Question

Let an $AC$ circuit contains a capacitor of capacitance $C=100\mu \text{F}$ only. If the instantaneous current in the circuit is $i=1.57\sin 100\pi t$. The expression for the instantaneous voltage, $E$ at this instantacross the capacitor will be

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

Answer 1

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

Since, it is a purely capacitive circuit, so voltage will lag current by $\pi /2$ i.e.,

$\therefore E=E_0\sin (\omega t-\pi /2)$

Given, peak current in the circuit, $i_o=1.57\text{A}$

Now, capacitive reactance

$X_C=\frac{1}{\omega C}=\frac{1}{100\pi \times 100\times {10}^{-6}}$

$=\frac{1}{\pi \times {10}^{-2}}=\frac{100}{\pi }\Omega$

So, peak volatge, $E_0=i_0\times X_C=1.57\times \frac{100}{3.14}=50\text{V}$

$\therefore E=50\sin (100t-\pi /2)$

Hence, option $\left(A\right)$ is correct.