1. **Problem statement:** Find the current $I(t)$ in amperes at time $t$ seconds after the switch is closed in an RL circuit with voltage $E=60$, inductance $L=1$, resistance $R=30$, and initial current $I(0)=0$. 2. **Formula and rules:** The differential equation for the current in an RL circuit is: $$L\frac{dI}{dt} + RI = E$$ where $L$ is inductance, $R$ is resistance, $E$ is constant voltage. 3. **Solve the differential equation:** Rewrite: $$\frac{dI}{dt} + \frac{R}{L}I = \frac{E}{L}$$ Substitute values: $$\frac{dI}{dt} + 30I = 60$$ 4. **Find integrating factor:** $$\mu(t) = e^{\int 30 dt} = e^{30t}$$ Multiply both sides: $$e^{30t}\frac{dI}{dt} + 30e^{30t}I = 60e^{30t}$$ This is: $$\frac{d}{dt}(Ie^{30t}) = 60e^{30t}$$ 5. **Integrate both sides:** $$Ie^{30t} = \int 60e^{30t} dt = 60 \cdot \frac{e^{30t}}{30} + C = 2e^{30t} + C$$ 6. **Solve for $I(t)$:** $$I = \frac{2e^{30t} + C}{e^{30t}} = 2 + Ce^{-30t}$$ 7. **Apply initial condition $I(0)=0$:** $$0 = 2 + C e^{0} = 2 + C \Rightarrow C = -2$$ 8. **Final solution:** $$I(t) = 2 - 2e^{-30t} = 2(1 - e^{-30t})$$ This means the current starts at 0 and approaches 2 amperes as $t$ increases.