---
title: "Series R-L with sinusoidal source: particular solution"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:rl-series-sinusoidal-particular-solution"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/rl-series-sinusoidal-particular-solution
site: BCT Engineering
---

# Series R-L with sinusoidal source: particular solution

Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution

- Asked in: 2068 Baisakh, 8 marks
- Video: [TRANSIENT RESPONSE OF RL SERIES CIRCUIT WITH SINUSOIDAL EXCITATION|ELECTRICAL @KKCEE218#YOUTUBE](https://www.youtube.com/watch?v=GSFiX9gB8mI) — KK CONCEPTS OF ELECTRICAL ENGINEERING
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/rl-series-sinusoidal-particular-solution
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

In a series R-L circuit the applied voltage is v(t) = 10 sin(10⁴t + π/6) with R = 2 Ω, L = 0.01 H. v(t) is applied at t = 0. Obtain the particular solution for current i(t) through the circuit. Assume zero initial current through the inductor. [Use classical method].

## Short answer

The particular solution is the steady-state sinusoidal response. Using phasors, impedance magnitude is 100.02 Ohms and angle 88.85 degrees. Current magnitude is 0.09998 A, lagging voltage by 88.85 degrees. i_p(t) = 0.09998 sin(10^4 t - 58.85 degrees) A.

## Answer

**Given:** v(t) = 10 sin(10⁴t + π/6), R = 2 Ω, L = 0.01 H, i(0) = 0.

#### Differential equation

```text
L di/dt + R i = v(t)
0.01 di/dt + 2 i = 10 sin(10⁴ t + 30°)
```

#### Particular solution (steady-state phasor method)

```text
X_L = ωL = 10⁴ × 0.01 = 100 Ω
Z = R + jX_L = 2 + j100,   |Z| = 100.02 Ω,   θ = tan⁻¹(100/2) = 88.85°
I_m = V_m/|Z| = 10/100.02 = 0.09998 A
```

The current lags the voltage by θ:

```text
i_p(t) = 0.09998 sin(10⁴ t + 30° − 88.85°)
       = 0.09998 sin(10⁴ t − 58.85°) A
```

(0.09998 ≈ 0.1 A.)

Check by substitution: take i_p = I sin(ωt + β) with I = 0.09998 and β = −1.0272 rad. Then L ω I cos(…) + R I sin(…) reproduces 10 sin(ωt + 30°) ✓.

#### Complete solution (for completeness)

Complementary solution: s = −R/L = −200, so i_c = K e^(−200t).

```text
i(0) = 0:  0.09998 sin(−58.85°) + K = 0  →  K = 0.0856
i(t) = 0.09998 sin(10⁴ t − 58.85°) + 0.0856 e^(−200t) A
```

This was verified numerically (for example i(1 ms) = 0.1137 A).

**Answer: particular solution i_p(t) = 0.09998 sin(10⁴t − 58.85°) A ≈ 0.1 sin(10⁴t − 58.85°) A.**

## Check yourself

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

1. How is the particular solution found for a sinusoidal source?
   A. By the phasor method with Z = R + jwL
   B. By setting i = K e^(-200t)
   C. By integrating the source
   D. By using the time constant only
   Answer: A. The forced response is the sinusoidal steady state. (Key point: Method)

2. What are X_L and |Z|?
   A. 0.01 ohm and 2 ohm
   B. 100 ohm and 100.02 ohm
   C. 100 ohm and 2 ohm
   D. 10^4 ohm and 10^4 ohm
   Answer: B. X_L = 10^4 x 0.01 = 100; |Z| = sqrt(4 + 10^4). (Key point: Impedance)

3. What is the phase angle of i_p?
   A. 30 + 88.85 = 118.85 degrees
   B. 30 degrees
   C. 30 - 88.85 = -58.85 degrees; current lags
   D. -88.85 degrees
   Answer: C. Current lags voltage by tan^-1(X_L/R). (Key point: Phase)

4. What is the role of the 0.0856 e^(-200t) term?
   A. It is part of the particular solution
   B. It is the steady state
   C. It is the forced response
   D. It is the transient needed so that i(0) = 0, not part of the particular solution
   Answer: D. The particular solution is only the sinusoid. (Key point: Natural response)

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