of the 10 A from the current source passes through the inductor. - **So:** \[ i_L(0^-) = 10~A \] --- ### **Step 3: Analyze the Circuit for \( t \geq 0 \) (After Switching)** **Switch 1 is CLOSED, Switch 2 is OPEN:** - Now, the left branch (80 V, 10 Ω, 250 μF) is connected to the right branch (316 mH, 1 kΩ), but the current source is disconnected (Switch 2 open). - The circuit is now a **series RLC circuit**: \[ 80~V \rightarrow 10~\Omega \rightarrow 250~\mu F \rightarrow 316~mH \rightarrow 1~k\Omega \] But we are interested in the current through the **inductor** after switching, \( i_L(t) \). --- ### **Step 4: Write the Loop Equation for \( t \geq 0 \)** Let’s define current \( i(t) \) as the current through the inductor and the series circuit after \( t=0 \): \[ 80 - 10i - v_C - 1000i - 316 \cdot 10^{-3}\frac{di}{dt} = 0 \] Where \( v_C \) is the voltage across the capacitor. But, for a **series RLC circuit**, the KVL loop is: \[ V_{source} = V_R + V_L + V_C \] \[ 80 = (10 + 1000)i + 316 \cdot 10^{-3}\frac{di}{dt} + v_C \] But we need everything in terms of \( i \): \[ v_C = \frac{1}{C} \int i(t)dt \] But a more straightforward approach is to write the **differential equation** for an RLC series circuit: \[ V_{source} = Ri + L\frac{di}{dt} + \frac{1}{C}\int i(t) dt \] Or, differentiating both sides: \[ 0 = R\frac{di}{dt} + L\frac{d^2i}{dt^2} + \frac{i}{C} \] But our circuit has two resistors: \[ R_{total} = 10~\Omega + 1~k\Omega = 1010~\Omega \] \[ L = 316~\text{mH} = 0.316~H \] \[ C = 250~\mu F = 250 \times 10^{-6}~F \] Rewrite the equation: \[ 0 = 1010\frac{di}{dt} + 0.316\frac{d^2i}{dt^2} + \frac{i}{250 \times 10^{-6}} \] --- ### **Step 5: Initial Conditions for \( t = 0 \)** - **Current through inductor is continuous:** \[ i(0^+) = i(0^-) = 10~A \] - **Voltage across capacitor is also continuous.** But, before switching, the capacitor is not connected to any source (open circuit), so its voltage is 0: \[ v_C(0^+) = 0~V \] --- ### **Step 6: Write the Differential Equation (Homogeneous)** \[ 0.316\frac{d^2i}{dt^2} + 1010\frac{di}{dt} + \frac{i}{250 \times 10^{-6}} = 0 \] \[ 0.316\frac{d^2i}{dt^2} + 1010\frac{di}{dt} + 4000i = 0 \] Divide by 0.316: \[ \frac{d^2i}{dt^2} + \frac{1010}{0.316}\frac{di}{dt} + \frac{4000}{0.316}i = 0 \] \[ \frac{d^2i}{dt^2} + 3196.2\frac{di}{dt} + 12658.2i = 0 \] --- ### **Step 7: Characteristic Equation** \[ s^2 + 3196.2s + 12658.2 = 0 \] Solve for \( s \): \[ s = \frac{-3196.2 \pm \sqrt{3196.2^2 - 4 \cdot 12658.2}}{2} \] Calculate discriminant: \[ D = 3196.2^2 - 4 \cdot 12658.2 = 10263034.44 - 50632.8 = 10212301.64 \] \[ \sqrt{10212301.64} \approx 3195.67 \] So, \[ s_1 = \frac{-3196.2 + 3195.67}{2} = \frac{-0.53}{2} \approx -0.265 \] \[ s_2 = \frac{-3196.2 - 3195.67}{2} = \frac{-6391.87}{2} \approx -3195.94 \] --- ### **Step 8: General Solution** \[ i(t) = A_1 e^{-0.265 t} + A_2 e^{-3195.94 t} \] --- ### **Step 9: Apply Initial Conditions** 1. **At \( t = 0 \):** \[ i(0) = A_1 + A_2 = 10 \] 2. **Find \( \frac{di}{dt} \) at \( t = 0 \):** From KVL at \( t=0 \): \[ 80 = 1010i(0) + 0.316 \cdot \frac{di}{dt}\bigg|_{t=0} + v_C(0) \] But \( v_C(0) = 0 \): \[ 80 = 1010 \times 10 + 0.316 \cdot \frac{di}{dt}(0) \] \[ 80 = 10100 + 0.316 \cdot \frac{di}{dt}(0) \] \[ 0.316 \cdot \frac{di}{dt}(0) = 80 - 10100 = -10020 \] \[ \frac{di}{dt}(0) = \frac{-10020}{0.316} \approx -31747.5~A/s \] Now, take the derivative of the general solution: \[ \frac{di}{dt} = -0.265A_1 e^{-0.265 t} - 3195.94A_2 e^{-3195.94 t} \] At \( t=0 \): \[ \frac{di}{dt}(0) = -0.265A_1 - 3195.94A_2 = -31747.5 \] --- ### **Step 10: Solve for \( A_1 \) and \( A_2 \)** Two equations: 1. \( A_1 + A_2 = 10 \) 2. \( -0.265A_1 - 3195.94A_2 = -31747.5 \) Substitute \( A_1 = 10 - A_2 \): \[ -0.265(10 - A_2) - 3195.94A_2 = -31747.5 \] \[ -2.65 + 0.265A_2 - 3195.94A_2 = -31747.5 \] \[ 0.265A_2 - 3195.94A_2 = -31747.5 + 2.65 \] \[ -3195.675A_2 = -31744.85 \] \[ A_2 = \frac{-31744.85}{-3195.675} \approx 9.93 \] \[ A_1 = 10 - 9.93 = 0.07 \] --- ### **Step 11: Final Solution** \[ \boxed{ i_L(t) = 0.07\,e^{-0.265 t} + 9.93\,e^{-3195.94 t} \quad \text{A}, \quad t \geq 0 } \] --- ### **Interpretation** - The term \( e^{-3195.94t} \) decays **very rapidly** (within milliseconds), so for \( t \gg 0 \), \( i_L(t) \approx 0.07\,e^{-0.265 t} \) A. - **Initial current** is 10 A, as expected. - **Initial rate of change** matches the calculated value. --- ## **Summary Table** | Step | Action | Value/Equation | |-------------|-------------------------------------|--------------------------------------------| | \( i_L(0^-) \) | Inductor current at \( t = 0^- \) | \( 10~A \) | | \( v_C(0^-) \) | Capacitor voltage at \( t = 0^- \)| \( 0~V \) | | Diff. Eq. | Series RLC equation | \( 0.316\frac{d^2i}{dt^2} + 1010\frac{di}{dt} + 4000i = 0 \) | | Solution | General solution | \( i(t) = A_1 e^{-0.265 t} + A_2 e^{-3195.94 t} \) | | \( A_1, A_2 \) | Constants | \( A_1 = 0.07,~A_2 = 9.93 \) | | Final Answer| | \( \boxed{i_L(t) = 0.07\,e^{-0.265 t} + 9.93\,e^{-3195.94 t}~\text{A}} \) | If you need the voltage across any element or more details, let me know!

Let's carefully analyze and solve the problem step-by-step, ensuring clarity and correctness. --- ### **Problem Summary:** - **Switches** 1 and 2 are **synchronized**: when one opens, the other closes. - **Initially** (for \( t < 0 \)), **Switch 1 is open**, **Switch 2 is closed**. - At \( t=0 \), **Switch 1 closes**, **Switch 2 opens**. - **Initial inductor current:** \( i_L(0^-) \) (just before switching). - **Goal:** Find \( i_L(t) \) for \( t \geq 0 \). --- ### **Step 1: Steady State for \( t < 0 \)** - **Switch 1:** open - **Switch 2:** closed In this state: - The **10 A current source** feeds the inductor through the 1 kΩ resistor. - **Inductor in steady state:** acts as a short circuit (for DC). - **Initial inductor current:** \[ i_L(0^-) = 10\,A \] - The capacitor (250 μF) is isolated with no initial voltage (assuming initial charge is zero). --- ### **Step 2: Initial conditions at \( t=0 \)** - **Inductor current:** remains continuous: \[ i_L(0^+) = i_L(0^-) = 10\,A \] - **Capacitor voltage:** \( v_C(0^-) = 0\,V \). No sudden change because capacitor voltage is continuous. --- ### **Step 3: Circuit for \( t \ge 0 \)** After switching: - **Switch 1:** closed - **Switch 2:** open The circuit now: - The **80 V source** is connected through the resistor network and capacitor to the inductor. - **Resistors:** 10 Ω (series with capacitor), 10 Ω (series with inductor), totaling 20 Ω. - **Capacitor:** 250 μF. - **Inductor:** 316 mH. --- ### **Step 4: Write the differential equation** The series R-L-C circuit: \[ \text{Sum of voltages} = 0 \] Applying KVL: \[ 80\,V = R_{total} \cdot i(t) + L \frac{di}{dt} + v_C(t) \] Where: - \( R_{total} = 10\,\Omega + 10\,\Omega = 20\,\Omega \), - \( L = 0.316\,H \), - \( C = 250\, \mu F = 250 \times 10^{-6}\,F \). Express \( v_C(t) \): \[ v_C(t) = \frac{1}{C} \int i(t) dt \] Differentiating the entire equation: \[ 0 = R_{total} \frac{di}{dt} + L \frac{d^2 i}{dt^2} + \frac{i}{C} \] Rearranged: \[ L \frac{d^2 i}{dt^2} + R_{total} \frac{di}{dt} + \frac{i}{C} = 0 \] Plugging in values: \[ 0.316 \frac{d^2 i}{dt^2} + 20 \frac{di}{dt} + \frac{i}{250 \times 10^{-6}} = 0 \] \[ 0.316 \frac{d^2 i}{dt^2} + 20 \frac{di}{dt} + 4000\, i = 0 \] Divide through by 0.316: \[ \frac{d^2 i}{dt^2} + 63.29 \frac{di}{dt} + 12658.2\, i = 0 \] --- ### **Step 5: Characteristic equation** \[ s^2 + 63.29 s + 12658.2 = 0 \] Discriminant: \[ D = (63.29)^2 - 4 \times 12658.2 \approx 4004.3 - 50632.8 = -46628.5 \] Since discriminant is negative, the solution is **underdamped**: \[ s = -\frac{63.29}{2} \pm j \frac{\sqrt{46628.5}}{2} \] Calculate: \[ \alpha = 31.645\, \text{s}^{-1} \] \[ \omega_d = \frac{\sqrt{46628.5}}{2} \approx \frac{216}{2} = 108\, \text{rad/sec} \] --- ### **Step 6: General solution** \[ i(t) = e^{-\alpha t} \left( A \cos \omega_d t + B \sin \omega_d t \right) + i_{ss} \] Since the circuit is driven to a steady DC current: \[ i_{ss} = \frac{V}{R_{total}} = \frac{80}{20} = 4\,A \] The **initial inductor current** is 10 A, so: \[ i(0) = 10 = A + i_{ss} \Rightarrow A = 6\,A \] --- ### **Step 7: Find \( B \) using initial derivative** Derivative: \[ \frac{di}{dt} = -\alpha e^{-\alpha t} (A \cos \omega_d t + B \sin \omega_d t) + e^{-\alpha t} (-A \omega_d \sin \omega_d t + B \omega_d \cos \omega_d t) \] At \( t=0 \): \[ \frac{di}{dt}(0) = -\alpha A + B \omega_d \] Calculate \( \frac{di}{dt}(0) \): From KVL at \( t=0 \): \[ 80 = R_{total} i(0) + L \frac{di}{dt}(0) + v_C(0) \] Since \( v_C(0)=0 \): \[ 80 = 20 \times 10 + 0.316 \times \frac{di}{dt}(0) \] \[ 80 = 200 + 0.316 \times \frac{di}{dt}(0) \] \[ \frac{di}{dt}(0) = \frac{80 - 200}{0.316} = -120 / 0.316 \approx -379.75\, \text{A/sec} \] Now: \[ -379.75 = -31.645 \times 6 + B \times 108 \] \[ -379.75 = -189.87 + 108 B \] \[ 108 B = -379.75 + 189.87 = -189.88 \] \[ B \approx -1.76\, \text{A} \] --- ### **### **Final expression:** \[ \boxed{ \boxed{ i_L(t) = 4 + e^{-31.645 t} \left( 6 \cos 108 t - 1.76 \sin 108 t \right) \quad \text{A} } } \] --- ### **Summary:** - **Initial current:** 10 A - **Steady-state current:** 4 A - **Transient oscillations:** damped at \( \alpha \approx 31.6\, \text{s}^{-1} \), frequency \( \approx 108\, \text{rad/sec} \). This solution matches the physical behavior and initial conditions. --- **Note:** The earlier detailed solution had some approximations. This final version uses precise steps aligning with circuit analysis principles and gives a clear, correct expression for \( i_L(t) \).