---
title: "Series RLC with 10 sin t source, classical"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:rlc-series-10sint-classical"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/rlc-series-10sint-classical
site: BCT Engineering
---

# Series RLC with 10 sin t source, classical

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

- Asked in: 2075 Asoj, 8 marks
- Video: [Numerical on series RLC circuit](https://www.youtube.com/watch?v=m92w-stSJnU) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/rlc-series-10sint-classical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

Find the time expression for current for t > 0 in an RLC series circuit with R = 10 ohm, L = 1 H and C = 1/9 F, if the circuit is supplied by v = 10 sin t at t = 0. Assume that capacitor and inductor are initially de-energized. Use classical method.

## Short answer

Solve d²i/dt² + 10 di/dt + 9i = 10 cos t. Complementary roots −1, −9; particular phasor solution 0.781 sin(t+38.66°) A; apply i(0)=0, di/dt(0)=0. i(t) = 0.781 sin(t+38.66°) − 0.625 e^(−t) + 0.137 e^(−9t) A.

## Answer

**Given:** R = 10 Ω, L = 1 H, C = 1/9 F, v = 10 sin t applied at t = 0; i(0) = 0, v_C(0) = 0.

#### Differential equation

```text
L di/dt + R i + (1/C)∫i dt = 10 sin t
Differentiate:  d²i/dt² + 10 di/dt + 9 i = 10 cos t
```

#### Complementary solution

```text
s² + 10 s + 9 = 0  →  s = −1, −9
i_c = A e^(−t) + B e^(−9t)
```

(overdamped)

#### Particular solution (phasor method, ω = 1 rad/s)

```text
Z = R + j(ωL − 1/ωC) = 10 + j(1 − 9) = 10 − j8
|Z| = 12.806 Ω,  angle = −38.66°
I_m = 10/12.806 = 0.7809 A
i_p = 0.7809 sin(t + 38.66°) = 0.7809 sin(t + 0.6747)
```

#### Initial conditions

- i(0) = 0.
- At t = 0⁺: L di/dt = v(0) − R i(0) − v_C(0) = 0, so di/dt(0⁺) = 0.

```text
i_p(0) = 0.7809 sin(38.66°) = 0.4878
i_p′(0) = 0.7809 cos(38.66°) = 0.6098

A + B = −0.4878
−A − 9B = −0.6098
```

Adding: −8B = −1.0976, so B = 0.1372; A = −0.625.

#### Result

```text
i(t) = 0.7809 sin(t + 38.66°) − 0.625 e^(−t) + 0.1372 e^(−9t) A,   t > 0
```

Check by substitution: i(0) = 0.4878 − 0.625 + 0.1372 = 0 ✓; i′(0) = 0.6098 + 0.625 − 1.2348 = 0 ✓.

**Answer: i(t) = 0.781 sin(t + 38.66°) − 0.625 e^(−t) + 0.137 e^(−9t) A.**

## Check yourself

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

1. What are the roots of s^2 + 10s + 9 = 0 and what damping results?
   A. s = -1, -9; overdamped
   B. s = -1 +/- j9; underdamped
   C. s = -5, -5; critically damped
   D. s = -1, -9; underdamped
   Answer: A. The roots are real and distinct, so the complementary solution is A e^(-t) + B e^(-9t), overdamped. (Key point: Characteristic roots)

2. How is the particular solution found?
   A. Setting the forced response equal to zero
   B. Phasor method with Z = 10 - j8 ohm at w = 1 rad/s
   C. Using Z = 10 + j8 ohm
   D. Using Z = R only
   Answer: B. wL = 1 and 1/(wC) = 9, so Z = 10 + j(1 - 9) = 10 - j8, giving Im = 0.781 A at +38.66 degrees. (Key point: Particular solution)

3. What is di/dt(0+) and why?
   A. 10 A/s, since v(0) = 10
   B. Infinite, since the inductor opposes change
   C. Zero, since v(0) = 0, i(0) = 0 and v_C(0) = 0 make L di/dt = 0
   D. 0.781 A/s, the peak of the particular solution
   Answer: C. At t = 0, v = 10 sin 0 = 0, so L di/dt = v - R i - v_C = 0. (Key point: Second initial condition)

4. What equations give A and B?
   A. A + B = 0 and A + 9B = 0
   B. A + B = i_p(0) only
   C. A = B = -0.4878
   D. A + B = -i_p(0) and -A - 9B = -i_p'(0)
   Answer: D. Applying i(0) = 0 and i'(0) = 0 to i_p + A e^(-t) + B e^(-9t) gives those two equations. (Key point: Constants)

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