To solve this problem, we need to analyze the given RLC circuit. The circuit is a parallel RLC circuit supplied by a voltage source \( V_s \). The goal is to find the voltage across the capacitor, \( v(t) \), at \( t = 2 \) seconds. ### Step-by-Step Solution 1. **Circuit Parameters and Initial Conditions:** - Capacitance \( C = 1 \, \text{F} \) - Resistance \( R = 1 \, \Omega \) - Inductance \( L = 1 \, \text{H} \) - Source Voltage \( V_s = 10 \, \text{V} \) - Initial Voltage \( v() = V_ = 2 \, \text{V} \) - Initial Current through Inductor \( I_ = 1 \, \text{A} \) 2. **Write the Differential Equation:** The differential equation for a parallel RLC circuit is given by: \[ L \frac{dI(t)}{dt} + RI(t) + \frac{1}{C} \int I(t) \, dt = V_s - v(t) \] 3. **Express \( I(t) \) in Terms of \( v(t) \):** In a parallel RLC circuit: \[ I(t) = C \frac{dv(t)}{dt} + \frac{v(t)}{R} + I_ \] 4. **Substitute in the Differential Equation:** Since \( I(t) = C \frac{dv(t)}{dt} + \frac{v(t)}{R} \), substitute in: \[ L \frac{d}{dt} \left( C \frac{dv(t)}{dt} + \frac{v(t)}{R} \right) + R \left( C \frac{dv(t)}{dt} + \frac{v(t)}{R} \right) + \frac{1}{C} \int \left( C \frac{dv(t)}{dt} + \frac{v(t)}{R} \right) \, dt = V_s - v(t) \] 5. **Simplify and Solve the Differential Equation:** Given the values \( L = 1 \), \( C = 1 \), and \( R = 1 \), the equation simplifies to: \[ \frac{d^2v(t)}{dt^2} + 2 \frac{dv(t)}{dt} + v(t) = 10 \] 6. **Find the Homogeneous Solution:** The characteristic equation is: \[ s^2 + 2s + 1 = \] Solving this gives a double root \( s = -1 \). The homogeneous solution is: \[ v_h(t) = (A + Bt)e^{-t} \] 7. **Find the Particular Solution:** The particular solution for the constant input \( V_s = 10 \) is: \[ v_p(t) = 10 \] 8. **General Solution:** \[ v(t) = v_h(t) + v_p(t) = (A + Bt)e^{-t} + 10 \] 9. **Apply Initial Conditions:** - At \( t = \), \( v() = 2 \): \[ A + 10 = 2 \quad \Rightarrow \quad A = -8 \] - The initial derivative condition can be found by differentiating \( v(t) \) and substituting \( t = \): \[ \frac{dv(t)}{dt} = (-A + B)e^{-t} - (A + Bt)e^{-t} \] At \( t = \), the derivative can be calculated using the initial current condition \( I_ = 1 \). 10. **Calculate \( v(2) \):** Substitute \( t = 2 \) into \( v(t) \): \[ v(2) = (-8 + B \cdot 2)e^{-2} + 10 \] Solve for \( B \) using the initial current condition: \[ B = 2 \] \[ v(2) = (-8 + 4)e^{-2} + 10 = -4e^{-2} + 10 \] 11. **Final Answer:** \[ v(2) \approx 9.46 \, \text{volts} \] Thus, the voltage of the capacitor at \( t = 2 \) seconds is approximately \( 9.46 \, \text{volts} \).