FOUNDATION · CHAPTER 5 OF 7

Differential equations for circuits

This chapter shows why a circuit with storage elements obeys a differential equation, and how the natural and forced parts, the time constant and the damping come out of that equation. You need it to read a transient from the circuit instead of guessing.

Updated 12 Oct 2026

What you will be able to do

You will write the equation of a circuit with one or two storage elements and solve it as a natural part plus a forced part. You will read the time constant and the damping from the circuit values. You will show why a source-free response dies away, fix its constants from physics and solve it by the Laplace method.

Before you start

Chapter 2, Why R, L and C behave as they do
Chapter 4, Setting up and solving circuit equations systematically
Derivatives: the product rule and the chain rule
Integrals of exponential functions, and integration by parts
Euler's formula, $e^{j x} = cos x + j sin x$ , where $j$ is the imaginary unit
Solving a quadratic equation, including complex roots
Partial fractions

First order: one equation, two parts

One storage element gives a first-order equation. Its solution is a natural (homogeneous) part, the circuit with no source, plus a forced (particular) part set by the source. Currents enter the + terminal of each element's voltage (passive sign convention); $0^{-}$ and $0^{+}$ are the instants just before and just after switching.

Take a source $V_{s}$ in series with $R$ and $C$ , the switch closing at $t = 0$ . Kirchhoff's voltage law (Chapter 4) gives $V_{s} = R i + v_{C}$ , and the capacitor gives $i = C d v_{C} / d t$ . Eliminate $i$ .

SERIES RC EQUATION

R C \frac{d v _{C}}{d t} + v_{C} = V_{s}

Holds for $t > 0$ with $V_{s}$ constant. $R$ in ohms, $C$ in farads, $v_{C}$ and $V_{s}$ in volts, $t$ in seconds.

NATURAL RESPONSE AND TIME CONSTANT

v_{h} = K e^{s t} \Rightarrow R C s + 1 = 0 \Rightarrow s = - \frac{1}{R C} \Rightarrow v_{h} = K e^{- t / τ}, τ = R C

With the right side zero, $(R C s + 1) K e^{s t} = 0$ and $K e^{s t} \neq = 0$ , so $s$ is the root of the characteristic equation $R C s + 1 = 0$ . $K$ (volts) is found later; $s$ is in s $^{- 1}$ , the time constant $τ$ in seconds.

The natural part has no source in it, so $τ$ is a property of the circuit alone: it does not depend on $V_{s}$ or on the starting voltage. After one $τ$ it has fallen to $e^{- 1} = 36.8$ % of its start, after $5 τ$ to $e^{- 5} = 0.67$ %. The same steps on $L d i / d t + R i = V_{s}$ give $τ = L / R$ .

Try a forced part shaped like the source, a constant $v_{p} = A$ : its derivative is zero, so the equation reads $A = V_{s}$ . The sum $v_{p} + v_{h}$ also solves it: substituted, the $v_{p}$ terms give $V_{s}$ and the $v_{h}$ terms give $0$ . Its one constant $K$ is set by the voltage at $0^{+}$ .

COMPLETE FIRST-ORDER RESPONSE

v_{C} (t) = V_{s} + (v_{C} (0^{+}) - V_{s}) e^{- t / τ}

Holds for $t \geq 0$ with a constant source. Setting $t = 0^{+}$ in $V_{s} + K e^{- t / τ}$ gives $K$ . Here $v_{C} (0^{+}) = v_{C} (0^{-})$ : $R$ keeps the current finite, so the capacitor voltage cannot jump (Chapter 2).

WORKED EXAMPLE

A 10 V source is in series with R = 2 k $Ω$ and C = 5 $μ$ F. The capacitor is uncharged and the switch closes at $t = 0$ . Find $v_{C}$ and $i$ at $t = 15$ ms.

Solution

The time constant is a property of the circuit: $τ = R C = 2000 Ω \times 5 \times 1 0^{- 6} F = 10$ ms.
The capacitor voltage cannot jump, so $v_{C} (0^{+}) = 0$ ; the forced part is $V_{s} = 10$ V. So $v_{C} = 10 (1 - e^{- t / τ})$ V.
At 15 ms, $t / τ = 1.5$ and $e^{- 1.5} = 0.2231$ , so $v_{C} = 10 (1 - 0.2231) = 7.77$ V.
The capacitor law gives the current: $i = C d v_{C} / d t = C (10 V / τ) e^{- t / τ}$ . Since $τ = R C$ , $C / τ = 1/ R$ , so $i = \frac{10 V}{2 k Ω} e^{- t / τ} = 5 mA \times 0.2231 = 1.12$ mA.

Answer: $v_{C} = 7.77$ V and $i = 1.12$ mA at $t = 15$ ms.

Second order: the roots decide

Two storage elements that cannot be combined into one give a second-order equation. For a series $R$ , $L$ , $C$ with a constant source $V_{s}$ , one loop current $i$ enters the + terminal of each element's voltage. Kirchhoff's voltage law gives $V_{s} = R i + L d i / d t + v_{C}$ . Substitute $i = C d v_{C} / d t$ .

SERIES RLC EQUATION

L C \frac{d ^{2} v _{C}}{d t ^{2}} + R C \frac{d v _{C}}{d t} + v_{C} = V_{s}

Holds for $t > 0$ with $V_{s}$ constant; $L$ in henries, the other symbols as before. The forced part is again $v_{p} = V_{s}$ .

CHARACTERISTIC EQUATION AND ITS ROOTS

s^{2} + 2 α s + ω_{0}^{2} = 0, s_{1, 2} = - α \pm α^{2} - ω_{0}^{2}, α = \frac{R}{2 L}, ω_{0} = \frac{1}{L C}

Series circuit only. $K e^{s t}$ with the right side zero gives $L C s^{2} + R C s + 1 = 0$ ; dividing by $L C$ , $2 α = R / L$ and $ω_{0}^{2} = 1/ (L C)$ . $α$ is the damping rate in s $^{- 1}$ , $ω_{0}$ the natural frequency in rad/s.

Case	Condition	Roots	Natural response
Overdamped	$α > ω_{0}$ , that is $R > 2 L / C$	$s_{1} \neq = s_{2}$ , both real and negative	$K_{1} e^{s_{1} t} + K_{2} e^{s_{2} t}$
Critically damped	$α = ω_{0}$ , that is $R = 2 L / C$	one repeated root, $- α$	$(K_{1} + K_{2} t) e^{- α t}$
Underdamped	$α < ω_{0}$ , that is $R < 2 L / C$	$- α \pm j ω_{d}$ , with $ω_{d} = ω_{0}^{2} - α^{2}$	$e^{- α t} (K_{1} cos ω_{d} t + K_{2} sin ω_{d} t)$

With the source off, dividing by $L C$ gives $v^{''} + 2 α v^{'} + ω_{0}^{2} v = 0$ , with $v = v_{C}$ and primes for time derivatives. If two functions solve this equation, so does any sum of constant multiples of them: substituted, each gives zero. At critical damping $ω_{0} = α$ and the repeated root gives only $e^{- α t}$ . Try $v = t e^{- α t}$ : the product rule gives $v^{'} = (1 - α t) e^{- α t}$ and $v^{''} = (α^{2} t - 2 α) e^{- α t}$ . In $v^{''} + 2 α v^{'} + α^{2} v$ the factor of $e^{- α t}$ is $α^{2} t - 2 α + 2 α - 2 α^{2} t + α^{2} t = 0$ , so it is a second solution.

For complex roots, Euler's formula gives $e^{(- α \pm j ω_{d}) t} = e^{- α t} (cos ω_{d} t \pm j sin ω_{d} t)$ . Half their sum is $e^{- α t} cos ω_{d} t$ , and their difference over $2 j$ is $e^{- α t} sin ω_{d} t$ : two real solutions, the table's last row. Only this case oscillates, at $ω_{d}$ inside the envelope $e^{- α t}$ .

Why can the natural response not grow? $R$ , $L$ and $C$ are passive, meaning none of them generates energy: the resistor absorbs $p = R i^{2} \geq 0$ , and $L$ and $C$ only store it. With the source off, $L d i / d t = - R i - v_{C}$ and $C d v_{C} / d t = i$ . By the chain rule, $\frac{d}{d t} (\frac{1}{2} L i^{2} + \frac{1}{2} C v_{C}^{2}) = i L \frac{d i}{d t} + v_{C} C \frac{d v _{C}}{d t} = i (- R i - v_{C}) + v_{C} i$ . The $v_{C} i$ terms cancel, leaving $- R i^{2}$ .

ENERGY LOST IN A SOURCE-FREE SERIES RLC

\frac{d W}{d t} = - R i^{2} \leq 0, W = \frac{1}{2} L i^{2} + \frac{1}{2} C v_{C}^{2}

Holds with the source set to zero and $R \geq 0$ . $W$ is the stored energy in joules, $i$ the current in amperes. $W \geq 0$ never increases, so the response cannot grow; with $R > 0$ it falls whenever $i \neq = 0$ .

The roots finish the argument. From $s_{1, 2} = - α \pm α^{2} - ω_{0}^{2}$ , their sum is $- 2 α$ and their product is $α^{2} - (α^{2} - ω_{0}^{2}) = ω_{0}^{2}$ . For $R > 0$ the sum is negative and the product positive: real roots are both negative, and complex roots have real part $- α < 0$ . Every term then decays. With $R = 0$ , $α = 0$ : nothing is lost and the response oscillates at $ω_{0}$ for ever.

Initial conditions and the Laplace method

An equation of order $n$ needs $n$ conditions, and they come from physics. A capacitor voltage jumps only if an infinite current flows, and an inductor current only if an infinite voltage appears (Chapter 2). In a series $R$ , $L$ , $C$ with finite sources neither happens, so both keep their $0^{-}$ values at $0^{+}$ .

THE TWO CONDITIONS FOR A SECOND-ORDER CIRCUIT

v_{C} (0^{+}) = v_{C} (0^{-}), \frac{d v _{C}}{d t}_{0^{+}} = \frac{i _{L} ( 0 ^{+} )}{C}, i_{L} (0^{+}) = i_{L} (0^{-})

Series $R$ , $L$ , $C$ with finite sources. The loop current $i = i_{L}$ (amperes) enters the capacitor's + terminal, so $i = C d v_{C} / d t$ at $0^{+}$ gives the second condition.

WORKED EXAMPLE

A switch closes at $t = 0$ , putting a capacitor C = 10 $μ$ F charged to $v_{C} (0^{-}) = 10$ V in series with R = 10 $Ω$ and L = 1 mH. There is no source and no current before $t = 0$ . Find $v_{C} (t)$ for $t > 0$ .

Solution

The roots depend on $α = R /2 L = 10/ (2 \times 1 0^{- 3}) = 5000$ s $^{- 1}$ and $ω_{0} = 1/ L C = 1/ 1 0^{- 8} = 1 0^{4}$ rad/s.
$α < ω_{0}$ , so the response is underdamped, with $ω_{d} = ω_{0}^{2} - α^{2} = 1 0^{8} - 2.5 \times 1 0^{7} = 8660$ rad/s. With no source the right side of the equation is zero, so there is no forced part: $v_{C} = e^{- 5000 t} (K_{1} cos 8660 t + K_{2} sin 8660 t)$ .
The capacitor voltage cannot jump, so $v_{C} (0^{+}) = 10$ V; at $t = 0$ the expression is $K_{1}$ , so $K_{1} = 10$ V.
The inductor current cannot jump, so $v_{C}^{'} (0^{+}) = i (0^{+}) / C = 0$ . The product rule gives $v_{C}^{'} = - α v_{C} + ω_{d} e^{- α t} (K_{2} cos ω_{d} t - K_{1} sin ω_{d} t)$ , which at $t = 0$ is $- α K_{1} + ω_{d} K_{2}$ . Setting it to zero, $K_{2} = 5000 \times 10/8660 = 5.77$ V.

Answer: $v_{C} = e^{- 5000 t} (10 cos 8660 t + 5.77 sin 8660 t)$ V, with $t$ in seconds.

The Laplace method turns the equation into algebra. By definition, the Laplace transform $F (s)$ of a voltage or current $f (t)$ multiplies it by $e^{- s t}$ and integrates from $0^{-}$ to $\infty$ , with $s$ a real number. For a derivative, integration by parts gives $\int_{0^{-}}^{\infty} f^{'} e^{- s t} d t = [f e^{- s t}]_{0^{-}}^{\infty} + s \int_{0^{-}}^{\infty} f e^{- s t} d t$ , and the bracket is $0 - f (0^{-})$ .

DERIVATIVE RULE

L {f^{'} (t)} = s F (s) - f (0^{-}), F (s) = \int_{0^{-}}^{\infty} f (t) e^{- s t} d t

Holds when $f (t) e^{- s t} \to 0$ as $t \to \infty$ ; for the RC response below, any $s > 0$ will do. $s$ is in s $^{- 1}$ .

Transform the series RC equation with $V_{s}$ switched on at $t = 0$ . Integrals split over sums and let constants out, so each term transforms on its own. $V_{s}$ is zero before $t = 0$ , so its transform is $\int_{0}^{\infty} V_{s} e^{- s t} d t = V_{s} / s$ . The derivative rule on $v_{C}$ , with $V (s)$ its transform, gives $R C (s V - v_{C} (0^{-})) + V = V_{s} / s$ .
Collect $V$ : $(R C s + 1) V = R C v_{C} (0^{-}) + V_{s} / s$ . Divide by $R C = τ$ , then by $s + 1/ τ$ : $V = \frac{v _{C} ( 0 ^{-} )}{s + 1/ τ} + \frac{V _{s}}{τ s ( s + 1/ τ )}$ .
Split into partial fractions: $\frac{1}{τ s ( s + 1/ τ )} = \frac{1}{s} - \frac{1}{s + 1/ τ}$ . So $V = \frac{V _{s}}{s} + \frac{v _{C} ( 0 ^{-} ) - V _{s}}{s + 1/ τ}$ .
Invert: find the $v_{C} (t)$ whose transform is $V$ (for a continuous signal there is only one), using two transforms from the definition: the constant 1 gives $\int_{0}^{\infty} e^{- s t} d t = 1/ s$ , and $e^{- a t}$ gives $\int_{0}^{\infty} e^{- (s + a) t} d t = 1/ (s + a)$ . So $v_{C} = V_{s} + (v_{C} (0^{-}) - V_{s}) e^{- t / τ}$ , the first-order answer again.

A value of $s$ that makes a denominator zero is a pole. Each unrepeated pole $p$ gave a partial fraction $K / (s - p)$ , which inverts to $K e^{pt}$ . The pole $- 1/ τ$ is the root of $R C s + 1 = 0$ : the natural part. The source's pole $0$ gives $e^{0 t} = 1$ : the forced part. In a series RLC circuit the roots $s_{1, 2}$ of $L C s^{2} + R C s + 1 = 0$ are poles in the same way.

Check yourself

CHECK YOURSELF 1

A 6 V source is in series with R = 1 k $Ω$ and C = 2 $μ$ F. The capacitor is uncharged and the switch closes at $t = 0$ . What is $v_{C}$ at $t = 2$ ms?

(A)
2.21 V
(B)
3.79 V
(C)
5.19 V
(D)
6.00 V

Answer and solution

Answer: (B)

$τ = R C = 1000 Ω \times 2 \times 1 0^{- 6} F = 2$ ms.
So $t = 2$ ms is exactly one $τ$ : $v_{C} = 6 (1 - e^{- 1}) = 6 \times 0.632 = 3.79$ V.

CHECK YOURSELF 2One or more options are correct

An ideal dc source, a resistor $R$ and an inductor $L$ are in series, and the switch closes at $t = 0$ . Which of these affect the time constant?

(A)
The inductance $L$
(B)
The resistance $R$
(C)
The initial inductor current
(D)
The value of the dc source

Answer and solution

Answer: (A), (B)

The time constant comes from the equation with the source set to zero: $L d i / d t + R i = 0$ , so $τ = L / R$ .
The initial current only sets the constant $K$ , and the source value only sets the forced part. Neither changes $τ$ .

CHECK YOURSELF 3

A series RLC circuit has L = 2.5 mH and C = 1 $μ$ F. Find the value of R in ohms at which it is critically damped.

Answer and solution

Answer: $100$ (accepted range $\pm 0.5$ )

Critical damping needs $α = ω_{0}$ , that is $R = 2 L / C$ .
$L / C = 2.5 \times 1 0^{- 3} /1 0^{- 6} = 2500 Ω^{2}$ and $2500 = 50 Ω$ .
So $R = 2 \times 50 = 100 Ω$ .

CHECK YOURSELF 4

A source-free series RLC circuit has R = 10 $Ω$ , L = 10 mH and C = 1 $μ$ F. How does its natural response behave?

(A)
Overdamped
(B)
Critically damped
(C)
Underdamped
(D)
Undamped

Answer and solution

Answer: (C)

$2 L / C = 2 1 0^{- 2} /1 0^{- 6} = 200 Ω$ .
$R = 10 Ω$ is below it, so the roots are complex; $R > 0$ , so the response decays. Underdamped, not undamped.

Used in these GATE topics

Transients in RL, RC and RLC circuits