The voltage across a capacitor having capacitance C=100 pF is given by [latex] \nu(t)=V_0 \cos (2 \pi f t) [/latex] with [latex] \nu_0=5 V [/latex] 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.

The phasor [latex] \tilde{V} [/latex] for the voltage and the admittance Y of the capacitor are [latex] \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} [/latex] From (12.24), the phasor for the current is [latex] \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} [/latex] It follows that [latex]i(t)=1.57 \cos (2 \pi f t+\pi / 2) mA .[/latex] 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 [latex] \pi / 2 [/latex] . Equivalently, the voltage lags the current by [latex]\pi / 2[/latex].

Question 12.16: The voltage across a capacitor having capacitance C=100 pF ...

Introduction to Circuit Analysis and Design

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.

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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 $\pi / 2$ .