Question

A capacitor of capacitive reactance, $12\Omega$ is connected with an AC source having voltage, $V=3\sin (\omega t+\pi /6)$. Find the expression of instantaneous current in the circuit.

• A. $0.35\sin (\omega t+2\pi /3)$
• B. $0.25\sin (\omega t+2\pi /3)$
• C. $0.57\sin (\omega t-2\pi /3)$
• D. $0.15\sin (\omega t-2\pi /3)$

Answer 1

The correct option is B $0.25\sin (\omega t+2\pi /3)$

$I_0=\frac{V_0}{X_C}=\frac{3}{12}=0.25\text{A}$

For a purely capacitive circuit, current leads ahead the voltage by $\pi /2$.

Therefore, instantaneous current in the circuit,

$I=0.25\sin (\omega t+\pi /6+\pi /2)$

$\Rightarrow I=0.25\sin (\omega t+2\pi /3)$

Hence, option $\left(B\right)$ is the correct answer.