Alternating Current - Result Question 1
1. A $40 \mu F$ capacitor is connected to a $200 V, 50 Hz$ ac supply. The rms value of the current in the circuit is, nearly:
[2020]
(a) $2.05 A$
(b) $2.5 A$
(c) $25.1 A$
(d) $1.7 A$
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Solution:
- (b) Given :
Capacitance, $C=40 \mu F=40 \times 10^{-6} F$
Frequency, $f=50 Hz$
$\therefore \omega=2 \pi f=100 \pi$
$\varepsilon _{\text{rms }}=200 V$
$\therefore I _{\text{rms }}=\frac{\varepsilon _{\text{rms }}}{X_C}=\frac{\varepsilon _{\text{rms }}}{\frac{1}{C \omega}}$
$ =200 \times 40 \times 10^{-6} \times 2 \pi \times 50=2.5 A \text{. } $
