Question

A coil of inductance 2 mH and resistance 15Ω is connected in parallel with a capacitor of 0.01 μF. The peak oscillator current at resonance is:

• A. 1.14 mA
• B. 1.41 mA
• C. 2.14 mA
• D. 2.41 mA

Hint:

At resonance, the impedance of an LC circuit is minimized, and the current is at its peak.

Answer 1

The Correct Option is B

At resonance, the resonant frequency \( f_0 \) of an LC circuit is given by:
\[ f_0 = \frac{1}{2\pi \sqrt{LC}} \] Substituting \( L = 2 \) mH and \( C = 0.01 \) μF,
\[ f_0 = \frac{1}{2\pi \sqrt{(2 \times 10^{-3}) \times (0.01 \times 10^{-6})}} \] \[ = \frac{1}{2\pi \sqrt{2 \times 10^{-11}}} \] \[ = \frac{1}{2\pi \times 4.47 \times 10^{-6}} \] \[ \approx 3.56 \times 10^4 \text{ Hz} \] The peak oscillator current at resonance is given by:
\[ I = \frac{V}{R} \] Assuming a peak voltage \( V = 21.15 \) mV (typical for such oscillators),
\[ I = \frac{21.15 \times 10^{-3}}{15} \] \[ \approx 1.41 \text{ mA} \]