Question

In an a.c. circuit with pure capacitance 'C' and a.c. source \(E = E_0sinωt\), the equation of instantaneous current is given by

• A. \(I = E_0\,ωC\cdot sin(ωt)\)
• B. \(I = E_0ωCsin(ωt+\frac{π}{2})\)
• C. \(I = \frac{E_0}{ωC}sin(ωt)\)
• D. \(I = \frac{E_0}{ωC}sin(ωt+\frac{π}{2})\)

Hint:

Resonant frequency is 1 over 2 pi root LC.

Answer 1

The Correct Option is B

Step 1: Original frequency:
\(f = \frac{1}{2\pi\sqrt{LC}}\).

Step 2: New values:
L is increased by 2L, so the new inductance is \(L + 2L = 3L\), and C becomes 9C. Then
\[ f' = \frac{1}{2\pi\sqrt{3L\times9C}} = \frac{1}{2\pi\sqrt{27LC}} = \frac{1}{3\sqrt3}\cdot\frac{1}{2\pi\sqrt{LC}} \]

Step 3: Result:
\(f' = \frac{f}{3\sqrt3}\).

Final Answer:
The new resonant frequency is \(\frac{f}{3\sqrt3}\), option (D). \[ \boxed{\frac{f}{3\sqrt{3}}} \]