Question

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. \(\mathrm{I}=\mathrm{E}_{0}\omega \mathrm{csin}(\omega \mathrm{t})\)
• B. \(I=E_{0}\omega c\mathrm{sin}(\omega t+\frac{\pi}{2})\)
• C. \(\mathrm{I}=\frac{\mathrm{E}_{0}}{\omega \mathrm{c}}\mathrm{sin}(\omega \mathrm{t})\)
• D. \(\mathrm{I}=\frac{\mathrm{E}_{0}}{\omega \mathrm{c}}\mathrm{sin}(\omega \mathrm{t}+\frac{\pi}{2})\)

Hint:

In a pure capacitor the current leads the voltage by 90 degrees and the amplitude is E0 times omega C.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Charge on the capacitor is \(q=CE=CE_0\sin\omega t\).

Step 2: Current
\(I=\dfrac{dq}{dt}=CE_0\omega\cos\omega t\).

Step 3: Phase form
\(\cos\omega t=\sin\left(\omega t+\dfrac\pi2\right)\), so \(I=E_0\omega C\sin\left(\omega t+\dfrac\pi2\right)\).

Final Answer:
The current is \(I=E_0\omega C\sin(\omega t+\pi/2)\). \[ \boxed{\text{(B) }I=E_{0}\omega c\sin\left(\omega t+\dfrac{\pi}{2}\right)} \]