Question

A 6.95 uF capacitor is connected across an AC generator that produces a peak voltage of 99.41 V. What is the peak current, in milliamps, through the capacitor if the emf frequency is 88 Hz?

Answer 1

Given:

The capacitance is,

$\begin{gathered} C=6.95\text{ }\mu F \\ =6.95\times10^{-6}\text{ F} \end{gathered}$

The peak voltage is,

$V=99.41\text{ V}$

The frequency of the emf is,

$f=88\text{ Hz}$

To find:

The peak current

Explanation:

The peak current is,

$\begin{gathered} I=\frac{V}{X_C} \\ =\frac{V}{\frac{1}{2\pi fC}} \\ =2\pi fCV \end{gathered}$

Substituting the values we get,

$\begin{gathered} I=2\pi\times88\times6.95\times10^{-6}\times99.41 \\ =0.3820 \\ =382\times10^{-3} \\ =382\text{ mA} \end{gathered}$

Hence, the peak current is 382 mA.