Electric Circuits 1

Fig.1. RC-circuit.

 

Problem 1.

Determine $I(t)$ for
(a) a constant electromotive force,
(b) a periodic electromotive force,
if $I(0):=I_0$ is given.

Show Solution

From Kirchhoff's voltage law

we get $$RI(t)+\frac{1}{C}\int I(t)\,dt=E(t).$$ Differentiating the equation with respect to $t$ we find $$RI'(t)+\frac{1}{C}I(t)=E'(t).$$

Using the general formula for solution

we obtain $$I(t)=e^{-t/(RC)}\left(\frac{1}{R}\int e^{t/(RC)}E'(t)\,dt+c \right).$$ Since $I(0)=I_0$, we get $c=I_0$, and $$\bbox[lightblue,5px,border:2px solid red]{\color{#800000}{ \bf{I(t)=e^{-t/(RC)}\left(\frac{1}{R}\int_0^t e^{\tau/(RC)}E'(\tau)\,d\tau+I_0 \right).}}}$$ Case (a). Constant electromotive force.
If $E(t)$ is constant, then it follows simply $$I(t)=I_0 e^{-t/(RC)}.$$ Case (b). Periodic electromotive force $E(t)=E_0\sin(\omega t)$.
In this case $E'(t)=E_0\omega\cos(\omega t)$ and $$\int_0^t e^{\tau/(RC)}E'(\tau)\,d\tau=E_0\omega \int_0^t e^{\tau/(RC)}\cos(\omega \tau)\,d\tau.$$

Applying integration by parts

we obtain $$\begin{align*} \frac{E_0\,\omega}{R}\! \int_0^t \! e^{\tau/(RC)}\cos(\omega\, \tau)\,d\tau\! &= e^{t/(RC)} E_0\,\omega\,C \frac {CR\,\omega\sin ( \omega\,t) +\cos ( \omega\,t ) }{C^2R^2\omega^2\!+\!1} \\ &-\frac{E_0\,\omega\,C}{C^2R^2\omega^2\!+\!1}. \end{align*}$$ It follows $$I(t) =E_0\,\omega\,C \frac {CR\,\omega\sin ( \omega\,t) +\cos ( \omega\,t ) }{C^2R^2\omega^2\!+\!1} +e^{-t/(RC)}\left(I_0- \frac{E_0\,\omega\,C}{C^2R^2\omega^2\!+\!1} \right).$$ Here we can transform the first term further if we use that $$\frac{a\sin(x)+b\cos(x)}{\sqrt{a^2+b^2}}=\sin(x+\delta),\qquad(a,b\gt 0),$$ where $\tan(\delta)=\frac{b}{a}$.
Hence $$I(t)=\frac{E_0\,\omega\,C}{\sqrt{C^2R^2\omega^2\!+\!1}} \sin(\omega t+\delta)+e^{-t/(RC)}\left(I_0- \frac{E_0\,\omega\,C}{C^2R^2\omega^2\!+\!1} \right),$$ where $\tan(\delta)=\frac{1}{CR\,\omega}$.