Question

Let frequency ν = 50 Hz, and capacitance C = 100μF in an ac circuit containing a capacitor only. If the peak value of the current in the circuit is 1.57 A. The expression for the instantaneous voltage across the capacitor will be

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

Answer 1

Answer: (A)

E = 50 sin (100 πt – $\frac{\pi}{2}$)

Answer 2

Peak value of voltage $V_{0} = i_{0}X_{C} = \frac{i_{0}}{2\pi\nu C}$ ⇒

$\frac{1.57}{2 \times 3.14 \times 50 \times 100 \times 10^{- 6}} = 50V$Hence if equation of current $i = i_{0}\sin\omega t$ then in capacitive circuit voltage is $V = V_{0}\sin\left( \omega t - \frac{\pi}{2} \right)$

⇒$V = 50\left( \sin 2\pi \times 50t - \frac{\pi}{2} \right) = 50\sin\left( 100\pi t - \frac{\pi}{2} \right)$