Let's solve the problem step by step as requested: ### **Given Data:** - Voltage: \( E(t) = 40\sin(60t) \) volts - Inductance: \( L = 1 \) H - Resistance: \( R = 20 \) Ω - Initial current: \( I(0) = \frac{4}{5} \) A We are to solve for \( I(t) \). --- ## **Step 1: Find the differential equation and solve for \( I(t) \)** The circuit is an RL circuit, so by **Kirchhoff's law:** \[ E(t) = L \frac{dI}{dt} + R I \] Plug in the given values: \[ 40\sin(60t) = 1 \cdot \frac{dI}{dt} + 20I \] \[ \frac{dI}{dt} + 20I = 40\sin(60t) \] This is a first-order linear ODE. The integrating factor is \( e^{20t} \). **Solution Approach:** 1. **Homogeneous Solution:** \( I_h = Ce^{-20t} \) 2. **Particular Solution:** Assume \( I_p = A\sin(60t) + B\cos(60t) \) - Plug into the ODE and solve for \( A \) and \( B \). Calculate derivatives: \[ \frac{dI_p}{dt} = 60A\cos(60t) - 60B\sin(60t) \] Insert into ODE: \[ [60A\cos(60t) - 60B\sin(60t)] + 20[A\sin(60t) + B\cos(60t)] = 40\sin(60t) \] Group like terms: - Coefficient of \( \sin(60t) \): \( 20A - 60B \) - Coefficient of \( \cos(60t) \): \( 60A + 20B \) Set equal to \( 40\sin(60t) \): \[ 20A - 60B = 40 \] \[ 60A + 20B = 0 \] Solve the system: - From the second equation: \( 60A + 20B = 0 \implies A = -\frac{1}{3}B \) - Plug into the first: \( 20(-\frac{1}{3}B) - 60B = 40 \implies -\frac{20}{3}B - 60B = 40 \) \[ -\frac{20}{3}B - 60B = 40 \] \[ -\frac{20 + 180}{3}B = 40 \] \[ -\frac{200}{3}B = 40 \implies B = -\frac{3 \times 40}{200} = -\frac{120}{200} = -0.6 \] So, \( A = -\frac{1}{3}B = -\frac{1}{3}(-0.6) = 0.2 \) So the particular solution is: \[ I_p = 0.2\sin(60t) - 0.6\cos(60t) \] **General solution:** \[ I(t) = Ce^{-20t} + 0.2\sin(60t) - 0.6\cos(60t) \] **Use initial condition \( I(0) = \frac{4}{5} = 0.8 \):** \[ I(0) = C + 0.2(0) - 0.6(1) = C - 0.6 = 0.8 \implies C = 1.4 \] **Final expression:** \[ \boxed{I(t) = 1.4e^{-20t} + 0.2\sin(60t) - 0.6\cos(60t)} \] *(End of Step 1 Explanation: We derived the current equation using the ODE for the RL circuit, solved for constants, and applied the initial condition.)* --- ## **Step 2: Find the current at \( t = 0.2 \) seconds** Plug \( t = 0.2 \) into the equation: \[ I(0.2) = 1.4e^{-20 \times 0.2} + 0.2\sin(60 \times 0.2) - 0.6\cos(60 \times 0.2) \] Calculate values: - \( e^{-4} \approx 0.0183 \) - \( 60 \times 0.2 = 12 \) radians - \( \sin(12) \approx -0.5366 \) - \( \cos(12) \approx 0.8439 \) So: \[ I(0.2) = 1.4 \times 0.0183 + 0.2 \times (-0.5366) - 0.6 \times 0.8439 \] \[ = 0.0256 - 0.1073 - 0.5063 \] \[ = 0.0256 - 0.6136 \] \[ = -0.588 \] **Rounded to two decimal places:** \[ \boxed{I(0.2) \approx -0.59 \text{ A}} \] *(End of Step 2 Explanation: We substituted \( t = 0.2 \) into the function and evaluated each term to find the current at that instant.)* --- ## **Final Answers:** (a) \( I(t) = 1.4e^{-20t} + 0.2\sin(60t) - 0.6\cos(60t) \) A (b) \( I(0.2) \approx -0.59 \) A