Question
An $A C$ source generating a voltage $e=e 0$ sinwt is connected to a capacitor of capacitance $C$. Find the expression for the current i flowing through it. Plot a graph of e and i versus wt.
✓
Answer
Figure 13.12 shows an $A C$ source, generating a voltage $e=e_0 \sin \omega t$, connected to a capacitor of capacitance $C$. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the
capacitor is alternately charged and discharged. If $q$ is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is $V =\frac{q}{C} \therefore q = CV$. $q$ and $V$ are functions of time, with $V=e=e_0 \sin \omega t$. The instantaneous current in the circuit is $i =\frac{d q}{d t}=\frac{d}{d t}( CV )= C \frac{d v}{d t}= C \frac{d}{d t}\left( e _0 \sin \omega t \right)=\omega C e _0 \cos \omega t$ $\therefore i =\frac{e_0}{(1 / \omega C)} \sin \left(\omega t+\frac{\pi}{2}\right)=i_0 \sin \left(\omega t+\frac{\pi}{2}\right)$ where $i_0=\frac{e_0}{(1 / \omega C)}$ is the peak value of the current.
Table gives the values of $e$ and $i$ for different values of cot and Fig shows graphs of $e$ and $i$ versus $w$ t. i leads e by phase angle of $\frac{\pi}{2}$ rad.
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