# Series RLC Circuit Response to a Step Voltage

Use of Laplace transforms to study the response of an RLC circuit to a step voltage. Formulas for the current and all the voltages are developed and numerical examples are presented along with their detailed solutions.
An online calculator for step response of a series RLC circuit may be used check calculations done manually.



## Fomulas for the Current and Voltages in a Series RLC Circuit in Response to a Step Voltage

Problem
Find expressions for the current $i$ and voltages across the capacitor $C$ , the inductors $L$ and the resitor $R$ as functions of time in the ciruit below given that the voltage source $v_i = V_0 \; u(t)$, where $V_0$ is a constant and $u(t)$ is the unit step function . The initial current at $t = 0$ is equal to zero.

Solution to the Above Problem
Use Kirchhoff's law of voltages to write
$v_i - v_R - v_L - v_C = 0$       (I)
Use Ohm's law to write
$v_R = R \; i$
Relationship between voltage and charging current of a capacitor
$\displaystyle v_C = \dfrac{1}{C} \; \int i dt$
Relationship between voltage and charging current of an inductor
$\displaystyle v_L = L \; \dfrac{d i}{dt}$
Substitute $v_R$ , $v_L$ and $v_C$ by their expressions in equation (I)
$\displaystyle v_i - R i - L \dfrac{d i}{dt} - \dfrac{1}{C} \int i dt = 0$
Take the Laplace transform of both sides of the above equation
$\displaystyle \mathscr{L}\{ v_i - R i - L \dfrac{d i}{dt} - \dfrac{1}{C} \int i dt \} = \mathscr{L}\{ 0 \}$
Use the property of linearity of Laplace transform and aslo the fact that $\mathscr{L}\{ 0 \} = 0$ to rewrite the above as
$\displaystyle \mathscr{L}\{ v_i \} - R \mathscr{L}\{ i \} - L \mathscr{L} \left\{ \dfrac{d i}{dt} \right\} - \dfrac{1}{C} \mathscr{L} \left\{ \int i dt \right\} = 0$
Since $v_i(t) = V_0 \; u(t)$ where $V_0$ is a constant and $u(t)$ is the unit step function, $\mathscr{L}\{ v_i \} = \dfrac{V_0}{s}$
Let $\mathscr{L}\{ i\} = I(s)$
Use the property of the derivative and integral (see formulas and properties of Laplace transform) to write
$\mathscr{L} \left\{ \dfrac{d i}{dt} \right\} = s I(s) - i(0) = s I(s)$ since the initial current is equal to zero $i(0) = 0$
$\displaystyle \mathscr{L} \left\{ \int i dt \right\} = \dfrac{I(s)}{s}$
After substitution, our equation becomes
$\dfrac{V_0}{s} - R \; I(s) - L \; s \; I(s) - \dfrac{I(s)}{C s} = 0$
NOTE that we have transformed our initial differential equation from the $t$ (time) domain to the $s$ domain.
Multiply all terms in the above equation by $s$ and simplify
$V_0 - R \; s \; I(s) - L \; s^2 \; I(s) - \dfrac{I(s)}{C} = 0$
Factor $I(s)$ out and rewrite the above equation as
$I(s) (L \; s^2 + R \; s +\dfrac{1}{C}) = V_0$
Solve the above for $I(s)$ and rewrite as follows
$I(s) = \dfrac{V_0}{L} \times \dfrac{1}{s^2 + \dfrac{R}{L} s + \dfrac{1}{L C} }$
Complete the square in the denominator
$I(s) = \dfrac{V_0}{L} \times \dfrac{1}{ \left(s + \dfrac{R}{2 L} \right)^2 + \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 }$
Let $\alpha = \dfrac{R}{2L}$
and rewrite the above equation as
$I(s) = \dfrac{V_0}{L} \times \dfrac{1}{ \left(s + \alpha \right)^2 + \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 }$

We now consider 3 cases depending on the sign of the expression $\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2$
Case 1: $\dfrac{1}{L C} \gt \left(\dfrac{R}{2 L}\right)^2$ : The circuit is underdamped

Let $\omega = \sqrt {\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2}$ and rewrite $I(s)$ as
$I(s) = \dfrac{V_0}{\omega L} \times \dfrac{\omega}{ \left(s + \alpha \right)^2 + \omega^2 }$
Use formulas and properties of Laplace transform to find the inverse Laplace transform of $I(s)$ as
For $t \ge 0$ , $v_i (t) = V_0$ and we have the following
$i(t) = \dfrac{V_0}{\omega L} \; \sin (\omega t) \; e^{-\alpha t}$
$v_R(t) = R \; i(t) = \dfrac{R V_0}{\omega L} \sin (\omega t) e^{-\alpha t}$
$\quad \quad = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t}$
$v_L(t) = L \; \dfrac{d i}{dt} = V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$v_C(t) = v_i(t) - v_R(t) - v_L(t)$
$\quad \quad = V_0 - V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t} - V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$\quad \quad = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$

Numerical Applications - Example 1 - Underdamped Circuit
Let $V_0 = 1 \; V$ , $R = 10 \; \Omega$ , $L = 0.4 \; H$ and $C = 50 \;\mu F$
$\dfrac{1}{L C} = 50000$
$\left(\dfrac{R}{2 L}\right)^2 = 156.25$
Hence $\dfrac{1}{L C} \gt \left(\dfrac{R}{2 L}\right)^2$ ; the circuit is underdamped
$\alpha = \dfrac{R}{2L} = \dfrac{10}{2 \times 0.4} = 12.50$
$\omega = \sqrt {\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2} = \sqrt {\dfrac{1}{0.4 \times 50 \times 10^{-6}} - \left(\dfrac{10}{2 \times 0.4}\right)^2} = 223.26$
$i(t) = \dfrac{1}{223.26 \times 0.4} \; \sin (223.26 t) \; e^{-12.5 t}$
Simplify
$i(t) = 0.011 \; \sin (223.26 t) \; e^{-12.5 t}$
The voltages may be calculated as follows
$v_R(t) = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t}$
$\quad \quad = 0.11198 \; \sin (223.26 t) \; e^{-12.5 t}$
$v_L(t) = V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$\quad \quad = \left\{ \cos \left(223.26t\right) - 0.0559875 \sin (223.26t ) \right\}e^{-12.5t}$
$v_C(t) = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$\quad \quad = 1 - \left\{ \cos (223.26t) + 0.055988 \sin (223.26t) \right\} e^{-12.5t}$
You may use the calculator for step response of a series RLC circuit to check all the above calculations.

Case 2: $\dfrac{1}{L C} \lt \left(\dfrac{R}{2 L}\right)^2$ : The circuit is overdamped

Let $\beta = \sqrt { \left(\dfrac{R}{2 L}\right)^2 - \dfrac{1}{L C} }$ and rewrite $I(s)$ as
$I(s) = \dfrac{V_0}{\beta L} \times \dfrac{\beta}{ \left(s + \alpha \right)^2 - \beta^2 }$
Use formulas and properties of laplace transform to find the inverse Laplace transform of $I(s)$ as
For $t \ge 0$ , $v_i (t) = V_0$ and we have the following
$i(t) = \dfrac{V_0}{\beta L} \; \sinh (\beta t) \; e^{-\alpha t}$
$\quad \quad = \dfrac{V_0}{\beta L} \; \left\{ \dfrac{e^{\beta t} - e^{\beta t}} {2} \right\} \; e^{-\alpha t}$
$\quad \quad = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$v_R(t) = R \; i(t) = \dfrac{R V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$v_L(t) = L \; \dfrac{d i}{dt} = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\}$
$v_C(t) = v_i(t) - v_R(t) - v_L(t)$
$\quad \quad = V_0 - V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} - \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\}$

Numerical Applications - Example 2 - Overdamped Circuit
Let $V_0 = 1 \; V$ , $R = 200 \; \Omega$ , $L = 0.4 \; H$ and $C = 50 \;\mu F$
$\dfrac{1}{L C} = 50000$
$\left(\dfrac{R}{2 L}\right)^2 = 62500$
Hence $\dfrac{1}{L C} \lt \left(\dfrac{R}{2 L}\right)^2$ ; the circuit is overdamped
$\alpha = \dfrac{R}{2L} = \dfrac{200}{2 \times 0.4} = 250$
$\beta = \sqrt { \left(\dfrac{R}{2 L} - \dfrac{1}{L C}\right)^2} = \sqrt {\dfrac{1}{0.4 \times 50 \times 10^{-6}} - \left(\dfrac{200}{2 \times 0.4}\right)^2} = 111.80339$
$i(t) = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = 0.01118 \; \left\{ e^{ -138.2 t} - e^{ -361.8 t} \right\}$
$v_R(t) = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = 2.23613 \; \left\{ e^{ -138.2 t} - e^{ -361.8 t} \right\}$

$v_L (t) = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = -0.617754 e^{ -138.2 t} + 1.617246 e^{ -361.8 t}$

$v_C(t) = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\}$
$\quad \quad = 1 - 1.61806 e^{ -138.2 t} + 0.61806 e^{-361.8 t}$
You may use the calculator for step response of a series RLC circuit to check all the above calculations.

Case 3: $\dfrac{1}{L C} = \left(\dfrac{R}{2 L}\right)^2$ : The circuit is critically damped

$I(s)$ simplifies to
$I(s) = \dfrac{1}{ \left(s + \alpha \right)^2} \dfrac{V_0}{L}$
Use formulas and properties of laplace transform to find the inverse Laplace transform of $I(s)$ as
For $t \ge 0$ , $v_i (t) = V_0$ and we have the following
$i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t}$
$v_R(t) = R \; i(t) = \dfrac{R V_0}{ L} \; t \; e^{-\alpha t}$
$\quad \quad = 2 V_0 \alpha \; t \; e^{-\alpha t}$
$v_L(t) = L \; \dfrac{d i}{dt} = V_0 \left( 1 - \alpha t \right) e^{-at}$
$v_C(t) = v_i(t) - v_R(t) - v_L(t)$
$\quad \quad = V_0 - 2 V_0 \alpha \; t \; e^{-\alpha t} - V_0 e^{-at} \left( 1 - \alpha t \right)$
$\quad \quad = V_0 - V_0(1+\alpha t)e^{-\alpha t}$

Numerical Applications - Example 3 - Critically damped Circuit
Let $V_0 = 1 \; V$ , $R = 100 \; \Omega$ , $L = 0.4 \; H$ and $C = 160 \;\mu F$
$\dfrac{1}{L C} = 15625$
$\left(\dfrac{R}{2 L}\right)^2 = 15625$
Hence $\dfrac{1}{L C} = \left(\dfrac{R}{2 L}\right)^2$ ; the circuit is critically damped
$\alpha = \dfrac{R}{2L} = \dfrac{100}{2 \times 0.4} = 125$
$i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t}$
$i(t) = 2.5 \; t \; e^{- 125 t}$
$v_R(t) = 2 V_0 \alpha \; t \; e^{-\alpha t}$
$\quad \quad = 250 e^{ - 125 t}$
$v_L(t) = L \; \dfrac{d i}{dt} = V_0 ( 1 - \alpha t ) e^{-at}$
$\quad \quad = (1 - 125) e^{ - 125 t}$
$v_C (t) = V_0 - V_0(1+\alpha t)e^{-\alpha t}$
$\quad \quad = 1 - (1 + 125) e^{ - 125 t}$
You may use the calculator for step response of a series RLC circuit to check all the above calculations.