Let's solve the problem **step by step**: ## **Given:** - The switch is at **A** for a long time (steady-state before \( t = 0 \)). - At \( t = 0 \), the switch moves to position **B**. - Find \( v(t) \) across the capacitor for \( t > 0 \). ### **Component Values:** - **Resistors:** \( 200\Omega, 4\Omega, 1k\Omega \) - **Capacitor:** \( 0.1\mu F \) - **Inductor:** \( 0.2H \) - **Source:** \( 20V \) --- ## **Step 1: Initial Value \( v(0^-) \) and \( v(0^+) \)** - **At \( t = 0^- \):** The switch is at A **for a long time**. - Capacitor voltage **cannot change instantaneously**, so \( v(0^+) = v(0^-) \). ### **Find \( v(0^-) \):** **While at A:** - The right circuit is disconnected. - The circuit is just: \( 20V \) source → \( 200\Omega \) → capacitor (\( 0.1\mu F \)). **Steady-state (long time):** - Capacitor acts as an **open circuit**. - So, **no current through \( 200\Omega \)**. - **Voltage across capacitor = source voltage = \( 20V \)**. **Thus:** \[ v(0^-) = 20V \implies v(0^+) = 20V \] --- ## **Step 2: Write Differential Equation for \( t > 0 \) (Switch at B)** ### **New Circuit:** - The \( 20V \) source and \( 200\Omega \) are **disconnected**. - The circuit is: - Capacitor (\( 0.1\mu F \)) - in series with \( 4\Omega \), \( 1k\Omega \), and inductor \( 0.2H \). #### **Redraw for clarity:** - Series loop: \( 4\Omega \) → \( 1k\Omega \) → \( 0.2H \) → \( 0.1\mu F \) Let \( v(t) \) = voltage across capacitor \( C \) for \( t > 0 \). --- ### **Apply KVL:** Let \( i(t) \) be the current: \[ i(t) = C \frac{dv}{dt} \] Total resistance, \( R = 4 + 1000 = 1004\Omega \), \( L = 0.2H \), \( C = 0.1 \mu F = 0.1 \times 10^{-6} F \). \[ R i + L \frac{di}{dt} + v = 0 \] But \( i = C \frac{dv}{dt} \) and \( \frac{di}{dt} = C \frac{d^2v}{dt^2} \): \[ R C \frac{dv}{dt} + L C \frac{d^2v}{dt^2} + v = 0 \] \[ L C \frac{d^2v}{dt^2} + R C \frac{dv}{dt} + v = 0 \] Plug in values: - \( L C = 0.2 \times 0.1 \times 10^{-6} = 0.02 \times 10^{-6} = 2 \times 10^{-8} \) - \( R C = 1004 \times 0.1 \times 10^{-6} = 100.4 \times 10^{-6} = 1.004 \times 10^{-4} \) \[ 2 \times 10^{-8} \frac{d^2v}{dt^2} + 1.004 \times 10^{-4} \frac{dv}{dt} + v = 0 \] Divide both sides by \( 2 \times 10^{-8} \): \[ \frac{d^2v}{dt^2} + \frac{1.004 \times 10^{-4}}{2 \times 10^{-8}} \frac{dv}{dt} + \frac{1}{2 \times 10^{-8}} v = 0 \] \[ \frac{d^2v}{dt^2} + 5020 \frac{dv}{dt} + 5 \times 10^7 v = 0 \] --- ## **Step 3: Characteristic Equation** \[ s^2 + 5020 s + 5 \times 10^7 = 0 \] Solve for \( s \): \[ s = \frac{-5020 \pm \sqrt{5020^2 - 4 \times 5 \times 10^7}}{2} \] Calculate discriminant: \[ 5020^2 = 25,200,400 \] \[ 4 \times 5 \times 10^7 = 2 \times 10^8 = 200,000,000 \] \[ \text{Discriminant} = 25,200,400 - 200,000,000 = -174,799,600 \] \[ \sqrt{174,799,600} \approx 13,228 \] So the roots are complex: \[ s = \frac{-5020 \pm j13,228}{2} = -2510 \pm j6614 \] --- ## **Step 4: General Solution** \[ v(t) = e^{-2510 t} [A \cos(6614 t) + B \sin(6614 t)] \] --- ## **Step 5: Initial Conditions** - \( v(0^+) = 20V \) - Need \( \frac{dv}{dt}\big|_{t=0^+} \): Recall \( i(0^+) = C \frac{dv}{dt}\big|_{t=0^+} \). **At \( t = 0^+ \):** Apply KVL: \[ v_{R,\text{total}} + v_L + v_C = 0 \] \[ i(0^+) \cdot (1004) + 0.2 \cdot \frac{di}{dt}\bigg|_{t=0} + 20 = 0 \] But the inductor current **initially is zero** (since the loop was open before \( t=0 \)), so \( i(0^+) = 0 \). \[ i(0^+) = C \frac{dv}{dt}\bigg|_{t=0} \implies 0 = 0.1 \times 10^{-6} \frac{dv}{dt}\bigg|_{t=0} \implies \frac{dv}{dt}\bigg|_{t=0} = 0 \] --- ## **Step 6: Apply Initial Conditions** \[ v(0) = 20 = A \] \[ \frac{dv}{dt} = e^{-2510 t} [-2510 A \cos(6614 t) - 6614 A \sin(6614 t) - 2510 B \sin(6614 t) + 6614 B \cos(6614 t)] \] At \( t = 0 \): \[ \frac{dv}{dt}\bigg|_{t=0} = -2510A + 6614B = 0 \] \[ -2510 \times 20 + 6614 B = 0 \implies 6614 B = 50200 \implies B = \frac{50200}{6614} \approx 7.59 \] --- ## **Step 7: Final Solution** \[ \boxed{ v(t) = e^{-2510 t} \left[ 20 \cos(6614 t) + 7.59 \sin(6614 t) \right] } \] --- ## **Step 8: Final Value (as \( t\to\infty \))** As \( t\to\infty \), \( e^{-2510 t}\to 0 \): \[ v(\infty) = 0 \] --- ## **Step 9: Sketch** - At \( t=0 \), \( v = 20V \) - Decays to \( 0 \) with damped oscillation due to the complex roots. --- ## **Summary Table** | Step | Result | |------|--------| | **a. Initial Value** | \( v(0^+) = 20V \) | | **b. Differential Eqn** | \( \frac{d^2v}{dt^2} + 5020 \frac{dv}{dt} + 5 \times 10^7 v = 0 \) | | **Roots** | \( s = -2510 \pm j6614 \) | | **General Solution** | \( v(t) = e^{-2510 t} [A \cos(6614 t) + B \sin(6614 t)] \) | | **With ICs** | \( v(t) = e^{-2510 t} [20 \cos(6614 t) + 7.59 \sin(6614 t)] \) | | **Final Value** | \( v(\infty) = 0 \) | --- **If you need the sketch, let me know!**

Certainly! Here's a detailed explanation and guidance on sketching \( v(t) \): --- ### **Behavior of \( v(t) \):** The solution is: \[ v(t) = e^{-2510 t} \left[ 20 \cos(6614 t) + 7.59 \sin(6614 t) \right] \] - **Exponential decay** with a very rapid decay rate (~\( 2510 \text{ rad/sec} \)) - **Oscillations** at angular frequency \( \omega = 6614 \text{ rad/sec} \), which is very high, leading to rapid oscillations - **Initial value** at \( t=0 \): \( v(0) = 20V \) --- ### **Step-by-step for sketching:** 1. **Initial point:** - \( v(0) = 20V \) 2. **Oscillations:** - The sine and cosine terms oscillate rapidly (~\( 6614 \) rad/sec). - The period of oscillation: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{6614} \approx 0.00095\, \text{sec} \approx 0.95\, \text{ms} \] - So, about 1 ms per cycle. 3. **Decay:** - The amplitude decays exponentially with \( e^{-2510 t} \). - The **time constant**: \[ \tau = \frac{1}{2510} \approx 0.000398\, \text{sec} \approx 0.4\, \text{ms} \] - After about 1–2 ms, the voltage amplitude diminishes significantly. 4. **Shape:** - The waveform starts at 20V. - It oscillates rapidly with decreasing amplitude. - The envelope of decay shows the exponential envelope \( e^{-2510 t} \). --- ### **Sketch guidelines:** - **Plot time (x-axis):** from 0 to a few milliseconds (say, 5 ms). - **Plot voltage (y-axis):** from -20V to +20V (initial amplitude). - **Draw the initial point at (0, 20V).** - **Oscillations:** - Rapid sine wave oscillations at ~1 ms period. - The peaks diminish exponentially. - **Envelope:** - Draw an exponential decay curve from 20V down to near 0V as \( t \to \infty \). - **Waveform:** - Rapid oscillations within the envelope, diminishing over time. --- ### **Visual summary:** - The waveform starts at 20V. - Oscillates rapidly with decreasing amplitude. - Becomes negligible after a few milliseconds. - Approaches zero as \( t \to \infty \). --- ### **Note:** Given the very high frequency (~6614 rad/sec), the waveform appears as a rapidly damped oscillation, almost like a "ringing" that dies out quickly. --- **Would you like a hand-drawn sketch or a plotted graph?**

Certainly! Here's the plotted graph of \( v(t) \) based on the derived expression: \[ v(t) = e^{-2510 t} \left[ 20 \cos(6614 t) + 7.59 \sin(6614 t) \right] \] **Key features:** - **Initial value:** 20 V at \( t=0 \) - **Oscillation frequency:** approximately 6614 rad/sec (~1053 Hz), so period ~0.95 ms - **Decay rate:** exponential decay with time constant ~0.4 ms - **Behavior:** rapid oscillations with diminishing amplitude, approaching zero as \( t \to \infty \) --- ### **Graph Description:** - The waveform starts at 20 V. - It oscillates very quickly, crossing zero multiple times within a few milliseconds. - The amplitude of oscillations decreases exponentially, with the envelope dropping sharply. - After about 3–4 ms, the voltage approaches zero. --- ### **Sample Plot (Approximate):** ```plaintext Voltage (V) 20 |* * * * | \ / \ / \ / \ | \ / \ / \ / \ | \____/ \____/ \____/ \____ |------------------------------------------ 0 1ms 2ms 3ms 4ms 5ms ``` *Note:* The actual plot would show rapid oscillations with decreasing amplitude, more dense than the sketch above. --- **If you'd like, I can generate a precise plot using graphing tools or code.**