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Find R, L, C of series resonant circuit

Short answer

At resonance total resistance = 100/0.7 ≈ 142.86 Ω, so coil resistance R = 142.86 − 50 = 92.86 Ω. Capacitive reactance = 200/0.7 ≈ 285.71 Ω, giving C = 1/(2π·200·285.71) ≈ 2.785 μF. Inductive reactance equals that, so L = 285.71/(2π·200) ≈ 0.227 H.

A 50 Ω resistor is connected in series with a coil having resistance R and inductance L, a capacitor "C" and a 100 V variable frequency supply as shown in the figure below. At a frequency of 200 Hz, the maximum current of 0.7 Amp flows through the circuit and the voltage across the capacitor is 200 V. Determine the value of R, L, and C. [Figure: 100 V variable-frequency source in a single series loop with a 50 Ω resistor, a coil (R in series with L) and a capacitor C]

Assumptions: the supply is 100 V (rms), the 50 Ω resistor is in series with a coil (R, L) and a capacitor C. "Maximum current" at 200 Hz means the circuit is at resonance at 200 Hz.

Step 1: Total resistance (resonance)

At resonance Z = R_total, so

R_total = V/I = 100/0.7 = 142.86 Ω
R = R_total − 50 = 92.86 Ω       (coil resistance)

Step 2: Capacitive reactance

X_C = V_C/I = 200/0.7 = 285.71 Ω
C = 1/(2π f X_C) = 1/(2π × 200 × 285.71) = 2.785 μF

Step 3: Inductance

At resonance X_L = X_C = 285.71 Ω:

L = X_L/(2π f) = 285.71/(2π × 200) = 0.2274 H

Check

f0 = 1/(2π√(LC)) = 1/(2π√(0.2274 × 2.785×10⁻⁶)) = 200 Hz ✓
Q = X_L/R_total = 285.71/142.86 = 2.0, V_C = Q·V = 200 V ✓

Answer: R = 92.86 Ω, L = 0.227 H, C = 2.785 μF.

✓ Answer checked by Claude Opus 5.5 (high effort)

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