---
title: "Forced response depends on input, natural does not"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:forced-vs-natural-response-example"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/forced-vs-natural-response-example
site: BCT Engineering
---

# Forced response depends on input, natural does not

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

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Natural and Forced Response.](https://www.youtube.com/watch?v=6wglI7Na5jw) — πrate
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/forced-vs-natural-response-example
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

With a suitable example prove that the forced response of a network depends upon the nature of input excitation while the natural response never depends upon the input excitation.

## Short answer

Natural response form depends only on network elements (e.g., A e^(−2t) fixed by R/L). Forced response varies: 5 A (DC), 10e^(−t) (exponential), 4.47 sin(t−26.57°) (sinusoidal), proving it follows the input shape.

## Answer

**Definitions.** The natural (complementary) response is the solution of the homogeneous equation, with the source set to zero. Its form depends only on the network (R, L, C). The forced (particular) response is the part that follows the shape of the applied source.

#### Example: series R-L circuit
Take R = 2 Ω, L = 1 H with zero initial current. Equation: di/dt + 2i = v(t)/L = v(t).

**Natural response (same for every input).** Homogeneous equation di/dt + 2i = 0 gives i_n = A e^(−2t). It contains only the network constant R/L = 2.

**Case 1: DC input v = 10 V**

- Forced: i_f = 10/2 = 5 A (constant, same shape as the source)
- i = 5 + A e^(−2t); i(0) = 0 → A = −5
- i₁(t) = 5 − 5e^(−2t)

**Case 2: exponential input v = 10 e^(−t) V**

- Try i_f = K e^(−t): −K + 2K = 10 → K = 10, so i_f = 10e^(−t) (decays like the source)
- i = 10e^(−t) + A e^(−2t); i(0) = 0 → A = −10
- i₂(t) = 10e^(−t) − 10e^(−2t)

**Case 3: sinusoidal input v = 10 sin t V.** Forced response i_f = (10/√5) sin(t − 26.57°) = 4.47 sin(t − 26.57°). The natural term is again of the form A e^(−2t).

#### Comparison
| Input | Forced response | Natural response |
|---|---|---|
| 10 V (DC) | 5 A (constant) | A e^(−2t) |
| 10e^(−t) | 10e^(−t) | A e^(−2t) |
| 10 sin t | 4.47 sin(t − 26.57°) | A e^(−2t) |

The forced part changes with the source (constant, exponential, sinusoid). The natural part always has the exponent −R/L = −2 whatever the source, because it is the solution of the equation with the source removed. Only its constant A depends on the input, through the initial condition. This proves the statement.

Numerical check: Case 2 at t = 1 s gives i = 10(0.3679 − 0.1353) = 2.325 A, matching a numerical solution of di/dt + 2i = 10e^(−t).

## Check yourself

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

1. What determines the form of the natural response?
   A. The type of source applied
   B. Only the network elements (R, L, C), found with the source set to zero
   C. The amplitude of the input only
   D. The initial condition only
   Answer: B. The natural response solves the homogeneous equation, so its exponent depends only on the network (here -R/L = -2). (Key point: Definition of natural response)

2. For the RL circuit (R = 2, L = 1), what is the forced response for v = 10e^(-t)?
   A. 5 A
   B. 5 e^(-2t)
   C. 10 e^(-t)
   D. 4.47 sin(t - 26.57 deg)
   Answer: C. Trying K e^(-t): -K + 2K = 10, so K = 10; the forced response copies the source shape. (Key point: Forced response changes)

3. Across DC, exponential and sinusoidal inputs, what stays the same and what changes in the natural term A e^(-2t)?
   A. The exponent -2 is fixed by the network; only A changes through the initial condition
   B. The exponent changes with the input but A stays fixed
   C. Both the exponent and A are fixed
   D. The natural term vanishes for non-DC inputs
   Answer: A. The exponent is -R/L = -2 whatever the source; A is set so that i(0) = 0 with the particular forced term. (Key point: What the input changes)

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