An ac source of peak voltage 220 V and frequency 50 Hz is connected across a capacitor, and the peak current is 2 A. The capacitance of the capacitor is about
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 30\,\mu\mathrm{F} \)
In a purely capacitive ac circuit, the capacitive reactance limits the current amplitude, \( i_m=v_m/X_C \). Here \( X_C=1/\omega C \) and \( \omega=2\pi\nu \).
The capacitive reactance is:
\( \begin{aligned} X_C &= \frac{v_m}{i_m} \\ &= \frac{220\,\mathrm{V}}{2\,\mathrm{A}} \\ &= 110\,\Omega \end{aligned} \)
The capacitance is then:
\( \begin{aligned} C &= \frac{1}{2\pi\nu X_C} \\ &= \frac{1}{2\times3.14\times50\,\mathrm{Hz}\times110\,\Omega} \\ &\approx 2.9\times10^{-5}\,\mathrm{F} \end{aligned} \)
That is about \( 29\,\mu\mathrm{F} \), or nearly \( 30\,\mu\mathrm{F} \).