Q143 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
In an a.c. circuit with pure capacitance 'C' and a.c. source $E = E_0\sin\omega t$, the equation of instantaneous current is given by
- A$I = E_0\,\omega C \cdot \sin(\omega t)$
- Correct answerB$I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$
- C$I = \dfrac{E_0}{\omega C}\sin(\omega t)$
- D$I = \dfrac{E_0}{\omega C}\sin\left(\omega t + \dfrac{\pi}{2}\right)$
Solution
The instantaneous voltage across the capacitor is given by $E = E_0\sin\omega t$.
The charge on the capacitor at any instant is $q = CE = CE_0\sin\omega t$.
The instantaneous current $I$ is the rate of flow of charge, given by $I = \dfrac{dq}{dt}$.
Differentiating $q$ with respect to $t$, we get:
$I = \dfrac{d}{dt}(CE_0\sin\omega t) = E_0\omega C\cos\omega t$
Using the trigonometric identity $\cos\theta = \sin\left(\theta + \dfrac{\pi}{2}\right)$, the current can be written as:
$I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$
Answer: $I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$
The charge on the capacitor at any instant is $q = CE = CE_0\sin\omega t$.
The instantaneous current $I$ is the rate of flow of charge, given by $I = \dfrac{dq}{dt}$.
Differentiating $q$ with respect to $t$, we get:
$I = \dfrac{d}{dt}(CE_0\sin\omega t) = E_0\omega C\cos\omega t$
Using the trigonometric identity $\cos\theta = \sin\left(\theta + \dfrac{\pi}{2}\right)$, the current can be written as:
$I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$
Answer: $I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$
Hence, the correct answer is option: 'B'.