Question

An inductor of inductance \(2 μ\text{H}\) is connected in series with a resistance, a variable capacitor and an a.c. source of 10 kHz. The value of capacitance for which maximum current is drawn in to the circuit is \(\frac{1}{x} \text{F}\), where the value of x is (Take \(π^2 = 10\))

• A. \(8000\)
• B. \(600\)
• C. \(400\)
• D. \(1600\)

Hint:

Maximum current needs resonance, so the frequency equals 1 over 2 pi root LC.

Answer 1

The Correct Option is A

Step 1: Resonance
Maximum current occurs when \(X_L = X_C\), i.e. \(f = \frac{1}{2\pi\sqrt{LC}}\).

Step 2: Solve for C
\(C = \frac{1}{4\pi^2f^2L}\).

Step 3: Numbers
With \(\pi^2 = 10\), \(f = 10^4\) Hz and \(L = 2\times10^{-6}\) H: \(4\pi^2f^2L = 40\times10^8\times2\times10^{-6} = 8000\).

Step 4: Result
\(C = \frac1{8000}\ \text{F}\), so \(x = 8000\). Option (A).

Final Answer:
x is 8000. \[ \boxed{\text{(A)}\ 8000} \]