---
title: "Find R, L, C of series resonant circuit"
subject: "Electrical Circuit Theory"
chapter: 1
key: "semester_3/electrical_circuit_theory:series-resonance-find-r-l-c"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/series-resonance-find-r-l-c
site: BCT Engineering
---

# Find R, L, C of series resonant circuit

Electrical Circuit Theory · Chapter 1: Network Analysis of AC circuit & Dependent Sources

- Asked in: 2068 Baisakh, 6 marks
- Video: [Series RLC Circuits, Resonant Frequency, Inductive Reactance & Capacitive Reactance - AC Circuits](https://www.youtube.com/watch?v=2GvqQvohP2k) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/series-resonance-find-r-l-c
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-1

## Question

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]

## 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.

## Answer

**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

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

#### Step 2: Capacitive reactance

```text
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:

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

#### Check

```text
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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At resonance, how is the total resistance found and what is the coil resistance R?
   A. R_total = V/I = 142.86 Ω, so R = 92.86 Ω
   B. R = V/I = 142.86 Ω
   C. R = V_C/I = 285.7 Ω
   D. R = 142.86 + 50 Ω
   Answer: A. At resonance Z = R_total; subtract the separate 50 Ω resistor to get the coil resistance. (Key point: Total resistance)

2. How is X_C found?
   A. X_C = V/I = 142.9 Ω
   B. X_C = V_C/I = 200/0.7 = 285.7 Ω
   C. X_C = V_C × I
   D. X_C = 200/100
   Answer: B. The same current flows through C, so X_C = V_C/I. (Key point: Capacitive reactance)

3. How is L found at resonance?
   A. L = R/(2πf)
   B. L = 1/(2πf X_C)
   C. X_L = X_C, so L = 285.7/(2π × 200) = 0.227 H
   D. L = 142.9/(2π×200)
   Answer: C. At resonance X_L = X_C = 285.7 Ω. (Key point: Inductance)

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