Inductors: What Are They? (Worked Examples Included)

Contents

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Key learnings:

Inductor Definition: An inductor is a component that stores energy in a magnetic field when electric current flows through it.
Series Inductors: When inductors are connected in series, their total inductance is affected by mutual inductance.
Parallel Inductors: Inductors in parallel can either aid or oppose each other, influencing the overall inductance.
Voltage Across an Inductor: The voltage across an inductor is proportional to how quickly the current through it changes.
Example Problems: Example problems help in understanding how to calculate inductance in series and parallel configurations.

What is an Inductor?

Series and parallel inductor networks combine two or more coils to obtain an equivalent inductance. Each inductor is a two-terminal passive electrical element that stores energy in the form of a magnetic field when electric current flows. Depending on its use, it may be called a coil, choke or reactor.

For uncoupled inductors, inductance adds in series and follows the reciprocal rule in parallel. The component itself is a coil of conducting material, usually insulated copper, wound around air, a non-magnetic former or a ferromagnetic material.

Inductors are made across a wide range of values, including microhenry values where 1 µH equals 10^-6 H. A ferrite or iron core can concentrate the magnetic field and raise inductance, although the core also sets saturation, loss and frequency limits.

Under Faraday’s law of electromagnetic induction, a changing current produces a changing magnetic flux and an induced electromotive force. The resulting voltage is proportional to the rate of change of current.

Inductance (L) relates induced voltage to the rate at which current changes. For the same current, a larger inductance stores more magnetic energy.

How Do Inductors Work?

An inductor responds to a change in current by developing a voltage whose polarity opposes that change. The following circuit illustrates this behaviour.

inductor working — Inductor Working in a Circuit

The circuit contains a lamp, an inductor and a switch connected to a battery. Without the inductor branch, the lamp follows the switch directly. The inductor adds a current transient.

The inductor winding has much less resistance than the lamp. After DC conditions settle, most current takes the low-resistance coil branch and the lamp becomes dim.

When the switch closes, the lamp first glows brightly and then dims as current builds in the coil. When the switch opens, stored magnetic energy briefly drives current through the lamp before it goes out.

Current in the coil establishes a magnetic field. As that field changes, it induces a voltage with the polarity required by Lenz’s law to oppose the change in current.

The induced voltage opposes both an increase and a decrease in current. Once the DC current reaches a steady value, the ideal inductive voltage falls to zero.

When the switch opens, the collapsing magnetic field maintains current for a short time. That discharge current keeps the lamp glowing briefly.

The inductor stores magnetic energy while current flows. In an ideal circuit, its current cannot change instantaneously because an instantaneous change would require unbounded voltage.

Inductor Circuit Symbol

The following images show the schematic symbol for an inductor.

Inductor Equation

Voltage Across an Inductor

For a constant ideal inductance, voltage is proportional to the rate of change of current:

$\begin{align*} v_L = L \frac{di_L}{dt} \end{align*}$

where $v_L$ is the instantaneous voltage across the inductor in volts,

$L$ is inductance in henries,

$\frac{di_L}{dt}$ is the rate of change of current in amperes per second.

The sign of the voltage depends on the reference direction and whether current is rising or falling.

For steady d.c. current, $\frac{di_L}{dt}$ is zero and the voltage across an ideal inductor is zero. A real winding still has resistance, so only the ideal inductor becomes a short circuit in DC steady state.

Current Through an Inductor

Integrating the voltage relation expresses inductor current as:

$\begin{align*} i_L = \frac{1}{L} \int v_L dt \end{align*}$

The integration limits account for the current history or initial condition from $-\infty \,\, to \,\, t(0^-)$ .

If switching occurs at t=0, the current equation becomes:

$\begin{align*} i_L = \frac{1}{L} \int v_L dt \end{align*}$

Split the integral into $-\infty \,\, to \,\, 0$ and $0 \,\, to \,\,t$ . The symbol $0^-$ is the instant before switching, while $0^+$ is the instant after switching. Therefore:

$\begin{align*} i_L = \frac{1}{L} \int_{-\infty}^{t} v_L dt \end{align*}$

Therefore,

$\begin{align*} i_L = \frac{1}{L} \int_{-\infty}^{0^-} v_L dt + \frac{1}{L} \int_{0^-}^{t} v_L dt \end{align*}$

The term $\frac{1}{L} \int_{-\infty}^{0^-} v_L dt$ represents the current $i_L$ established before switching. Denote its initial value by $i_L(0^-)$ .

$\begin{align*} i_L = i_L(0^-) + \frac{1}{L} \int_{0^-}^{t} v_L dt \end{align*}$

At $t=0^+$ , write:

$\begin{align*} i_L (0^+)= i_L(0^-) + \frac{1}{L} \int_{0^-}^{0^+} v_L dt \end{align*}$

For finite voltage over an interval that shrinks to zero, the integral from $0^-$ to $0^+$ is zero.

Therefore,

$\begin{align*} i_L(0^-) = i_L(0^+) \end{align*}$

Therefore, ideal inductor current is continuous through the switching instant: its value immediately before switching equals its value immediately after.

Inductor at t=0

At $t = 0$ , an inductor does not automatically produce a voltage of $\infty$ . Such a result would require a finite current change during zero time $dt$ . Its current instead remains continuous. In a zero-current DC energisation model it may initially resemble an open circuit, while at $t = \infty$ an ideal inductor resembles a short circuit.

If the inductor carries initial current I₀, its current at $t=0^+$ is $I_0$ . A constant-current source is a useful equivalent at that instant, while the ideal inductor becomes a short circuit in DC steady state at $t=\infty$ .

Series and Parallel Inductors

Uncoupled inductors combine like resistors: inductances add in series and their reciprocals add in parallel. Magnetic coupling changes those rules. Consider coils 1 and 2 with self-inductance $L_1$ and $L_2$ , plus mutual inductance M in henries.

Equivalent inductance depends on whether the coils are in series or parallel and whether their mutual fluxes aid or oppose.

Inductors in Series Formula

How to add inductors in series

Two mutually coupled inductors have two series phasing arrangements.

In a series-aiding or cumulative connection, the mutual flux contributions act in the same direction.
In a series-opposition or differential connection, one winding is reversed so the mutual flux contributions oppose.

Let inductor 1 have self-inductance $L_1$ and inductor 2 have self-inductance $L_2$ . Their mutual inductance is M.

Series-aiding (Cumulative) Connection (mutually induced emf assists the self-induced EMFs)

The following diagram shows the series-aiding, or cumulative, connection.

With this dot orientation, the mutual terms add to the self-induced EMFs.

For each coil:

Self-induced EMF in inductor 1: $e_s_1 = -L_1\frac{di}{dt}$
Mutually induced EMF in inductor 1: $e_m_1 = -M\frac{di}{dt}$
Self-induced EMF in inductor 2: $e_s_2 = -L_2\frac{di}{dt}$
Mutually induced EMF in inductor 2: $e_m_2 = -M\frac{di}{dt}$

The total induced EMF is:

$\begin{align*} e=-(L_1\frac{di}{dt}+M\frac{di}{dt}+L_2\frac{di}{dt}+M\frac{di}{dt}) \end{align*}$

(1) $\begin{equation*} e = -(L_1+L_2+2M) \frac{di}{dt} \end{equation*}$

If $L_eq$ is the equivalent series-aiding inductance, its induced EMF is:

(2) $\begin{equation*} e = -L_e_q_. \frac{di}{dt} \end{equation*}$

Comparing equations (1) and (2) gives:

(3) $\begin{equation*} L_e_q_. = L_1 + L_2 + 2M \end{equation*}$

Mutual inductance therefore raises the cumulative series value by 2M.

For uncoupled coils, M = 0 and the expression reduces to:

$\begin{align*} L_e_q_. = L_1 + L_2 \end{align*}$

Series Opposition (Differential) Connection (mutually induced emf opposes the self-induced EMFs)

The next diagram reverses one winding so the mutual flux contributions oppose.

The mutual EMF terms now have the opposite sign to the self-induced terms:

Self-induced EMF in inductor 1: $e_s_1 = -L_1\frac{di}{dt}$
Mutually induced EMF in inductor 1: $e_m_1 = +M\frac{di}{dt}$
Self-induced EMF in inductor 2: $e_s_2 = -L_2\frac{di}{dt}$
Mutually induced EMF in inductor 2: $e_m_2 = +M\frac{di}{dt}$

The total induced EMF is:

$\begin{align*} e=-(L_1\frac{di}{dt}-M\frac{di}{dt}+L_2\frac{di}{dt}-M\frac{di}{dt}) \end{align*}$

(4) $\begin{equation*} e = -(L_1+L_2-2M) \frac{di}{dt} \end{equation*}$

If $L_e_q$ is the equivalent series-opposition inductance, its induced EMF is:

(5) $\begin{equation*} e = -L_e_q_. \frac{di}{dt} \end{equation*}$

Comparing equations (4) and (5) gives:

(6) $\begin{equation*} L_e_q_. = L_1 + L_2 - 2 M \end{equation*}$

Mutual inductance therefore lowers the differential series value by 2M.

For uncoupled coils, M = 0 and the expression reduces to:

$\begin{align*} L_e_q_. = L_1 + L_2 \end{align*}$

Example 1

Two coils have self-inductances of 10 mH and 15 mH with mutual inductance of 10 mH. Find their equivalent inductance in series aiding.

Solution:

Given: L₁ = 10 mH, L₂ = 15 mH and M = 10 mH.

Use the series-aiding formula:

$\begin{align*} \begin{split} & L_e_q_. = L_1 + L_2 + 2M \\ & = 10 + 15 + 2(10) \\ & = 10 + 15 + 20 \\ & L_e_q_. = 45\,\,mH \end{split} \end{align*}$

The series-aiding equivalent inductance is 45 mH.

Example 2

Use the same coil values to find the equivalent inductance in series opposition.

Solution:

Given: L₁ = 10 mH, L₂ = 15 mH and M = 10 mH.

Use the series-opposition formula:

$\begin{align*} \begin{split} & L_e_q_. = L_1 + L_2 - 2M \\ & = 10 + 15 - 2(10) \\ & = 10 + 15 - 20 \\ & = 25 - 20 \\ & L_e_q_. = 5\,\,mH \end{split} \end{align*}$

The series-opposition equivalent inductance is 5 mH.

Inductors in Parallel Formula

How to add inductors in parallel

Coupled inductors also have two parallel phasing arrangements:

The mutually induced EMF assists the self-induced EMFs in parallel aiding.
The mutually induced EMF opposes the self-induced EMFs in parallel opposition.

Parallel-aiding (Cumulative) Connection(mutually induced emf assists the self-induced EMFs)

In the parallel-aiding connection, mutual EMF assists self-induced EMF as shown below.

Let i₁ and i₂ be the branch currents through L₁ and L₂, with I as the total current.

Thus,

(7) $\begin{equation*} i = i_1 + i_2 \end{equation*}$

Therefore,

(8) $\begin{equation*} \frac{di}{dt} = \frac{di_1}{dt} + \frac{di_2}{dt} \end{equation*}$

Each inductor has a self-induced EMF and a mutually induced EMF.

Because both branches share the same terminals, their voltages are equal.

Therefore,

(9) $\begin{equation*} L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} = L_2 \frac{di_2}{dt} + M \frac{di_1}{dt} \end{equation*}$

$\begin{align*} L_1 \frac{di_1}{dt} - M \frac{di_1}{dt} = L_2 \frac{di_2}{dt} - M \frac{di_2}{dt} \end{align*}$

$\begin{align*} \frac{di_1}{dt} (L_1 - M) = \frac{di_2}{dt} (L_2 - M) \end{align*}$

(10) $\begin{equation*} \frac{di_1}{dt} = (\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} \end{equation*}$

Substitute equation (9) into equation (8):

$\begin{align*} \frac{di}{dt} = (\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} + \frac{di_2}{dt} \end{align*}$

(11) $\begin{equation*} \frac{di}{dt} = (1 + \frac{L_2 - M}{L_1 - M}) + \frac{di_2}{dt} \end{equation*}$

If $L_e_q.$ is the equivalent parallel inductance, its induced EMF is:

(12) $\begin{equation*} e = L_e_q_. \frac{di}{dt} \end{equation*}$

This equals the EMF across either coil:

$\begin{align*} L_e_q_. \frac{di}{dt} = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} \end{align*}$

(13) $\begin{equation*} \frac{di}{dt} = \frac{1}{L_e_q_.} [L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}] \end{equation*}$

Substitute $\frac{di_1}{dt}$ from equation (10) into equation (13):

$\begin{align*} \frac{di}{dt} = \frac{1}{L_e_q_.} [L_1 (\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} + M \frac{di_2}{dt}] \end{align*}$

(14) $\begin{equation*} \frac{di}{dt} = \frac{1}{L_e_q_.} [L_1 (\frac{L_2 - M}{L_1 - M}) + M] \frac{di_2}{dt} \end{equation*}$

Equate equation (11) with equation (14):

$\begin{align*} 1+(\frac{L_2 - M}{L_1 - M}) \frac{di_2}{dt} = \frac{1}{L_e_q_.}[L_1 (\frac{L_2 - M}{L_1 - M}) + M]\frac{di_2}{dt} \end{align*}$

$\begin{align*} \frac{L_1+L_2 - 2M}{L_1 - M} = \frac{1}{L_e_q_.} [\frac{L_1L_2- L_1M+L_1M - M^2}{L_1 - M}] \end{align*}$

$\begin{align*} \frac{L_1+L_2 - 2M}{L_1 - M} = \frac{1}{L_e_q_.} [\frac{L_1L_2 - M^2}{L_1 - M}] \end{align*}$

(15) $\begin{equation*} L_e_q_. = \frac{L_1L_2 - M^2}{L_1+L_2 - 2M} \end{equation*}$

This is the equivalent inductance for the parallel-aiding, or cumulative, connection.

For uncoupled coils, M = 0 and the expression reduces to the usual parallel rule:

$\begin{align*} L_e_q_. = \frac{L_1L_2 - (0)^2}{L_1+L_2-2(0)} = \frac{L_1L_2}{L_1+L_2} = \frac{product}{sum} \end{align*}$

Parallel Opposition (Differential) Connection (mutually induced emf opposes the self-induced EMFs)

In parallel opposition, the mutually induced EMF opposes the self-induced EMFs.

The following diagram shows the parallel-opposition, or differential, connection.

inductor connected in parallel opposition

Following the same branch-voltage method gives:

(16) $\begin{equation*} L_e_q_. = \frac{L_1L_2 - M^2}{L_1+L_2 + 2M} \end{equation*}$

This is the equivalent inductance for parallel opposition.

With M = 0, the formula reduces to the uncoupled parallel rule:

$\begin{align*} L_e_q_. = \frac{L_1L_2 - (0)^2}{L_1+L_2+2(0)} = \frac{L_1L_2}{L_1+L_2} = \frac{product}{sum} \end{align*}$

Example 1

Two coils have self-inductances of 5 mH and 10 mH with mutual inductance of 5 mH. Find their equivalent inductance in parallel aiding.

Solution:

Given: L₁ = 5 mH, L₂ = 10 mH and M = 5 mH.

Use the parallel-aiding formula:

$\begin{align*} \begin{split} & L_e_q_. = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}.... if \,\, fluxes \,\, aid \\ & = \frac{5 * 10 - (5)^2}{5 + 10 - 2(5)} \\ & = \frac{50 - 25}{15 - 10} \\ & = \frac{25}{5} \\ & L_e_q_. = 5\,\,mH \end{split} \end{align*}$

The parallel-aiding equivalent inductance is 5 mH.

Example 2

Use the same coil values to find the equivalent inductance in parallel opposition.

Solution:

Given: L₁ = 5 mH, L₂ = 10 mH and M = 5 mH.

Use the parallel-opposition formula:

$\begin{align*} \begin{split} & L_e_q_. = \frac{L_1 L_2 - M^2}{L_1 + L_2 + 2M}.... if \,\, fluxes \,\, oppose \\ & = \frac{5 * 10 - (5)^2}{5 + 10 + 2(5)} \\ & = \frac{50 - 25}{15 + 10} \\ & = \frac{25}{25} \\ & L_e_q_. = 1\,\,mH \end{split} \end{align*}$

The parallel-opposition equivalent inductance is 1 mH.

Coupling Inductors

Two inductors are magnetically coupled when some of the flux from one coil links turns in the other. That shared flux produces mutual inductance.

Coupling transfers energy between circuits through their shared field. Examples include a two-winding transformer, an autotransformer and an induction motor.

Let coupled coils 1 and 2 have self-inductances L₁ and L₂. Their mutual inductance is M.

The dot orientation sets whether mutual terms add as L₁+ M and L₂+ M or subtract as L₁– M and L₂– M in the branch equations.

When the mutual fluxes aid, the branch terms become L₁+ M for coil 1 and L₂+ M for coil 2.
When the mutual fluxes oppose, the branch terms become L₁– M for coil 1 and L₂– M for coil 2.

Mutual Inductance Formula

A changing current in either coil induces an EMF in the other coil.

Mutual inductance measures the voltage induced in one coil for a given rate of current change in the other.

The induced voltage’s polarity follows the dot convention and Lenz’s law. It may aid or oppose the self-induced voltage in a chosen connection.

Mutual inductance (M) can also be expressed as one coil’s linked flux per unit current in the other.

In equation form:

$\begin{align*} M = \frac{N_2 \phi_1_2}{I_1} \end{align*}$

where:

$I_1$ is the current in the first coil,

$\phi_1_2$ is the flux linking the second coil,

$N_2$ is the number of turns on the second coil.

The mutual inductance is 1 henry when a current change of 1 ampere per second in one coil induces 1 volt in the other.

Coefficient of Coupling

The coefficient of coupling (k) is the fraction of magnetic flux from one coil that links the other coil.

Its value describes how tightly the coils share flux.

The coupling coefficient is:

$\begin{align*} k = \frac{M}{\sqrt{L_1L_2}} \end{align*}$

where:

L₁ is the first coil’s self-inductance,

L₂ is the second coil’s self-inductance,

M is the mutual inductance between the coils.

For fixed self-inductances, a larger k gives a larger M and indicates that more flux is shared.

If all modelled flux from one coil links the other, k is 1 (100% coupling) and the coils are ideally coupled.
If half the relevant flux links the other coil, k is 0.5 (50% coupling).
If no flux links the other coil, k is 0 and the coils are magnetically uncoupled.

For passive coupled coils, k lies from 0 to 1 inclusive. Geometry, spacing, orientation and the magnetic path determine the actual value, which must be measured or taken from device data.

Inductive Coupling Example Problem

Example 1

Calculate the mutual inductance of two coils with k = 1 and self-inductances of 9 H and 4 H.

Solution:

Given: L₁ = 9 H, L₂ = 4 H and k = 1.

Use the coupling formula:

$\begin{align*} \begin{split} & M = k \sqrt{L_1 L_2} \\ & = 1 \sqrt{9 * 4} \\ & = \sqrt{36} \\ & M = 6\,\,mH \end{split} \end{align*}$

The mutual inductance is 6 H.

Example 2

Calculate the mutual inductance for the circuit shown below.

Solution:

Given: L₁ = 12 mH, L₂ = 3 mH and k = 0.8.

Use the coupling formula:

$\begin{align*} \begin{split} & M = k \sqrt{L_1 L_2} \\ & = 0.8 \sqrt{12 * 3} \\ & = 0.8\sqrt{36} \\ & M = 4.8\,\,mH \end{split} \end{align*}$

The mutual inductance is 4.8 mH.

Example 3

Two coupled coils have series-aiding and series-opposition values of 16 mH and 8 mH. Find their mutual inductance M.

Solution:

Given: $L_e_q_.$ = 16 mH or 8 mH.

Apply the two series formulas:

$\begin{align*} L_e_q_. = L_1 + L_2 \pm 2M \end{align*}$

So,

(17) $\begin{equation*} 16 = L_1 + L_2 + 2M \end{equation*}$

and

(18) $\begin{equation*} 8 = L_1 + L_2 - 2M \end{equation*}$

Subtract equation (18) from equation (17):

$\begin{align*} 4 M = 8 \end{align*}$

$\begin{align*} M = 2 \,\, mH \end{align*}$

The mutual inductance is 2 mH.

What Are Inductors Used For

Inductors serve these circuit functions:

Choking, filtering, smoothing and attenuating high-frequency noise
Chokes that pass DC while presenting reactance to AC in radio and electronic equipment
Hartley oscillators, where tapped or series-connected inductance works with a capacitor to form tuned oscillators or LC resonance tank circuits
Colpitts oscillators, where a capacitive divider and inductor form a resonant circuit. These tuned circuits support radio-frequency generation or selection.
Energy storage and transfer in power electronics converters (DC-DC or AC-DC), including switch-mode power supplies
Output-current ripple filtering in power converters, with inductance selected against ripple, size, loss and transient response
Impedance matching, where inductive reactance can tune a capacitive load as part of a frequency-specific matching network
Reactors that limit the rate of current rise or fault current in an electrical transmission system

Types of Inductors

Core material, winding form and construction give rise to the following inductor types.

Air core inductor
Variable core inductor
Iron core inductor
Powdered iron core inductor
Ferrite core inductor
Ferromagnetic core inductor
Radio-frequency inductor
Toroidal core inductor
Multilayer ceramic inductor
Film inductor
Coupled inductor
Molded inductor

Resistance of Inductors

An ideal inductor has zero resistance. A real winding has DC resistance, while its core and winding geometry add frequency-dependent loss.

The Impedance of Inductors (Inductive Reactance of Inductors)

The ideal impedance of an inductor is purely imaginary. Its inductive reactance is the magnitude of this frequency-dependent opposition:

$\begin{align*} Z_L = j \omega L \end{align*}$

where:

$Z_L$ is the ideal inductor impedance,

$\omega$ is angular frequency, equal to $2 \pi f$ ,

$L$ is inductance.

Inductive reactance rises in direct proportion to frequency. At zero frequency, the ideal inductive part is zero.

The factor j means ideal inductor impedance is reactive rather than positive resistance. A real part’s impedance also includes winding resistance, parasitic capacitance and core loss, especially near self-resonance.

Inductors in DC Circuits vs AC Circuits

An inductor responds to DC transients and steady AC in different ways.

Inductors in DC Circuits

An ideal inductor becomes a short circuit only after a DC transient has settled. A real inductor retains its winding resistance.

During the switching transient, current changes and the inductor develops voltage. Once ideal DC current is constant, its rate of change and induced voltage are zero.

The voltage relation is:

$\begin{align*} v_L = L \frac{di_L}{dt} \end{align*}$

where

$v_L$ is instantaneous voltage in volts,

$L$ is inductance in henries,

$\frac{di_L}{dt}$ is the rate of change of current in amperes per second.

For steady d.c. current, $\frac{di_L}{dt}$ is zero, so the voltage across an ideal inductor is zero.

The ideal short-circuit model therefore applies to DC steady state, after the transient.

Inductors in AC Circuits

With AC, current changes continuously and the inductor develops a continuous reactive voltage.

For a fixed inductance, reactance rises with frequency and reduces current for a fixed applied voltage.

The image shows an inductor connected across an AC source. Its terminal voltage equals the applied source voltage, subject to the chosen polarity reference.

inductor connected across ac — Inductor Connected Across an AC Supply

For sinusoidal current, induced voltage is largest when current crosses zero because the current slope is then greatest. In an ideal inductor, voltage leads current by 90 degrees.

Do Inductors Have Polarity

A single unpolarised inductor can carry current in either direction. With coupled windings, dot markings identify relative winding polarity and determine whether mutual terms aid or oppose.

Under Lenz’s law, induced voltage has the polarity that opposes the current change that produced it.

How Do Inductors Store Energy

An ideal inductor stores energy in its magnetic field and returns that energy without net loss. Real inductors also dissipate power through winding resistance and core loss.

Input energy in a real coil has two destinations:

Winding resistance converts part of the input to I²R heat.
The balance builds the magnetic field, subject to core and other losses.

Consider inductance L with winding resistance R connected to a d.c. source through switch S. Closing the switch causes current to rise gradually toward its steady value.

The changing current creates a self-induced EMF. The source supplies energy to build the magnetic field while resistance converts some energy to heat.

When the switch opens, the field collapses and releases stored energy. The circuit may return that energy to a source, transfer it to another component or dissipate it in resistance.

A lifted mass provides a mechanical analogy. Raising mass ‘m’ through height ‘h’ stores potential energy “m*g*h”. Work is required to lift it, while holding the idealised mass at that height does not add stored energy.

Allowing the mass to fall releases its potential energy. In the same way, decreasing inductor current releases magnetic energy into the connected circuit.

The Magnitude of Energy Stored in the Magnetic Field

Derive the stored magnetic energy from voltage, current and time.

At an instant, let current i change at the rate $\frac{di}{dt}$ .

The inductor voltage is $e = L \frac{di}{dt}$ .

The instantaneous power is:

$\begin{align*} \begin{split} & p = e * i \\ & = L \frac{di}{dt} * i \\ & = Li \frac{di}{dt} \end{split} \end{align*}$

During a short interval dt, the energy increment is:

$\begin{align*} \begin{split} & dW = p * dt \\ & = Li \frac{di}{dt} * dt \\ & = Li di \end{split} \end{align*}$

Integrating from zero current to final current I gives:

$\begin{align*} \begin{split} & \int dW = \int_{0}^{I} L\,\,i\,\,di \\ & W = L [\frac{i^2}{2}]_{0}^{I} \\ & W = \frac{1}{2} LI^2 \,\,joule \end{split} \end{align*}$

where L is inductance in henries,

I is current in amperes.

This equation gives the energy stored by a linear inductor.

An inductor holds magnetic energy whenever its current is nonzero. As current falls to zero, that energy returns to the source, transfers elsewhere or becomes heat in circuit resistance.

Example

Two coils have self-inductances of 3 H and 2 H with mutual inductance of 2 H. A current of 6 A flows through their series connection. Calculate stored energy for (i) cumulative and (ii) differential connection, then find the coupling coefficient.

Solution:

Given: L₁ = 3 H, L₂ = 2 H, M = 2 H, I = 6 A.

(i) Cumulative connection:

$\begin{align*} \begin{split} & L_e_q_. = L_1 + L_2 + 2M \\ & = 3 + 2 + 2(2) \\ & = 5 + 4 \\ & L_e_q_. = 9 \,\, H \\ \end{split} \end{align*}$

$\begin{align*} \begin{split} & Energy \,\, Stored \,\, W = \frac{1}{2} L_e_q_. I^2 \,\, joule \\ & = \frac{1}{2} * 9 * (6)^2 \\ & W = 162 \,\, joule \end{split} \end{align*}$

(ii) Differential connection:

$\begin{align*} \begin{split} & L_e_q_. = L_1 + L_2 - 2M \\ & = 3 + 2 - 2(2) \\ & = 5 - 4 \\ & L_e_q_. = 1 \,\, H \\ \end{split} \end{align*}$

$\begin{align*} \begin{split} & Energy \,\, Stored \,\, W = \frac{1}{2} L_e_q_. I^2 \,\, joule \\ & = \frac{1}{2} * 1 * (6)^2 \\ & W = 18 \,\, joule \end{split} \end{align*}$

(iii) Coupling coefficient:

$\begin{align*} \begin{split} & K = \frac{M}{\sqrt{L_1 L_2}} \\ & = \frac{2}{\sqrt{3*2}} \\ & = \frac{2}{\sqrt6} \\ & = 0.816 \end{split} \end{align*}$