Question
A $15.0 \mu F$ capacitor is connected to a $220\ V, 50\ Hz$ source. Find the capacitive reactance and the current (rms and peak)
in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.

Answer

Data $: C =15 \mu F =15 \times 10-6 F , V _{ rms }=220 V , f =50 Hz$
The capacitive reactance $=\frac{1}{2 \pi f C}$
$ =\frac{1}{2(3.142)(50)\left(15 \times 10^{-6}\right)}=\frac{100 \times 100}{(3.142)(15)}$
$=212.2 \Omega$
$i_{ rms }=\frac{V_{ rms }}{\text { capacitive reactance }}=\frac{220}{212.2}$
$=1.037 A$
$i_{\text {peak }}=i_{\text {rms }} \sqrt{2}=(1.037)(1.414)=1.466 A$
If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

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