Electrical Circuit Analysis

Chapter 6
Capacitors, Inductors and Energy Storage

Bachelors Program in Electrical Engineering
Electrical Circuit Analysis   •   Bachelor

Capacitors, Inductors and Energy Storage

Introduction

Resistors dissipate electrical energy, whereas capacitors and inductors can store energy and later return it to a circuit. This distinction makes capacitors and inductors fundamental to filters, power converters, timing circuits, communication systems, electric drives and transient protection.

Their stored energy also gives circuits memory. A capacitor voltage or inductor current immediately after switching generally depends on the circuit state immediately before switching. Correct transient analysis therefore requires both circuit laws and initial conditions.

This chapter treats ideal, linear, time-invariant capacitors and inductors unless stated otherwise. Practical limitations—including losses, parasitic effects, saturation and safety—are considered later.

Learning Objectives

After completing this chapter, students should be able to:

  • derive capacitor and inductor voltage–current relationships;
  • apply passive sign convention to power and energy calculations;
  • determine energy stored in capacitors and inductors;
  • identify voltage and current continuity constraints;
  • calculate initial conditions before and after switching;
  • construct instantaneous and steady-state equivalent circuits;
  • analyze first-order RC and RL transients;
  • combine capacitors or inductors in series and parallel; and
  • account for important nonideal and safety considerations.

1. Energy-Storing Circuit Elements

An ideal capacitor stores energy in an electric field. An ideal inductor stores energy in a magnetic field. Neither consumes net energy over a complete operating cycle if all stored energy is returned to the circuit.

Under the passive sign convention, current enters the terminal marked positive for the referenced voltage. The instantaneous absorbed power is

p(t)=v(t)i(t)

where p is power in watts (W), v is voltage in volts (V), and i is current in amperes (A). Positive power means that the element absorbs energy; negative power means that it delivers energy.

Energy absorbed between times t1 and t2 is

W1\to2=\intt1t2p(t)\,dt

where W is measured in joules (J).

2. The Capacitor

2.1 Charge–voltage relationship

A capacitor consists physically of two conductors separated by an insulating material or dielectric. Equal and opposite charges accumulate on the conductors.

For an ideal linear capacitor,

q(t)=C vC(t)

where:

  • q is charge on the positively referenced plate, in coulombs (C);
  • C is capacitance, in farads (F);
  • vC is voltage from the positive terminal to the negative terminal, in volts.

Because 1\,F=1\,C/V, the equation is dimensionally consistent.

Capacitance depends on geometry and material. For an ideal parallel-plate capacitor, neglecting fringing,

C=\frac{\varepsilon A}{d}

where \varepsilon is dielectric permittivity in F/m, A is overlapping plate area in m², and d is plate separation in m.

2.2 Derivation of the current–voltage relationship

Current is the time rate of charge transfer:

iC(t)=\frac{dq(t)}{dt}

For a constant capacitance,

iC(t)=C\frac{dvC(t)}{dt}

This equation assumes passive sign convention: iC enters the terminal assigned positive vC. It also assumes that C is constant. If capacitance varies with time or voltage, the more general relationship is

iC(t)=\frac{d[C(t)vC(t)]}{dt}

For a constant capacitor, integration gives

vC(t)=vC(t0)+\frac{1}{C}\intt0tiC(\tau)\,d\tau

Thus capacitor voltage depends on its previous voltage and accumulated charge after t0. The variable \tau is a dummy integration variable.

A constant capacitor voltage produces zero current. Consequently, an ideal capacitor behaves as an open circuit in DC steady state—not necessarily during switching or while voltage is changing.

2.3 Energy stored in a capacitor

Using p=vCiC and iC=C\,dvC/dt,

pC(t)=C vC(t)\frac{dvC(t)}{dt}

The change in stored energy as voltage increases from v1 to v2 is

WC(v2)-WC(v1)=\intv1v2C v\,dv
=\frac{1}{2}C(v22-v12)

Taking zero voltage as the zero-energy reference gives

WC=\frac{1}{2}C vC2

The stored energy is nonnegative and is independent of voltage polarity. Reversing voltage reverses charge but does not make stored energy negative.

For a nonlinear capacitor described by q=q(v), the general stored energy is

WC(v)=\int0v\xi\,\frac{dq(\xi)}{d\xi}\,d\xi

The familiar Cv2/2 formula applies only when C is constant.

Worked Example 1: Capacitor current and energy

A 20\,\mu F capacitor has voltage

vC(t)=2+400t\quad \text{V},\qquad 0\le t\le5\text{ ms}

The current is

iC=C\frac{dvC}{dt}=(20\times10-6)(400)=8.0\text{ mA}

At t=5\,ms, the voltage is 2+400(0.005)=4\,V. Initial and final energies are

WC(0)=\frac{1}{2}(20\times10-6)(22)=40\,\mu J
WC(5\text{ ms})=\frac{1}{2}(20\times10-6)(42)=160\,\mu J

The capacitor absorbs 120\,\mu J during the interval.

3. The Inductor

3.1 Flux linkage and current

An inductor stores energy in the magnetic field established by current through a conducting winding. Its electromagnetic state is described by flux linkage \lambda, measured in weber-turns (Wb·turn).

For an ideal linear inductor,

\lambda(t)=L iL(t)

where L is inductance in henries (H) and iL is current in amperes. Since 1\,H=1\,V\!\cdot\!s/A, inductance relates voltage-time integral to current.

Faraday’s law gives the referenced terminal voltage:

vL(t)=\frac{d\lambda(t)}{dt}

For constant inductance,

vL(t)=L\frac{diL(t)}{dt}

This expression uses passive sign convention: current enters the terminal assigned positive voltage. Integration gives

iL(t)=iL(t0)+\frac{1}{L}\intt0tvL(\tau)\,d\tau

A constant inductor current produces zero voltage. Therefore, an ideal inductor behaves as a short circuit in DC steady state, provided its current has become constant.

If inductance depends on current, a more general magnetic relationship must be used:

vL=\frac{d\lambda(iL)}{dt}=\frac{d\lambda}{diL}\frac{diL}{dt}

The differential inductance is d\lambda/diL and need not equal \lambda/iL.

3.2 Energy stored in an inductor

For a constant linear inductor,

pL=vLiL=L iL\frac{diL}{dt}

Integrating from current i1 to i2 gives

WL(i2)-WL(i1)=\inti1i2L i\,di
=\frac{1}{2}L(i22-i12)

Taking zero current as the zero-energy reference,

WL=\frac{1}{2}L iL2

For a nonlinear magnetic element,

WL(i)=\int0i \lambda(\eta)\,d\eta

where \eta is a dummy current variable. Magnetic hysteresis may prevent all input energy from being recoverable, so this conservative energy expression must be applied cautiously to real ferromagnetic devices.

Worked Example 2: Inductor voltage and energy

A 50\,mH inductor carries

iL(t)=3e-200t\quad \text{A},\qquad t\ge0

Its voltage under passive sign convention is

vL(t)=L\frac{diL}{dt}
=(0.05)(-600e-200t)=-30e-200t\text{ V}

The negative voltage indicates that the inductor is delivering energy as its current decreases. Its initial stored energy is

WL(0)=\frac{1}{2}(0.05)(32)=0.225\text{ J}

As t\to\infty, the current and stored energy approach zero.

4. Continuity and Initial Conditions

Let t0^- denote the instant immediately before switching and t0^+ the instant immediately after switching.

4.1 Capacitor-voltage continuity

From the integral capacitor relationship,

vC(t0^+)-vC(t0^-)=\frac{1}{C}\intt0^-t0^+iC(t)\,dt

If capacitor current remains finite during an infinitesimally short switching interval, the integral is zero. Hence

vC(t0^+)=vC(t0^-)

Capacitor voltage cannot jump under finite current. A voltage jump is mathematically possible only if the capacitor receives an impulse of current carrying finite charge.

Capacitor current is not generally continuous; it may change abruptly when the surrounding circuit changes.

4.2 Inductor-current continuity

Similarly,

iL(t0^+)-iL(t0^-)=\frac{1}{L}\intt0^-t0^+vL(t)\,dt

If inductor voltage remains finite over the switching instant,

iL(t0^+)=iL(t0^-)

Inductor current cannot jump under finite voltage. A current jump requires a voltage impulse with finite volt-second area. Inductor voltage itself may change abruptly.

4.3 Determining pre-switch conditions

If a circuit has been connected to constant independent sources for a sufficiently long time, it is in DC steady state. Then:

  • replace each ideal capacitor by an open circuit;
  • replace each ideal inductor by a short circuit;
  • solve for vC(0^-) and iL(0^-);
  • apply continuity to obtain vC(0^+) and iL(0^+);
  • analyze the post-switch circuit.

The open- and short-circuit substitutions are valid only for DC steady state. They must not be used merely because the time is labeled 0^-.

5. Equivalent Circuits at a Switching Instant

At t0^+, a capacitor with initial voltage V0 may be represented instantaneously by an ideal voltage source of value V0 with exactly the same terminal polarity. An inductor with initial current I0 may be represented instantaneously by an ideal current source of value I0 in the same reference direction. These models enforce the initial states; they are not valid replacements for all later times.

Switching-instant equivalent circuits Physical element before/at switching Equivalent at t₀⁺ Capacitor a b +− v_C(0⁻)=V₀ measured a to b a b +− V₀ Inductor a b i_L(0⁻)=I₀ a b I₀ from a to b
Capacitor and inductor switching-instant equivalent circuits with terminal-aligned polarity and current references
Figure: Capacitor and inductor switching-instant equivalent circuits with terminal-aligned polarity and current references

Figure: Instantaneous equivalents. Capacitor polarity is explicitly defined from terminal a to terminal b; the inductor current direction is likewise aligned with its terminals.

A useful distinction is therefore:

ConditionCapacitor modelInductor model
DC steady stateOpen circuitShort circuit
At t0^+ with known stateVoltage source vC(t0^-)Current source iL(t0^-)
General time-varying operationi=C\,dv/dtv=L\,di/dt

Topology can create idealized inconsistencies. For example, directly connecting capacitors having unequal voltages predicts impulsive current in a zero-resistance network. Likewise, interrupting an ideal inductor current with no alternative path predicts unbounded voltage. Real resistance, arcing, parasitic capacitance and switching speed limit these effects.

6. First-Order RC and RL Responses

6.1 General first-order form

A first-order circuit contains one independent energy-storage state after all algebraic constraints are considered. Its natural response has the form

x(t)=x(\infty)+[x(0^+)-x(\infty)]e-t/\tau,\qquad t\ge0

Here x is capacitor voltage or inductor current, x(\infty) is the final value, and \tau is the time constant.

This form applies to linear, time-invariant first-order circuits with constant sources after switching. It is not a universal solution for nonlinear or time-varying circuits.

6.2 RC circuits

For a capacitor seeing a Thévenin resistance Rth, with independent sources deactivated when finding resistance,

\tauRC=RthC

Independent voltage sources are replaced by shorts and independent current sources by opens. Dependent sources remain active; a test source may be required to determine Rth.

The capacitor voltage is

vC(t)=vC(\infty)+[vC(0^+)-vC(\infty)]e^{-t/(RthC)}

Worked Example 3: RC switching response

A 50\,\mu F capacitor initially has vC(0^-)=-3\,V. At t=0, it is connected through 4.0\,k\Omega to a 12\,V source, with the capacitor reference polarity matching the source.

Continuity gives vC(0^+)=-3\,V. The final voltage and time constant are

vC(\infty)=12\text{ V},\qquad \tau=(4000)(50\times10-6)=0.200\text{ s}

Therefore,

vC(t)=12-15e-t/0.200\text{ V}

The initial current is

iC(0^+)=\frac{12-(-3)}{4000}=3.75\text{ mA}

At t=0.400\,s=2\tau,

vC=12-15e-2\approx9.97\text{ V}

6.3 RL circuits

For an inductor seeing Thévenin resistance Rth,

\tauRL=\frac{L}{Rth}

The inductor current is

iL(t)=iL(\infty)+[iL(0^+)-iL(\infty)]e^{-tRth/L}

Worked Example 4: RL current decay

A 200\,mH inductor carrying 1.5\,A is switched into a 40\,\Omega resistor with no independent source. Assuming an ideal inductor,

\tau=\frac{0.200}{40}=5.0\text{ ms}

and

iL(t)=1.5e^{-t/(5.0\text{ ms})}\text{ A}

At t=10\,ms, iL=1.5e-2\approx0.203\,A. The initial magnetic energy is

WL(0)=\frac{1}{2}(0.200)(1.52)=0.225\text{ J}

With no other losses, this energy is ultimately dissipated in the resistor.

7. Combining Capacitors and Inductors

For ideal uncoupled elements, equivalent values follow from Kirchhoff’s laws.

ConnectionCapacitorsUncoupled inductors
ParallelCeq=\sum Ck1/Leq=\sum 1/Lk
Series1/Ceq=\sum 1/CkLeq=\sum Lk

These formulas require that the elements genuinely share the specified series or parallel relationship. Series capacitors can also possess trapped initial charge at internal nodes, so a single equivalent capacitor may not preserve every internal voltage. Inductor formulas assume negligible mutual inductive coupling. Coupled inductors require mutual-inductance terms.

For two capacitors in series,

Ceq=\frac{C1C2}{C1+C2}

For two uncoupled inductors in parallel,

Leq=\frac{L1L2}{L1+L2}

8. Frequency-Domain and Simulation Models

With zero initial conditions, sinusoidal or Laplace-domain impedances are

ZC(s)=\frac{1}{sC},\qquad ZL(s)=sL

where s is the complex frequency variable in s⁻¹. For sinusoidal steady state, set s=j\omega, where \omega is angular frequency in rad/s.

Initial conditions must not be discarded in switching analysis. Using the one-sided Laplace transform,

IC(s)=sCVC(s)-CvC(0^-)
VL(s)=sLIL(s)-LiL(0^-)

These equations motivate initial-condition source models used in circuit analysis software. Numerical simulators commonly replace derivatives with discrete-time companion models. Smaller time steps generally improve transient resolution but increase computation. Abrupt ideal switching can produce convergence difficulties; realistic parasitic resistance and capacitance often improve both physical fidelity and numerical stability.

9. Applied Case: Relay-Coil Turn-Off Protection

Background

This is an illustrative engineering case. A low-side transistor drives a 24\,V relay coil modeled by R=60\,\Omega and L=120\,mH. A flyback diode is connected across the coil and is reverse-biased during normal operation. Its forward drop during turn-off is approximated as 0.8\,V.

Problem

When the transistor opens, the coil current cannot instantly become zero. Without an alternative path, the inductor would generate a potentially destructive voltage across the transistor.

Analysis

After a long energized interval, the inductor is a short circuit in DC steady state. Therefore,

I0=\frac{24}{60}=0.400\text{ A}

The stored magnetic energy is

WL=\frac{1}{2}(0.120)(0.4002)=9.60\text{ mJ}

Immediately after turn-off, continuity requires iL(0^+)=0.400\,A. The diode becomes forward-biased and provides a recirculation path. With constant diode drop, Kirchhoff’s voltage law gives

L\frac{di}{dt}+Ri+VD=0

The current while the diode conducts is

i(t)=\left(I0+\frac{VD}{R}\right)e-Rt/L-\frac{VD}{R}

This equation applies until the current reaches zero. Substituting the values gives the conduction duration

toff=\frac{L}{R}\ln\left(\frac{I0+VD/R}{VD/R}\right)
toff=\frac{0.120}{60}\ln\left(\frac{0.400+0.8/60}{0.8/60}\right)
\approx6.87\text{ ms}

Approach and outcome

The flyback diode limits the inductive overvoltage and protects the transistor. The initial 9.60\,mJ is dissipated in the winding resistance and diode. The trade-off is slower current decay and therefore slower relay release. A Zener or transient-voltage suppressor can permit a higher controlled voltage, causing faster release, but the transistor must be rated for the resulting voltage and energy.

Learning points

  • Inductor current continuity determines the immediate post-switch current.
  • Stored energy must be safely transferred or dissipated.
  • Protection-component voltage, current, energy and thermal ratings all matter.
  • A protection method that minimizes voltage may not minimize release time.

10. Practical Limitations and Engineering Practice

Real components deviate from ideal models.

Capacitor nonidealities

  • Equivalent series resistance causes heating and voltage loss.
  • Leakage resistance allows stored charge to decay.
  • Equivalent series inductance limits high-frequency performance.
  • Capacitance may depend on voltage, temperature and frequency.
  • Electrolytic capacitors usually require correct polarity.
  • Dielectric absorption can cause voltage to reappear after discharge.

Inductor nonidealities

  • Winding resistance dissipates power.
  • Core losses include hysteresis and eddy-current losses.
  • Magnetic saturation reduces incremental inductance at high current.
  • Parasitic capacitance creates self-resonance.
  • Coupling to nearby circuits can produce electromagnetic interference.

Safety, governance and sustainability

Stored energy remains hazardous even after external power is removed. High-voltage capacitors require discharge resistors, insulated enclosures, warning labels and verified safe discharge procedures. Shorting a charged capacitor directly can cause arcing, high current and component damage.

Inductive loads require appropriately rated flyback diodes, snubbers or surge suppressors. Component selection should comply with applicable insulation, creepage, clearance, fire-safety and electromagnetic-compatibility requirements. Engineers should also consider capacitor lifetime, material sourcing, recyclability and proper disposal of electronic waste.

Recommended workflow

  1. Define voltage polarities and current directions explicitly.
  2. Analyze the pre-switch circuit and calculate stored states.
  3. Apply continuity only under the stated finite-excitation conditions.
  4. Construct the post-switch circuit with correct initial-condition equivalents.
  5. Determine final values and the resistance seen by the storage element.
  6. Calculate the time constant and transient response.
  7. Check units, limiting values and energy balance.
  8. Verify voltage, current, power, thermal and safety ratings.

Chapter Summary

Capacitors and inductors store energy in electric and magnetic fields, respectively. For constant linear elements, their governing relationships are iC=C\,dvC/dt and vL=L\,diL/dt. Their stored energies are WC=CvC2/2 and WL=LiL2/2.

A capacitor voltage is continuous unless an impulse current supplies finite charge; an inductor current is continuous unless an impulse voltage supplies finite flux linkage. These constraints determine post-switch initial conditions. In DC steady state, capacitors act as open circuits and inductors as short circuits. At the switching instant, a charged capacitor can be represented by a voltage source with matching terminal polarity, while an energized inductor can be represented by a current source with matching direction.

Linear first-order RC and RL circuits exhibit exponential responses governed by \tau=RC and \tau=L/R. Practical designs must also account for losses, parasitic effects, saturation, component ratings, transient protection and residual stored-energy hazards.

Key Terms

Capacitance: The ratio of charge to voltage for a linear capacitor.

Capacitor: A circuit element that stores energy in an electric field.

Charge: Electrical quantity measured in coulombs.

Continuity condition: A constraint preventing an instantaneous state change under finite excitation.

DC steady state: A condition in which circuit variables are constant with time.

Dielectric: Insulating material between capacitor conductors.

Equivalent series resistance: A capacitor’s effective internal series loss resistance.

Flux linkage: Magnetic flux linked by a winding, measured in weber-turns.

Inductance: The ratio of flux linkage to current for a linear inductor.

Inductor: A circuit element that stores energy in a magnetic field.

Initial condition: The capacitor voltage or inductor current at the beginning of an analysis interval.

Natural response: Response caused solely by initially stored energy.

Passive sign convention: Reference convention in which current enters the positive-voltage terminal.

Saturation: Magnetic condition in which increasing current produces relatively little additional flux.

State variable: A variable, such as capacitor voltage or inductor current, describing stored energy.

Time constant: Parameter governing the exponential rate of a first-order response.

Transient response: Time-varying behavior following a switching event or disturbance.

Flyback diode: A diode that provides a safe current path for an inductive load during turn-off.

Review Questions

Short Answer Questions

  1. State the voltage–current relationship of a constant linear capacitor under passive sign convention.
  2. Why does an ideal inductor behave as a short circuit in DC steady state?
  3. Under what condition can capacitor voltage change instantaneously?
  4. What do t0^- and t0^+ represent?
  5. Distinguish stored energy from instantaneous power.

Long Answer Questions

  1. Derive the energy stored in a linear capacitor from its current–voltage relationship.
  2. Derive the energy stored in a linear inductor and interpret the sign of its power during discharge.
  3. Explain how pre-switch analysis, continuity and post-switch equivalent circuits are combined in transient analysis.
  4. Derive the complete response of a source-driven series RC circuit with an arbitrary initial capacitor voltage.
  5. Compare ideal and practical capacitors and inductors, including losses, frequency limitations and safety implications.

Analytical / Application Questions

  1. A 100\,\mu F capacitor initially at 5\,V is connected through 2\,k\Omega to a 15\,V source. Determine its voltage and current at t=0^+ and t=0.4\,s.
  2. A 0.5\,H inductor carrying 2\,A discharges through 25\,\Omega. Determine the initial energy, time constant and time required for current to fall to 0.2\,A.
  3. Design a conceptual turn-off protection scheme for a DC solenoid. Explain how the selected clamp voltage affects semiconductor stress, energy dissipation and release time.

Multiple Choice Questions with Answers

Question 1

For a constant linear capacitor under passive sign convention, which relationship is correct?

A. vC=C\,diC/dt

B. iC=C\,dvC/dt

C. iC=vC/C

D. vC=CiC

Correct Answer: B. iC=C\,dvC/dt

Question 2

A 10\,\mu F capacitor is charged to 20\,V. How much energy does it store?

A. 0.2\,mJ

B. 1.0\,mJ

C. 2.0\,mJ

D. 4.0\,mJ

Correct Answer: C. 2.0\,mJ

Question 3

An ideal 0.25\,H inductor carries 4\,A. Its stored energy is:

A. 0.5\,J

B. 1.0\,J

C. 2.0\,J

D. 4.0\,J

Correct Answer: C. 2.0\,J

Question 4

If capacitor current remains finite at a switching instant, which statement is correct?

A. Capacitor current must be continuous.

B. Capacitor voltage must become zero.

C. Capacitor voltage is continuous.

D. Capacitor voltage equals its final value immediately.

Correct Answer: C. Capacitor voltage is continuous.

Question 5

In DC steady state, an ideal inductor is replaced by which equivalent?

A. An open circuit

B. A short circuit

C. A voltage source

D. A finite resistor equal to its inductance

Correct Answer: B. A short circuit

Question 6

A 40\,\mu F capacitor is connected to a resistance of 5\,k\Omega. What is the RC time constant?

A. 0.008\,s

B. 0.020\,s

C. 0.200\,s

D. 2.00\,s

Correct Answer: C. 0.200\,s

Question 7

A 150\,mH inductor sees a resistance of 30\,\Omega. What is the RL time constant?

A. 2\,ms

B. 5\,ms

C. 20\,ms

D. 200\,ms

Correct Answer: B. 5\,ms

Question 8

Immediately after switching, an inductor with initial current I0 may be represented instantaneously by:

A. An open circuit

B. A voltage source of value LI0

C. A current source of value I0 in the same reference direction

D. A resistor of value L/I0

Correct Answer: C. A current source of value I0 in the same reference direction

Question 9

A capacitor voltage changes at a constant rate of 300\,V/s. If C=20\,\mu F, what current enters its positive terminal?

A. 1.5\,mA

B. 6.0\,mA

C. 15\,mA

D. 60\,mA

Correct Answer: B. 6.0\,mA

Question 10

Why is a flyback diode placed across a DC relay coil?

A. To make the steady-state coil current zero

B. To increase coil inductance during energization

C. To provide a path for continuous inductor current and limit turn-off voltage

D. To make inductor current change instantaneously

Correct Answer: C. To provide a path for continuous inductor current and limit turn-off voltage

Electrical Circuit Analysis
Chapter 6 of 15