Electrical Transients

LessonAbout 2 h

Electrical transient is the temporary response of a circuit during the transition period between two steady-state operating conditions.

1. Transient Period

Transient period is the time interval during which circuit currents and voltages adjust from an initial steady state to a new final steady state following a switching operation.

Initiated by circuit switching, component change, or disturbance
Governed by energy storage in inductors and capacitors
Terminates when steady-state conditions are re-established

2. RL DC Series Circuit

RL DC series circuit consists of a resistor and an inductor connected in series across a direct-current voltage source.

Inductor opposes sudden changes in circuit current
Initial current at switching instant is zero (\(i(0) = 0\))
Final steady-state current is limited only by resistance (\(I = E / R\))

RL DC Series Current

\[ i = \dfrac{E}{R}\left[1 - e^{-\frac{R}{L}t}\right] \]

Where: \(i\) = instantaneous current (A), \(E\) = supply voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)

EE Board Exam April 1979 / EE Board Exam October 1982

Example. In an RL circuit with \(E = 200\ \text{V}\), \(R = 20\ \Omega\), and \(L = 1\ \text{H}\), find \(i\) when \(t = 0.02\ \text{second}\) assuming \(i = 0\) at \(t = 0\).

Solution.

\[ i = \dfrac{E}{R}\left[1 - e^{-\frac{R}{L}t}\right] = \dfrac{200}{20}\left[1 - e^{-\frac{20}{1}(0.02)}\right] = 3.3\ \text{A} \]

Answer: \(3.3\ \text{A}\)

2.1 Time Constant

Time constant of an RL circuit is the time required for current to reach \(63.2\%\) of its final steady-state value.

Symbolized as \(\text{TC}\) or \(\tau\)
Equal to ratio of inductance to resistance
Current reaches over \(99\%\) steady state after \(5\ \text{TC}\)

RL Circuit Time Constant

\[ \text{TC} = \dfrac{L}{R} \]

Where: \(\text{TC}\) = time constant (s), \(L\) = inductance (H), \(R\) = resistance (\(\Omega\))

EE Board Exam October 1990

Example. A \(6.0\ \text{H}\) coil whose resistance is \(12\ \Omega\) is connected in series with a \(24\ \Omega\) resistor and a \(144\ \text{V}\) battery. Determine the time constant of the coil.

Solution.

\[ \text{TC} = \dfrac{L}{R} = \dfrac{6.0}{12} = 0.50\ \text{s} \]

Answer: \(0.50\ \text{s}\)

2.2 Instant Voltage Drop

Instantaneous voltage drop represents the voltage across individual circuit elements at any time \(t\).

Resistor voltage increases exponentially with current
Inductor voltage is maximum at \(t = 0\) and decays to zero
Sum of voltage drops equals supply voltage (\(E_R + E_L = E\))

RL Instantaneous Voltage Drops

\[ E_R = E\left[1 - e^{-\frac{R}{L}t}\right] \]

\[ E_L = E e^{-\frac{R}{L}t} \]

Where: \(E_R\) = resistor voltage drop (V), \(E_L\) = inductor voltage drop (V), \(E\) = supply voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)

Example. A coil having a resistance of \(10\ \Omega\) and inductance of \(10\ \text{H}\) is connected in series with a \(15\ \Omega\) resistor across a \(200\ \text{V}\) DC source. Determine the voltage across the coil at \(t = 0.05\ \text{s}\).

Solution.

\[ i = \dfrac{200}{10 + 15}\left[1 - e^{-\frac{25}{10}(0.05)}\right] = 0.94\ \text{A} \]

\[ E_{\text{coil}} = E - i R_{15} = 200 - (0.94)(15) = 185.9\ \text{V} \]

Answer: \(185.9\ \text{V}\)

2.3 Instant Power

Instantaneous power is the rate of energy dissipation or storage in circuit elements during the transient state.

Resistor power \(P_R\) dissipates as heat energy
Inductor power \(P_L\) stores energy in magnetic field
Total power \(P_t\) delivered by source equals \(P_R + P_L\)

RL Instantaneous Powers

\[ P_R = \dfrac{E^2}{R}\left[1 - e^{-\frac{R}{L}t}\right]^2 \]

\[ P_L = \dfrac{E^2}{R}\left[e^{-\frac{R}{L}t} - e^{-\frac{2R}{L}t}\right] \]

\[ P_t = \dfrac{E^2}{R}\left[1 - e^{-\frac{R}{L}t}\right] \]

Where: \(P_R\) = resistor power (W), \(P_L\) = inductor power (W), \(P_t\) = total power (W), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)

3. RC DC Series Circuit

RC DC series circuit consists of a resistor and a capacitor connected in series across a direct-current voltage source.

Capacitor opposes sudden changes in terminal voltage
Initial charging current is maximum (\(I_0 = E / R\))
Final steady-state current decays to zero (\(i(\infty) = 0\))

RC DC Series Current Expression

\[ i = k e^{-\frac{t}{RC}} \]

Where: \(i\) = instantaneous current (A), \(k\) = arbitrary constant, \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F), \(t\) = time (s)

3.1 Time Constant

Time constant of an RC circuit is the time required for capacitor voltage to reach \(63.2\%\) of supply voltage or current to drop to \(36.8\%\) of initial value.

Equal to product of resistance and capacitance (\(\text{TC} = RC\))
Determines rate of charge accumulation and voltage buildup
Capacitor fully charged after \(5\ \text{TC}\)

RC Circuit Time Constant

\[ \text{TC} = R C \]

Where: \(\text{TC}\) = time constant (s), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F)

Example. In an RC series circuit with \(R = 1000\ \Omega\) and \(C = 66.667\ \mu\text{F}\) connected across \(110\ \text{V}\), find the current when \(t = \text{TC}\).

Solution.

\[ i = \dfrac{E}{R} e^{-\frac{t}{\text{TC}}} = \dfrac{110}{1000} e^{-1} = 0.0405\ \text{A} \]

Answer: \(0.0405\ \text{A}\)

3.2 Instant Current: No Initial Charge, With Initial Charge

Instantaneous charging current depends on whether the capacitor holds an initial charge \(Q_0\) prior to switching.

Without initial charge (\(Q_0 = 0\)), initial current is \(E / R\)
With initial charge \(Q_0\), initial voltage is \(E \pm (Q_0 / C)\)
Sign is positive for discharging and negative for charging

RC Instantaneous Current

\[ i = \dfrac{E}{R} e^{-\frac{t}{RC}} \]

\[ i = \left(\dfrac{E \pm \dfrac{Q_0}{C}}{R}\right) e^{-\frac{t}{RC}} \]

Where: \(i\) = instantaneous current (A), \(E\) = supply voltage (V), \(Q_0\) = initial charge (C), \(C\) = capacitance (F), \(R\) = resistance (\(\Omega\)), \(t\) = time (s)

Example. A resistance of \(100\ \text{k}\Omega\) is connected in series with a \(100\ \mu\text{F}\) capacitor across a \(125\ \text{V}\) DC source. Determine current \(1\ \text{s}\) after closing the switch.

Solution.

\[ i = \dfrac{E}{R} e^{-\frac{t}{RC}} = \dfrac{125}{100000} e^{-\frac{1}{(100000)(100 \times 10^{-6})}} = 1.13\ \text{mA} \]

Answer: \(1.13\ \text{mA}\)

3.3 Instant Voltage

Instantaneous voltage drop describes voltage across resistor and capacitor during charging.

Resistor voltage starts at \(E\) and decays exponentially
Capacitor voltage starts at zero and rises exponentially to \(E\)
Sum of instantaneous voltages satisfies Kirchhoff’s Voltage Law (\(E_R + E_C = E\))

RC Instantaneous Voltages

\[ E_R = E e^{-\frac{t}{RC}} \]

\[ E_C = E\left[1 - e^{-\frac{t}{RC}}\right] \]

Where: \(E_R\) = resistor voltage drop (V), \(E_C\) = capacitor voltage (V), \(E\) = supply voltage (V), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F), \(t\) = time (s)

EE Board Exam October 1991

Example. A series RC circuit consisting of \(R = 2\ \text{M}\Omega\) and \(C = 5\ \mu\text{F}\) is connected across a \(100\ \text{V}\) DC source at \(t = 0\). Determine the voltage across the resistor \(5\ \text{seconds}\) later.

Solution.

\[ E_R = E e^{-\frac{t}{RC}} = 100 e^{-\frac{5}{(2 \times 10^6)(5 \times 10^{-6})}} = 60.65\ \text{V} \]

Answer: \(60.65\ \text{V}\)

3.4 Instant Power

Instantaneous power in an RC circuit dictates energy dissipation in resistor and electric field storage in capacitor.

Resistor power \(P_R\) decays exponentially at twice the rate (\(2t / RC\))
Capacitor power \(P_C\) rises to a peak and then decays to zero
Total power \(P_t\) equals power supplied by source

RC Instantaneous Powers

\[ P_R = \dfrac{E^2}{R} e^{-\frac{2t}{RC}} \]

\[ P_C = \dfrac{E^2}{R}\left[e^{-\frac{t}{RC}} - e^{-\frac{2t}{RC}}\right] \]

\[ P_t = \dfrac{E^2}{R} e^{-\frac{t}{RC}} \]

Where: \(P_R\) = resistor power (W), \(P_C\) = capacitor power (W), \(P_t\) = total power (W), \(E\) = supply voltage (V), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F), \(t\) = time (s)

3.5 Instant Charge: Initial, No Initial Charge

Instantaneous charge is the electric charge stored on capacitor plates at time \(t\).

Proportional to instantaneous capacitor voltage (\(Q = C E_C\))
Without initial charge, \(Q\) increases exponentially to \(C E\)
For source-free discharge (\(E = 0\)), \(Q\) decays exponentially from \(Q_0\)

RC Instantaneous Charge

\[ Q = C E\left[1 - e^{-\frac{t}{RC}}\right] \]

\[ Q = C E + (Q_0 - C E) e^{-\frac{t}{RC}} \]

\[ Q = Q_0 e^{-\frac{t}{RC}} \]

Where: \(Q\) = instantaneous charge (C), \(Q_0\) = initial charge (C), \(C\) = capacitance (F), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(t\) = time (s)

EE Board Exam October 1992

Example. A \(120\ \mu\text{F}\) capacitor has an initial charge of \(600\ \mu\text{C}\). It is discharged through a \(40\ \Omega\) resistor. How long will it take to discharge to \(150\ \mu\text{C}\)?

Solution.

\[ Q = Q_0 e^{-\frac{t}{RC}} \implies 150 \times 10^{-6} = 600 \times 10^{-6} e^{-\frac{t}{(40)(120 \times 10^{-6})}} \]

\[ t = -0.0048 \ln(0.25) = 6.65\ \text{ms} \]

Answer: \(6.65\ \text{ms}\)

4. RLC Series DC Circuit

RLC series DC circuit contains a resistor, an inductor, and a capacitor connected in series across a direct-current source.

Behavior governed by second-order differential equation
Response depends on relative values of \(R\), \(L\), and \(C\)
Auxiliary equation roots determine damping state

RLC Series Auxiliary Equation

\[ m^2 + \dfrac{R}{L} m + \dfrac{1}{LC} = 0 \]

\[ m_{1,2} = -\dfrac{R}{2L} \pm \sqrt{\left(\dfrac{R}{2L}\right)^2 - \dfrac{1}{LC}} \]

Where: \(m_1, m_2\) = roots of auxiliary equation, \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(C\) = capacitance (F)

4.1 Overdamped

Overdamped case occurs when resistance is large enough that roots of auxiliary equation are real and distinct.

Condition: \((R / 2L)^2 > 1 / LC\)
Current rises to a peak and returns to zero without oscillation
Roots \(m_1\) and \(m_2\) are negative real numbers

Overdamped RLC DC Current

\[ \left(\dfrac{R}{2L}\right)^2 > \dfrac{1}{LC} \]

\[ i = C_1 e^{m_1 t} + C_2 e^{m_2 t} \]

Where: \(i\) = current (A), \(m_1, m_2\) = real distinct roots, \(C_1, C_2\) = constants from initial conditions, \(t\) = time (s)

Example. A series RLC circuit with \(R = 1000\ \Omega\), \(L = 1\ \text{H}\), and \(C = 6.25\ \mu\text{F}\) is connected across \(24\ \text{V}\) DC. Find the current at \(t = 0.01\ \text{s}\).

Solution.

\[ m_{1,2} = -500 \pm \sqrt{500^2 - 1.6 \times 10^5} \implies m_1 = -200, m_2 = -800 \]

\[ i = 0.04 e^{-200(0.01)} - 0.04 e^{-800(0.01)} = 5.4\ \text{mA} \]

Answer: \(5.4\ \text{mA}\)

4.2 Critically Damped

Critically damped case occurs when roots of auxiliary equation are real and equal.

Condition: \((R / 2L)^2 = 1 / LC\)
Fastest non-oscillatory return of transient current to zero
Repeated root \(m = -R / 2L\)

Critically Damped RLC DC Current

\[ \left(\dfrac{R}{2L}\right)^2 = \dfrac{1}{LC} \]

\[ m = -\dfrac{R}{2L} \]

\[ i = (C_1 + C_2 t) e^{m t} \]

Where: \(i\) = current (A), \(m\) = repeated root, \(C_1, C_2\) = constants, \(t\) = time (s)

Example. A series circuit with \(R = 200\ \Omega\), \(L = 0.1\ \text{H}\), and \(C = 10\ \mu\text{F}\) is connected across \(100\ \text{V}\) DC. Determine current at \(t = 0.001\ \text{s}\).

Solution.

\[ m = -\dfrac{200}{2(0.1)} = -1000 \qquad\qquad C_1 = 0, C_2 = \dfrac{E}{L} = 1000 \]

\[ i = (1000)(0.001) e^{-1000(0.001)} = 0.368\ \text{A} \]

Answer: \(0.368\ \text{A}\)

4.3 Underdamped: General Equation of the Rate of Change of Current

Underdamped case occurs when resistance is small, causing roots to be complex conjugates and current to oscillate.

Condition: \((R / 2L)^2 < 1 / LC\)
Transient current exhibits damped sinusoidal oscillations
Initial rate of change of current depends on voltage and initial charge

Underdamped RLC DC Current and Rate of Change

\[ \alpha = -\dfrac{R}{2L} \qquad\qquad \beta = \sqrt{\dfrac{1}{LC} - \left(\dfrac{R}{2L}\right)^2} \]

\[ i = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t) \]

\[ \dfrac{di}{dt} = \dfrac{E \pm \dfrac{Q_0}{C}}{L} \]

Where: \(\alpha\) = attenuation constant, \(\beta\) = damped angular frequency (rad/s), \(\frac{di}{dt}\) = initial rate of change of current (A/s), \(Q_0\) = initial charge (C), \(L\) = inductance (H)

Example. A series circuit with \(R = 40\ \Omega\), \(L = 0.1\ \text{H}\), \(C = 50\ \mu\text{F}\) is connected across \(100\ \text{V}\) DC at \(t = 0\). Solve for current at \(t = 0.02\ \text{s}\).

Solution.

\[ \alpha = -200, \beta = 400 \implies i = 2.5 e^{-200(0.02)} \sin(400 \times 0.02) = 6.4\ \text{mA} \]

Answer: \(6.4\ \text{mA}\)

5. RL Series AC Circuit

RL series AC circuit transient occurs when an alternating sinusoidal voltage source \(e = E_m \sin(\omega t + \theta)\) is switched onto an RL circuit.

Current consists of steady-state AC component and transient DC offset
DC offset component decays exponentially with time constant \(L / R\)
Switching angle \(\theta\) determines initial magnitude of transient offset

RL Series AC Transient Current

\[ Z = \sqrt{R^2 + (\omega L)^2} \qquad\qquad \delta = \tan^{-1}\left(\dfrac{\omega L}{R}\right) \qquad\qquad I_m = \dfrac{E_m}{Z} \]

\[ i = I_m \sin(\omega t + \theta - \delta) - I_m \sin(\theta - \delta) e^{-\frac{R}{L}t} \]

Where: \(i\) = instantaneous current (A), \(I_m\) = peak current (A), \(E_m\) = peak voltage (V), \(Z\) = impedance (\(\Omega\)), \(\theta\) = switching angle (\(^\circ\)), \(\delta\) = power factor angle (\(^\circ\)), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)

EE Board Exam October 1999

Example. Given \(L \dfrac{di}{dt} + R i = 100 \sin(377 t)\) with \(L = 0.1\ \text{H}\), \(R = 10\ \Omega\), and \(i(0) = 0\). Determine the current at \(t = 0.01\ \text{s}\).

Solution.

\[ Z = 10 + j37.7 = 39 \angle 75.144^\circ\ \Omega \qquad\qquad I_m = \dfrac{100}{39} = 2.564\ \text{A} \]

\[ i = 2.564 \sin(377 \times 0.01 - 75.144^\circ) - 2.564 \sin(-75.144^\circ) e^{-\frac{10}{0.1}(0.01)} = 2.53\ \text{A} \]

Answer: \(2.53\ \text{A}\)

6. RC Series AC Circuit

RC series AC circuit transient occurs when an alternating voltage \(e = E_m \sin(\omega t + \theta)\) is applied across an RC circuit.

Current contains steady-state sinusoidal component and decaying exponential term
Decay rate depends on time constant \(RC\)
Transient current vanishes as circuit reaches AC steady state

RC Series AC Transient Current

\[ Z = \sqrt{R^2 + \left(\dfrac{1}{\omega C}\right)^2} \qquad\qquad \delta = \tan^{-1}\left(\dfrac{1}{\omega R C}\right) \qquad\qquad I_m = \dfrac{E_m}{Z} \]

\[ i = I_m \sin(\omega t + \theta + \delta) + \left[\dfrac{E_m \sin\theta}{R} - I_m \sin(\theta + \delta)\right] e^{-\frac{t}{RC}} \]

Where: \(i\) = instantaneous current (A), \(I_m\) = peak current (A), \(E_m\) = peak voltage (V), \(Z\) = impedance (\(\Omega\)), \(\theta\) = switching angle (\(^\circ\)), \(\delta\) = phase angle (\(^\circ\)), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F)

Example. A series RC circuit with \(R = 30\ \Omega\) and \(C = 250\ \mu\text{F}\) is connected across \(e = 100 \sin(100 t)\). Determine current at \(t = 0.01\ \text{s}\).

Solution.

\[ Z = 30 - j40 = 50 \angle -53.13^\circ\ \Omega \qquad\qquad I_m = \dfrac{100}{50} = 2\ \text{A} \]

\[ i = 2 \sin(100(0.01) + 53.13^\circ) - 2 \sin(53.13^\circ) e^{-\frac{0.01}{30(250 \times 10^{-6})}} = 1.45\ \text{A} \]

Answer: \(1.45\ \text{A}\)

7. RLC Series AC Circuit

RLC series AC circuit transient represents complete second-order system response energized by a sinusoidal AC voltage source.

Combination of natural transient response and forced AC steady-state response
Damping behavior determined by circuit parameters \(R\), \(L\), and \(C\)
Steady-state current amplitude determined by AC impedance \(Z\)

RLC Series AC Parameters

\[ Z = \sqrt{R^2 + \left(\dfrac{1}{\omega C} - \omega L\right)^2} \qquad\qquad \delta = \tan^{-1}\left|\dfrac{\dfrac{1}{\omega C} - \omega L}{R}\right| \qquad\qquad I_m = \dfrac{E_m}{Z} \]

Where: \(Z\) = impedance (\(\Omega\)), \(I_m\) = peak current (A), \(E_m\) = peak voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(C\) = capacitance (F), \(\omega\) = angular frequency (rad/s)

7.1 Overdamped

Overdamped RLC AC response occurs when \((R / 2L)^2 > 1 / LC\).

Transient component consists of two decaying real exponentials
No natural oscillations occur during transient phase
Current smoothly approaches steady-state AC waveform

Overdamped RLC AC Current

\[ i = C_1 e^{m_1 t} + C_2 e^{m_2 t} + I_m \sin(\omega t + \theta + \delta) \]

Where: \(i\) = current (A), \(m_1, m_2\) = roots, \(C_1, C_2\) = constants, \(I_m\) = steady-state peak current (A)

7.2 Critically Damped

Critically damped RLC AC response occurs when \((R / 2L)^2 = 1 / LC\).

Transient component decays rapidly without oscillation
Repeated root \(m = -R / 2L\)
Transitions into forced sinusoidal response in minimum time

Critically Damped RLC AC Current

\[ i = (C_1 + C_2 t) e^{m t} + I_m \sin(\omega t + \theta + \delta) \]

Where: \(i\) = current (A), \(m\) = repeated root, \(C_1, C_2\) = constants, \(I_m\) = peak current (A)

7.3 Underdamped: Equation for the Change of Current

Underdamped RLC AC response occurs when \((R / 2L)^2 < 1 / LC\).

Transient component exhibits damped natural oscillations
Natural frequency \(\beta\) superimposes on forced frequency \(\omega\)
Initial rate of current change depends on instantaneous AC voltage

Underdamped RLC AC Current and Rate of Change

\[ i = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t) + I_m \sin(\omega t + \theta + \delta) \]

\[ \dfrac{di}{dt} = \dfrac{E_m \sin\theta \pm \dfrac{Q_0}{C}}{L} \]

Where: \(i\) = current (A), \(\alpha\) = attenuation constant, \(\beta\) = damped angular frequency (rad/s), \(\frac{di}{dt}\) = initial rate of change of current (A/s), \(Q_0\) = initial charge (C), \(L\) = inductance (H)

EE Board Exam April 1999

Example. A series circuit has \(R = 10\ \Omega\), \(L = 0.1\ \text{H}\), and \(C = 0.0001\ \text{F}\) across \(e = 100 \sin(377 t)\). Solve for the amplitude of the resulting steady-state current.

Solution.

\[ X_L = 377(0.1) = 37.7\ \Omega \qquad\qquad X_C = \dfrac{1}{377(0.0001)} = 26.525\ \Omega \]

\[ Z = 10 + j(37.7 - 26.525) = 15 \angle 48.176^\circ\ \Omega \qquad\qquad I_m = \dfrac{100}{15} = 6.67\ \text{A} \]

Answer: \(6.67\ \text{A}\)

8. Transient Analysis by Laplace Transform

Laplace transform method transforms differential equations into algebraic equations in complex frequency domain \(s\) to simplify circuit transient analysis.

Introduced by French mathematician Pierre-Simon Laplace
Converts time-domain differential equations into \(s\)-domain algebraic equations
Incorporates initial conditions directly into algebraic formulation
Solution obtained by taking inverse Laplace transform back to time domain

Laplace Circuit Equations

\[ E_R(s) = R I(s) \]

\[ E_L(s) = L \left[s I(s) - i(0)\right] \]

\[ E_C(s) = \dfrac{1}{C}\left[\dfrac{I(s)}{s} \pm \dfrac{Q_0}{s}\right] \]

Where: \(I(s)\) = Laplace current, \(i(0)\) = initial current (A), \(Q_0\) = initial charge (C), \(s\) = complex frequency variable

Example. Given \(R i + L \dfrac{di}{dt} = 10 e^{-10 t}\) with \(R = 10\ \Omega\), \(L = 1\ \text{H}\), and \(i(0) = 0\). Solve for \(i(t)\) at \(t = 0.01\ \text{s}\).

Solution.

\[ 10 I(s) + 1(s I(s) - 0) = \dfrac{10}{s + 10} \implies I(s) = \dfrac{10}{(s + 10)^2} \implies i(t) = 10 t e^{-10 t} \]

\[ i(0.01) = 10(0.01) e^{-10(0.01)} = 90.5\ \text{mA} \]

Answer: \(90.5\ \text{mA}\)

9. Laplace Transform Formulas Applicable to Electric Circuit Analysis

Laplace transform table for electric circuits provides time-domain to complex frequency-domain conversions for electrical signals, components, and circuit equations.

Time Domain \(f(t)\) / Circuit Element	Complex Frequency Domain \(F(s)\) / \(s\)-Domain Model
Constant \(k\)	\(\dfrac{k}{s}\)
Polynomial \(k t^n\)	\(\dfrac{k n!}{s^{n+1}}\)
Exponential decay \(k e^{\mp at}\)	\(\dfrac{k}{s \mp a}\)
Damped polynomial \(k t^n e^{\mp at}\)	\(\dfrac{k n!}{(s \mp a)^{n+1}}\)
Sinusoidal function \(k \sin\omega t\)	\(\dfrac{k \omega}{s^2 + \omega^2}\)
Cosine function \(k \cos\omega t\)	\(\dfrac{k s}{s^2 + \omega^2}\)
Sine with phase shift \(\sin(\omega t + \theta)\)	\(\dfrac{s \sin\theta + \omega \cos\theta}{s^2 + \omega^2}\)
Cosine with phase shift \(\cos(\omega t + \theta)\)	\(\dfrac{s \cos\theta - \omega \sin\theta}{s^2 + \omega^2}\)
Damped sine wave \(e^{-at}\sin\omega t\)	\(\dfrac{\omega}{(s+a)^2 + \omega^2}\)
Damped cosine wave \(e^{-at}\cos\omega t\)	\(\dfrac{s+a}{(s+a)^2 + \omega^2}\)
Resistor voltage \(e_R(t) = R i(t)\)	\(E_R(s) = R I(s)\)
Inductor voltage \(e_L(t) = L \dfrac{di}{dt}\)	\(E_L(s) = L\left[s I(s) - i(0)\right]\)
Capacitor voltage \(e_C(t) = \dfrac{1}{C}\int i(t)\,dt\)	\(E_C(s) = \dfrac{1}{C}\left[\dfrac{I(s)}{s} \pm \dfrac{Q_0}{s}\right]\)
Resistor \(s\)-domain impedance	\(Z_R(s) = R\)
Inductor \(s\)-domain impedance	\(Z_L(s) = s L\)
Capacitor \(s\)-domain impedance	\(Z_C(s) = \dfrac{1}{s C}\)

Laplace Transformed Circuit Equations

\[ E_R(s) = R I(s) \]

\[ E_L(s) = L s I(s) - L i(0) \]

\[ E_C(s) = \dfrac{I(s)}{s C} \pm \dfrac{Q_0}{s C} \]

Where: \(E_R(s), E_L(s), E_C(s)\) = transformed element voltages (V), \(I(s)\) = transformed current (A), \(i(0)\) = initial inductor current (A), \(Q_0\) = initial capacitor charge (C), \(s\) = complex frequency variable

Example. An RL circuit governed by \(R i + L \dfrac{di}{dt} = 10 e^{-10 t}\) has \(R = 10\ \Omega\), \(L = 1\ \text{H}\), and zero initial current \(i(0) = 0\). Solve for \(i(t)\) at \(t = 0.01\ \text{s}\) using Laplace transform.

Solution.

\[ 10 I(s) + 1[s I(s) - 0] = \dfrac{10}{s + 10} \implies I(s) = \dfrac{10}{(s + 10)^2} \]

\[ i(t) = 10 t e^{-10 t} = 10(0.01) e^{-10(0.01)} = 0.0905\ \text{A} = 90.5\ \text{mA} \]

Answer: \(90.5\ \text{mA}\)

Example. An RC circuit with \(R = 1000\ \Omega\) and \(C = 100\ \mu\text{F}\) is energized by a \(100\ \text{V}\) DC source with an initial capacitor charge \(Q_0 = 2000\ \mu\text{C}\). Find the \(s\)-domain current expression \(I(s)\).

Solution.

\[ 1000 I(s) + \dfrac{1}{100 \times 10^{-6}}\left[\dfrac{I(s)}{s} - \dfrac{2000 \times 10^{-6}}{s}\right] = \dfrac{100}{s} \]

\[ 1000 I(s) + \dfrac{10000 I(s)}{s} = \dfrac{120}{s} \implies I(s) = \dfrac{0.12}{s + 10} \]

Answer: \(\dfrac{0.12}{s + 10}\)

EE Board Exam March 1998

Example. Determine the inverse Laplace transform of \(I(s) = \dfrac{200}{s^2 + 50s + 10625}\).

Solution.

\[ I(s) = \dfrac{200}{(s + 25)^2 + 100^2} = 2 \left[\dfrac{100}{(s + 25)^2 + 100^2}\right] \]

\[ i(t) = 2 e^{-25t} \sin 100t \]

Answer: \(2 e^{-25t} \sin 100t\)

10. Formula Summary

Formula Name	Formula	Variables
RL DC Series Current	\(i = \dfrac{E}{R}\left[1 - e^{-\dfrac{R}{L}t}\right]\)	\(i\) = current (A), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)
RL Circuit Time Constant	\(\text{TC} = \dfrac{L}{R}\)	\(\text{TC}\) = time constant (s), \(L\) = inductance (H), \(R\) = resistance (\(\Omega\))
RL Instantaneous Voltage Drops	\(E_R = E\left[1 - e^{-\dfrac{R}{L}t}\right]\) \(E_L = E e^{-\dfrac{R}{L}t}\)	\(E_R\) = resistor voltage (V), \(E_L\) = inductor voltage (V), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)
RL Instantaneous Powers	\(P_R = \dfrac{E^2}{R}\left[1 - e^{-\dfrac{R}{L}t}\right]^2\) \(P_L = \dfrac{E^2}{R}\left[e^{-\dfrac{R}{L}t} - e^{-\dfrac{2R}{L}t}\right]\) \(P_t = \dfrac{E^2}{R}\left[1 - e^{-\dfrac{R}{L}t}\right]\)	\(P_R\) = resistor power (W), \(P_L\) = inductor power (W), \(P_t\) = total power (W), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)
RC Circuit Time Constant	\(\text{TC} = R C\)	\(\text{TC}\) = time constant (s), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F)
RC Instantaneous Current	\(i = \dfrac{E}{R} e^{-\dfrac{t}{RC}}\) \(i = \left(\dfrac{E \pm \dfrac{Q_0}{C}}{R}\right) e^{-\dfrac{t}{RC}}\)	\(i\) = current (A), \(E\) = voltage (V), \(Q_0\) = initial charge (C), \(C\) = capacitance (F), \(R\) = resistance (\(\Omega\)), \(t\) = time (s)
RC Instantaneous Voltages	\(E_R = E e^{-\dfrac{t}{RC}}\) \(E_C = E\left[1 - e^{-\dfrac{t}{RC}}\right]\)	\(E_R\) = resistor voltage (V), \(E_C\) = capacitor voltage (V), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F), \(t\) = time (s)
RC Instantaneous Powers	\(P_R = \dfrac{E^2}{R} e^{-\dfrac{2t}{RC}}\) \(P_C = \dfrac{E^2}{R}\left[e^{-\dfrac{t}{RC}} - e^{-\dfrac{2t}{RC}}\right]\) \(P_t = \dfrac{E^2}{R} e^{-\dfrac{t}{RC}}\)	\(P_R\) = resistor power (W), \(P_C\) = capacitor power (W), \(P_t\) = total power (W), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F), \(t\) = time (s)
RC Instantaneous Charge	\(Q = C E\left[1 - e^{-\dfrac{t}{RC}}\right]\) \(Q = C E + (Q_0 - C E) e^{-\dfrac{t}{RC}}\) \(Q = Q_0 e^{-\dfrac{t}{RC}}\)	\(Q\) = charge (C), \(Q_0\) = initial charge (C), \(C\) = capacitance (F), \(E\) = voltage (V), \(R\) = resistance (\(\Omega\)), \(t\) = time (s)
RLC Series Auxiliary Equation	\(m^2 + \dfrac{R}{L} m + \dfrac{1}{LC} = 0\) \(m_{1,2} = -\dfrac{R}{2L} \pm \sqrt{\left(\dfrac{R}{2L}\right)^2 - \dfrac{1}{LC}}\)	\(m_1, m_2\) = auxiliary roots, \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(C\) = capacitance (F)
Overdamped RLC DC Current	\(\left(\dfrac{R}{2L}\right)^2 > \dfrac{1}{LC}\) \(i = C_1 e^{m_1 t} + C_2 e^{m_2 t}\)	\(i\) = current (A), \(m_1, m_2\) = roots, \(C_1, C_2\) = constants, \(t\) = time (s)
Critically Damped RLC DC Current	\(\left(\dfrac{R}{2L}\right)^2 = \dfrac{1}{LC}\) \(m = -\dfrac{R}{2L}\) \(i = (C_1 + C_2 t) e^{m t}\)	\(i\) = current (A), \(m\) = repeated root, \(C_1, C_2\) = constants, \(t\) = time (s)
Underdamped RLC DC Current and Rate of Change	\(\alpha = -\dfrac{R}{2L}\) \(\beta = \sqrt{\dfrac{1}{LC} - \left(\dfrac{R}{2L}\right)^2}\) \(i = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t)\) \(\dfrac{di}{dt} = \dfrac{E \pm \dfrac{Q_0}{C}}{L}\)	\(\alpha\) = attenuation constant, \(\beta\) = frequency (rad/s), \(i\) = current (A), \(\dfrac{di}{dt}\) = rate of change (A/s), \(Q_0\) = initial charge (C), \(L\) = inductance (H)
RL Series AC Transient Current	\(Z = \sqrt{R^2 + (\omega L)^2}\) \(\delta = \tan^{-1}\left(\dfrac{\omega L}{R}\right)\) \(I_m = \dfrac{E_m}{Z}\) \(i = I_m \sin(\omega t + \theta - \delta) - I_m \sin(\theta - \delta) e^{-\dfrac{R}{L}t}\)	\(i\) = current (A), \(I_m\) = peak current (A), \(E_m\) = peak voltage (V), \(Z\) = impedance (\(\Omega\)), \(\theta\) = switching angle (\(^\circ\)), \(\delta\) = power factor angle (\(^\circ\)), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(t\) = time (s)
RC Series AC Transient Current	\(Z = \sqrt{R^2 + \left(\dfrac{1}{\omega C}\right)^2}\) \(\delta = \tan^{-1}\left(\dfrac{1}{\omega R C}\right)\) \(I_m = \dfrac{E_m}{Z}\) \(i = I_m \sin(\omega t + \theta + \delta) + \left[\dfrac{E_m \sin\theta}{R} - I_m \sin(\theta + \delta)\right] e^{-\dfrac{t}{RC}}\)	\(i\) = current (A), \(I_m\) = peak current (A), \(E_m\) = peak voltage (V), \(Z\) = impedance (\(\Omega\)), \(\theta\) = switching angle (\(^\circ\)), \(\delta\) = phase angle (\(^\circ\)), \(R\) = resistance (\(\Omega\)), \(C\) = capacitance (F)
RLC Series AC Parameters	\(Z = \sqrt{R^2 + \left(\dfrac{1}{\omega C} - \omega L\right)^2}\) \(\delta = \tan^{-1}\left\|\dfrac{\dfrac{1}{\omega C} - \omega L}{R}\right\|\) \(I_m = \dfrac{E_m}{Z}\)	\(Z\) = impedance (\(\Omega\)), \(I_m\) = peak current (A), \(E_m\) = peak voltage (V), \(R\) = resistance (\(\Omega\)), \(L\) = inductance (H), \(C\) = capacitance (F), \(\omega\) = angular frequency (rad/s)
Overdamped RLC AC Current	\(i = C_1 e^{m_1 t} + C_2 e^{m_2 t} + I_m \sin(\omega t + \theta + \delta)\)	\(i\) = current (A), \(m_1, m_2\) = roots, \(C_1, C_2\) = constants, \(I_m\) = peak current (A)
Critically Damped RLC AC Current	\(i = (C_1 + C_2 t) e^{m t} + I_m \sin(\omega t + \theta + \delta)\)	\(i\) = current (A), \(m\) = repeated root, \(C_1, C_2\) = constants, \(I_m\) = peak current (A)
Underdamped RLC AC Current and Rate of Change	\(i = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t) + I_m \sin(\omega t + \theta + \delta)\) \(\dfrac{di}{dt} = \dfrac{E_m \sin\theta \pm \dfrac{Q_0}{C}}{L}\)	\(i\) = current (A), \(\alpha\) = attenuation constant, \(\beta\) = frequency (rad/s), \(\dfrac{di}{dt}\) = rate of change (A/s), \(Q_0\) = initial charge (C), \(L\) = inductance (H)
Laplace Circuit Equations	\(E_R(s) = R I(s)\) \(E_L(s) = L \left[s I(s) - i(0)\right]\) \(E_C(s) = \dfrac{1}{C}\left[\dfrac{I(s)}{s} \pm \dfrac{Q_0}{s}\right]\)	\(I(s)\) = Laplace current, \(i(0)\) = initial current (A), \(Q_0\) = initial charge (C), \(s\) = Laplace variable
Core Laplace Transform Pairs	\(\mathcal{L}\{k\} = \dfrac{k}{s}\) \(\mathcal{L}\{e^{-at}\} = \dfrac{1}{s + a}\) \(\mathcal{L}\{\sin\omega t\} = \dfrac{\omega}{s^2 + \omega^2}\) \(\mathcal{L}\{\cos\omega t\} = \dfrac{s}{s^2 + \omega^2}\) \(\mathcal{L}^{-1}\left\{\dfrac{k}{(s + a)^2 + k^2}\right\} = e^{-at}\sin kt\)	\(s\) = Laplace variable, \(a\) = constant, \(\omega, k\) = angular frequency (rad/s), \(t\) = time (s)