Question

Let $f=50\text{Hz}$, and $C=100\mu F$ in an AC circuit containing a capacitor only. If the peak value of the current in the circuit is 1.57 A at t = 0. The expression for the instantaneous voltage across the capacitor will be

• A. $C=50\sin (100\pi t–\frac{\pi }{2})$
• B. $C=100\sin \left(50\pi t\right)$
• C. $C=50\sin 100\pi t$
• D. $C=50\sin (100\pi t+\frac{\pi }{2})$

Answer 1

The correct option is A $C=50\sin (100\pi t–\frac{\pi }{2})$

Given,
$f=50\text{Hz}$
$C=100\mu F$
$I_0=1.57A$
$\omega =2\pi f$
$X_c=\frac{1}{\omega C}$
Then $V_c=I_0{X}_c\sin (\omega t-\frac{\pi }{2})$
$=1.57\times \frac{1}{2\pi 50\times 100\times {10}^{-6}}\sin (100\pi t-\frac{\pi }{2})$
$=50\sin (100\pi t-\frac{\pi }{2})$