In a circuit with resistance R and self-inductance L , a…
In a circuit with resistance $R$ and self-inductance $L$, a periodic electromotive force $E_1=a\sin\dfrac{2\pi}{T}t$ acts. Determine the current $i$ at every moment if $i=0$ when $t=0$.
Explanation of Method
With $\omega=2\pi/T$, the equation $L\,di/dt+Ri=a\sin(\omega t)$ is linear; integrate using the standard formula for $\int e^{kt}\sin(\omega t)\,dt$.
Step 1: Apply the integrating factor
With $\mu(t)=e^{Rt/L}$: $\dfrac{d}{dt}\!\left(e^{Rt/L}i\right)=\dfrac{a}{L}e^{Rt/L}\sin(\omega t)$.
Step 2: Integrate
Using $\displaystyle\int e^{kt}\sin(\omega t)\,dt=\dfrac{e^{kt}}{k^2+\omega^2}[k\sin(\omega t)-\omega\cos(\omega t)]$ with $k=R/L$:
\[i=\frac{a}{R^2+L^2\omega^2}\big[R\sin(\omega t)-L\omega\cos(\omega t)\big]+Ke^{-Rt/L}.\]
Step 3: Apply the initial condition
At $t=0$: $0=\dfrac{-aL\omega}{R^2+L^2\omega^2}+K\Rightarrow K=\dfrac{aL\omega}{R^2+L^2\omega^2}$.
Result
Final answer:
\[\boxed{\,i(t)=\frac{a}{R^2+L^2\omega^2}\Big[R\sin(\omega t)-L\omega\cos(\omega t)+L\omega e^{-Rt/L}\Big],\qquad \omega=\frac{2\pi}{T}.}\]
The current is a decaying transient plus a steady-state sinusoidal response with amplitude $a/\sqrt{R^2+L^2\omega^2}$ and phase lag $\phi=\arctan(L\omega/R)$.