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*10^(-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

Step-by-step explanation:

The peak current is,

`\begin{gathered} I=(V)/(X_C) \\ =(V)/((1)/(2\pi fC)) \\ =2\pi fCV \end{gathered}`

Substituting the values we get,

`\begin{gathered} I=2\pi*88*6.95*10^(-6)*99.41 \\ =0.3820 \\ =382*10^(-3) \\ =382\text{ mA} \end{gathered}`

Hence, the peak current is 382 mA.