Certainly! Here is the solution from the image rewritten in a structured and organized way: --- ### **Given:** - **EMF:** \( \varepsilon(t) = 60\,V \) - **Inductance:** \( L = 40\,H \) - **Resistance:** \( R = 12\,\Omega \) - **Initial Condition:** \( i(0) = 0 \) --- ### **Step 1: Write the KVL Equation for the RL Circuit** \[ \varepsilon = i(t)R + L\frac{di(t)}{dt} \] Substitute the given values: \[ 60 = 12i(t) + 40\frac{di(t)}{dt} \] Divide the equation by 4: \[ \frac{di(t)}{dt} + 3i(t) = 15 \] --- ### **Step 2: Solve the Differential Equation** This is a first-order linear differential equation. The integrating factor (\(\mu(t)\)) is: \[ \mu(t) = e^{\int 3 dt} = e^{3t} \] Multiply both sides by the integrating factor: \[ e^{3t}\frac{di}{dt} + 3e^{3t}i = 15e^{3t} \] This simplifies to: \[ \frac{d}{dt}(i(t)e^{3t}) = 15e^{3t} \] --- ### **Step 3: Integrate Both Sides** \[ i(t)e^{3t} = \int 15e^{3t}dt \] \[ i(t)e^{3t} = 5e^{3t} + C \] \[ i(t) = 5 + Ce^{-3t} \] --- ### **Step 4: Apply Initial Condition** Given \(i(0) = 0\): \[ 0 = 5 + C \implies C = -5 \] So, \[ \boxed{i(t) = 5\left(1 - e^{-3t}\right) \ \text{A}} \] --- **Final Answer:** \[ \boxed{i(t) = 5\left(1 - e^{-3t}\right) \ \text{A}} \] ---