Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for V=5 sin (100t) The values of L and R are shown in the figure. The capacitance of the capacitor ( C ) used is ____ μF.

Solution

Physics sir · 4 steps

Check V₀/R first: it gives exactly 50 mA, so the circuit is at resonance at ω = 100.

1. Spot resonance. The peak current is I₀ = V₀/Z. With only R in the way, I₀ = 5/100 = 0.05 A = 50 mA. That is the measured maximum, so Z = R and X_L = X_C.
2. Read the frequency. From V = 5 sin(100t), the angular frequency is ω = 100 rad s⁻¹. So the circuit resonates at this ω.

   ω L = 1/(ω C) ⇒ C = 1/(ω² L) = 1/((100)² × 2)

3. Put in numbers. C = 1/(2 × 10⁴) = 5 × 10⁻⁵ F.
4. So. C = 5 × 10⁻⁵ F = 50 μF.

The common slip: 1.27. Treating the 100 in sin(100t) as the frequency f and using ω = 2π × 100 gives about 1.27 μF. The number multiplying t is already ω.

The idea to keep. If the peak current equals V₀/R, the reactances have cancelled. That is resonance, so use ω² = 1/LC with the ω read straight off the source.

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