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## Q. 12.16

The voltage across a capacitor having capacitance C=100 pF  is given by $\nu(t)=V_0 \cos (2 \pi f t)$  with $\nu_0=5 V$  and f=500 kHz . Find the current  i(t)  through the capacitor and draw a phasor diagram depicting the current and voltage, with the voltage as the phase reference.

## Verified Solution

The phasor $\tilde{V}$ for the voltage and the admittance Y  of the capacitor are

\begin{aligned} &\tilde{V}=V_0 \angle 0=5 \angle 0 V, \\ &Y=j \omega C=j 314 \mu S =314 \mu S \angle(\pi / 2). \end{aligned}

From (12.24), the phasor for the current is

\begin{aligned} \tilde{I} &=Y \tilde{V}=(5 V \angle 0)[314 \mu S \angle(\pi / 2)] \\ &=1.57 mA \angle(\pi / 2). \end{aligned}

It follows that

$i(t)=1.57 \cos (2 \pi f t+\pi / 2) mA .$

Figure  12.10  shows a phasor diagram for the current and voltage, where the voltage is taken as the reference. The current leads the voltage by $\pi / 2$ . Equivalently, the voltage lags the current by $\pi / 2$.