An inductor of inductance 2 μ H is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit 1/xF, where the value of x is ___________.
(Take π=22/7)

Solution

Physics sir · 4 steps

Maximum current means resonance, so set the two reactances equal and solve for C.

1. Resonance condition. In a series LCR circuit the current is largest when the impedance is smallest. That happens when X_L = X_C, so only R is left.

   ω L = 1/(ω C) ⇒ C = 1/(ω² L)

2. Angular frequency. ω = 2π f = 2 × 22/7 × 7000 = 44000 rad s⁻¹, so ω² = 1.936 × 10⁹ s⁻².
3. Put in numbers. ω² L = 1.936 × 10⁹ × 2 × 10⁻⁶ = 3872. So C = 1/3872 F.
4. So. Comparing with 1/x F gives x = 3872.

The common slip: 98. Using the frequency f = 7000 in place of ω = 2π f gives f² L = 98, not 3872.

The idea to keep. Whenever a question asks for maximum current in a series LCR circuit, think resonance: ω² LC = 1. Always use ω = 2π f, not f.

Answer from the NTA final answer key. Solution by PrepWiser.