Inductors and Inductance Introduction
Mutual-Inductance
Faradayโs law tells us that an EMF can be induced around a closed path in a conductor if the
magnetic flux through the surface enclosed by the path changes.
๐ = โ๐ฮฆ๐ต
๐๐ก
We also previously learned that a magnetic field is established when a current is made to flow
in a conductor. For example, the solenoid, as you will recall, has a magnetic field directed as
shown below with a magnitude given as:
๐ต = ๐0๐๐, with ๐ = ๐ ๐โ
Where, ๐ is the length of the coil.
Using these two observations, the configuration below can be used to establish an EMF in a
second coil by applying a time varying current in a first coil.
The amount of magnetic flux created in the second coil per unit of current from the first is
called the mutual inductance and is defined as follows:
๐21 = N2ฮฆ๐ต,1
๐1
Where, ๐21 refers to the inductance of coil 2 with respect to coil 1.
Circling back to Faradayโs law, the EMF induced in the second coil is as follows:
๐2 = โ๐2
๐
๐๐ก(ฮฆ๐ต,1)
๐2 = โ๐2
๐
๐๐ก (
๐21๐1
N2)
๐2 = โ๐21
๐๐1
๐๐ก
Although we will not prove it here it turns out that ๐21 = ๐12, and therefore depending on
which coil is providing the time varying current we have.
Coil 1 induces an EMF in coil 2 Coil 2 induces an EMF in coil 1
๐2 = โ๐๐๐1
๐๐ก ๐1 = โ๐
๐๐2
๐๐ก
The concept of mutual inductance has many practical applications, one of which is that of an
electrical transformer. As an example, electrical power delivery relies on transformers to
deliver relatively low voltage energy to homes from much higher voltage power lines.
Self-Inductance
The concept of inductance can also be applied to a single isolated coil. When a time varying
current is produced in a coil the magnetic flux through that coil is changing, which in turn will
induce a current, (to oppose the change in flux - โLenzโs lawโ), in the coil itself. As an analogy
to the mutual inductance we can define self-inductance as below.
๐ฟ = Nฮฆ๐ต
๐
Where, ๐ is the number of turns in the coil.
Solving for the EMF due to self-inductance just as we did with mutual inductance we have:
๐๐ฟ = โ๐ฟ๐๐
๐๐ก
In an earlier section we introduced the capacitor, which is a device commonly used in electronic
circuits that stores energy in the form of an electric field. An inductor is a device that is also
commonly used in electronic circuits that stores energy in the form of a magnetic field. One of
the simplest types of inductors is a solenoid. The amount of charge stored in a capacitor per
unit voltage applied was referred to as the capacitance, ๐ถ = ๐ ๐โ . Similarly, as defined above,
the amount of magnetic flux produced per unit of current is called the inductance, ๐ฟ = Nฮฆ๐ต
๐.
Note that we were able to write the capacitance of a parallel plate capacitor in terms of its
geometry only. As it turns out, we can do the same for inductors. The solenoid below, as you
will recall, has a magnetic field directed as shown with a magnitude given as:
๐ต = ๐0๐๐, with ๐ = ๐ ๐โ
Where, ๐ is the length of the coil.
The magnetic flux through the opening surface of the coil is ฮฆ๐ต = ๐ต๐ด = ๐0๐๐๐ด. Therefore, the
inductance of the coil is as follows:
๐ฟ = ๐ฮฆ๐ต
๐
๐ฟ = ๐๐(๐0๐๐๐ด)
๐
๐ฟ = ๐0๐2๐ด๐
Inductors in Circuits
When studying RC circuits, we noticed there exists a transient time immediately after we apply
a voltage until the capacitor is fully charged. For the simple RC circuit shown below we found
the following behavior.
Charging a Capacitor in an RC Circuit
Voltage across Capacitor Current in Circuit
๐ฃ(๐ก) = ๐ (1 โ ๐โ1
๐ ๐ถ๐ก) ๐(๐ก) =
๐
๐ ๐โ
1๐ ๐ถ
๐ก
C
+ -
V
R
I
We can see that the voltage across the capacitor increases as more charge builds on the
capacitor plates creating a stronger electric field. Conversely the current decays until it finally
stops flowing and the supply voltage drops entirely across the capacitor. Now letโs analyze
what happens when we replace the capacitor with an inductor.
L
+ -
VB
R
I
Just as we did with the RC circuit, we apply Kirchhoffโs voltage rule around the loop.
๐๐ต โ ๐(๐ก)๐ โ ๐ฟ๐๐(๐ก)
๐๐ก= 0
The result is a fist order differential equation, just as we had with an RC circuit, which can be
similarly solved using the separation of variables technique.
๐๐ต โ ๐๐ = ๐ฟ๐๐
๐๐ก
1
๐๐ต โ ๐๐ =
1
๐ฟ
๐๐ก
๐๐
( 1
๐๐ต โ ๐๐ ) ๐๐ =
1
๐ฟ ๐๐ก
โซ ( 1
๐๐ต โ ๐๐ ) ๐๐
๐
0
= โซ1
๐ฟ ๐๐ก
๐ก
0
โ1
๐ ln(๐๐ต โ ๐๐ )|0
๐ = 1
๐ฟ๐ก
โ1
๐ (ln(๐๐ต โ ๐๐ ) โ ln(๐๐ต)) =
1
๐ฟ๐ก
ln (๐๐ต โ ๐๐
๐๐ต) = โ
๐
๐ฟ๐ก
1 โ๐๐
๐๐ต= ๐โ
๐ ๐ฟ
๐ก
๐(๐ก) = ๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)
In this case we find that the current grows over time until it reaches a steady state value of
๐๐ต ๐ โ . At this point the magnetic flux stops changing and the inductor acts as a simple wire,
(i.e. a short circuit), so that the current is limited by the resistor only.
The voltage across the inductor is then
๐๐ฟ(๐ก) = ๐ฟ๐๐(๐ก)
๐๐ก
๐๐ฟ(๐ก) = ๐ฟ๐
๐๐ก(
๐
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก))
๐๐ฟ(๐ก) = ๐ฟ๐
๐ (0 +
๐
๐ฟ๐โ
๐ ๐ฟ
๐ก)
๐๐ฟ(๐ก) = ๐๐โ ๐ ๐ฟ
๐ก
Which shows that the voltage across the inductor exponentially decays until the entire supply
voltage drops across the resistor only.
Finally, we can find the energy stored in the magnetic field of the inductor. Recall the energy
stored in a capacitor when a voltage, V, is applied across was found to be
๐๐ถ = 1
2๐ถ๐2
The energy stored in an inductor is completely analogous and can be found in a similar fashion.
In this case letโs take our differential equation from above and multiply through by the current.
๐๐ = ๐2๐ + ๐ฟ๐๐๐
๐๐ก
Since the power is given as ๐ = ๐๐, we can interpret this relationship as follows:
The battery is supplying energy to the circuit at a rate of ๐๐, which is being dissipated by the
resistor at a rate of ๐2๐ and being stored, (in the form of a magnetic field), by the inductor at a
rate of ๐ฟ๐๐๐
๐๐ก. Therefore, we have the following.
๐๐๐ฟ
๐๐ก= ๐๐ฟ
๐๐๐ฟ
๐๐ก= ๐ฟ๐
๐๐
๐๐ก
โซ ๐๐๐ฟ
๐๐ฟ
0
= โซ ๐ฟ๐ ๐๐๐
0
๐๐ฟ =1
2๐ฟ๐2
As you can see the energy stored in an inductor is completely analogous to the energy stored in
a capacitor.
As you may have noticed the transient behavior of an RL circuit is also analogous to that of the
RC circuit we previously studied. Below is a side by side comparison of the two types of
components.
Property RC Circuit RL Circuit
Time Constant ๐ = ๐ ๐ถ ๐ = ๐ฟ
๐
Component Voltage Behavior
Component Current Behavior
Charging Equations
๐๐ถ(๐ก) = ๐
๐ ๐โ
1๐ ๐ถ
๐ก
๐๐ถ(๐ก) = ๐ (1 โ ๐โ 1
๐ ๐ถ๐ก)
๐๐ฟ(๐ก) = ๐
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)
๐๐ฟ(๐ก) = ๐๐โ ๐ ๐ฟ
๐ก
General Voltage-Current
Relationship
๐๐ถ(๐ก) = ๐ถ๐๐๐ถ(๐ก)
๐๐ก
๐๐ถ(๐ก) =1
๐ถโซ ๐๐ถ(๐)๐๐
๐ก
๐ก0
+ ๐๐ถ(๐ก0)
๐๐ฟ(๐ก) = ๐ฟ๐๐(๐ก)
๐๐ก
๐๐ฟ(๐ก) =1
๐ฟโซ ๐๐ฟ(๐)๐๐
๐ก
๐ก0
+ ๐๐ฟ(๐ก0)
Energy Stored ๐๐ถ(๐ก) =
1
2๐ถ๐๐ถ(๐ก)2
๐๐ฟ(๐ก) = 1
2๐ฟ๐(๐ก)2
Summary for Inductance, Inductors, and RL Circuits
Mutual Inductance
The amount of magnetic flux created in a second coil per unit of current from a first coil is called the mutual inductance and is defined as follows:
๐21 = N2ฮฆ๐ต,1
๐1
The EMF induced in a second coil from a first coil is given as follows:
๐1/2 = โ๐๐๐2/1
๐๐ก
Where, ๐21 = ๐12 = ๐
Self-Inductance
When a time varying current is produced in a coil the magnetic flux through that coil is changing, which in turn will induce a current, (to oppose the change in flux - โLenzโs lawโ), in the coil itself. As an analogy to the mutual inductance we can define self-inductance as below.
๐ฟ = Nฮฆ๐ต
๐
Where, ๐ is the number of turns in the coil. The EMF due to self-inductance is given as follows:
๐๐ฟ = โ๐ฟ๐๐
๐๐ก
Inductor Energy
Energy stored in an inductor, in the form of a magnetic field, is given by the following:
๐๐ฟ(๐ก) = 1
2๐ฟ๐(๐ก)2
RL Circuits
When a voltage source, ๐๐ต, is applied to a resistor and inductor in series the current and voltage vary according to the following equations.
๐๐ฟ(๐ก) = ๐๐ต๐โ ๐ ๐ฟ
๐ก ๐๐ฟ(๐ก) = ๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)
RL Time Constant
๐ =๐ฟ
๐
Examples:
Question 1:
Coil 1 has ๐ฟ1 = 25 ๐๐ป and ๐1 = 100 turns. Coil 2 has ๐ฟ1 = 40 ๐๐ป and ๐2 = 200 turns. The
coils are fixed in place; their mutual inductance is ๐ = 3 ๐๐ป. A 6 ๐๐ด current in coild 1 is
changing at a rate of 4 ๐ด/๐ .
a.) What is the magnetic flux that links coil 2, ฮฆ2,1. b.) What is the magnitude of the EMF induced in coil 2? c.) What is the magnitude of the self-induced EMF in coil 1?
Solution 1:
Part a.)
The magnetic flux through coil 2 from coil 1 divided by the current from coil 1 is defined as the
mutual inductance, which we know to be 3 ๐๐ป.
๐ = N2ฮฆ2,1
๐1
ฮฆ2,1 = ๐๐1
N2
ฮฆ2,1 = (3๐ธโ3) โ (6๐ธโ3)
200
ฮฆ2,1 = 9๐ธโ8 ๐ โ ๐2
Part b.)
The EMF induced in coil 2 is proportional to the rate of change of the current in coil 1, and the
mutual inductance is the proportionality constant.
๐2 = ๐๐๐1
๐๐ก
๐2 = (3๐ธโ3) โ 4
๐2 = 0.012 ๐
Part c.)
The self-induced EMF in coil 1 is also proportional to the rate of change of the current in coil 1,
but with ๐ฟ as the proportionality constant.
๐๐ฟ = ๐ฟ๐๐1
๐๐ก
๐๐ฟ = (25๐ธโ3) โ 4
๐๐ฟ = 0.1 ๐
Question 2:
In the circuit shown below, with ๐ = 3ฮฉ and ๐ฟ = 4๐ป, the switch has been open for a long time.
At time ๐ก = 0, the switch is closed. Answer the following questions.
a.) Find the time when the current through the inductor is 50% of its maximum value. b.) Find the rate at which energy is being dissipated through the resistor and being stored in
the inductor when the elapsed time is one time constant. c.) Find the total energy stored in the inductor and the total energy dissipated from the resistor
when the elapsed time is five time constants.
L
+ -
VB
R
I
Solution 2:
Part a.)
The current through the inductor at any time ๐ก > 0 is given as.
๐(๐ก) = ๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)
Where, ๐๐ต
๐ is the maximum value as ๐ก โ โ.
Therefore, we can solve for the time the current reaches 0.5๐
๐ as follows:
0.5๐๐ต
๐ =
๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)
๐โ ๐ ๐ฟ
๐ก = (1 โ 0.5)
โ ๐
๐ฟ๐ก = ln(0.5)
๐ก = โ ๐ฟ
๐ ln(0.5)
๐ก = โ 3
4ln(0.5)
๐ก = 0.52 ๐
Part b.)
Recall, when deriving the energy stored in an inductor, we took our differential equation and
multiplied through by the current. When we did this, we ended up with a power relationship as
shown below.
๐๐ต๐(๐ก) = ๐(๐ก)2๐ + ๐ฟ๐(๐ก)๐๐(๐ก)
๐๐ก
๐๐ต(๐ก) = ๐๐ (๐ก) + ๐๐ฟ(๐ก)
Using the equation for the current we can find new expressions for ๐๐ (๐ก) and ๐๐ฟ(๐ก).
๐๐ (๐ก) = ๐(๐ก)2๐
๐๐ (๐ก) = [๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)]2
๐
๐๐ (๐ก) = ๐๐ต
2
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)2
๐๐ฟ(๐ก) = ๐ฟ๐(๐ก)๐๐(๐ก)
๐๐ก
๐๐ฟ(๐ก) = ๐ฟ โ๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก) [๐
๐ (0 +
๐
๐ฟ๐โ
๐ ๐ฟ
๐ก)]
๐๐ฟ(๐ก) = ๐ฟ โ๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก) [๐
๐ (
๐
๐ฟ๐โ
๐ ๐ฟ
๐ก)]
๐๐ฟ(๐ก) = ๐๐ต
๐ ๐โ
๐ ๐ฟ
๐ก (1 โ ๐โ ๐ ๐ฟ
๐ก)
Now we can find the power at ๐ก = 1๐ = ๐ฟ
๐
๐๐ ( ๐) = ๐๐ต
2
๐ (1 โ ๐โ
๐ ๐ฟ
โ๐ฟ๐ )
2
๐๐ ( ๐) = 212
4(1 โ ๐โ 1)2
๐๐ (๐) = 44.05 ๐
๐๐ฟ(๐) = ๐๐ต
2
๐ ๐โ
๐ ๐ฟ
โ๐ฟ๐ (1 โ ๐โ
๐ ๐ฟ
โ๐ฟ๐ )
๐๐ฟ(๐) = 212
4๐โ 1(1 โ ๐โ 1)
๐๐ฟ(๐) = 25.64 ๐
Part c.)
The total energy stored in the inductor at any time ๐ก is given as:
๐๐ฟ(๐ก) =1
2๐ฟ๐(๐ก)2
๐๐ฟ(๐ก) =1
2๐ฟ [
๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)]2
Therefore, the energy stored at ๐ก = 5๐ = 5๐ฟ
๐ is as follows:
๐๐ฟ(5๐) =1
2๐ฟ [
๐
๐ (1 โ ๐โ
๐ ๐ฟ
5๐ฟ๐ )]
2
๐๐ฟ(5๐) =3
2[21
4(1 โ ๐โ 5)]
2
๐๐ฟ(5๐) = 40.79 ๐ฝ
To find the total amount of energy dissipated by the resistor we need to integrate ๐๐ (๐ก) from
๐ก = 0 to ๐ก = 5๐. With ๐๐ (๐ก) = ๐(๐ก)2๐ from above we have the following integral, which is not
trivial.
๐ธ๐ = โซ (๐
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก))
2
๐ 5
๐ฟ๐
0
๐๐ก
However, using conservation of energy we also know that the total energy delivered by the
battery is equal to the amount dissipated from the resistor plus the amount stored in the
inductor.
๐ธ๐ต(๐ก) = ๐ธ๐ (๐ก) + ๐๐ฟ(๐ก)
And since we already know ๐๐ฟ(5๐), we can instead find ๐ธ๐ต(5๐) by integrating ๐๐ต(๐ก), which is a
less difficult integral to evaluate.
๐ธ๐ต = โซ ๐๐ต(๐ก)5
๐ฟ๐
0
๐๐ก
๐ธ๐ต = ๐๐ต โซ ๐(๐ก)5
๐ฟ๐
0
๐๐ก
๐ธ๐ต = ๐๐ต โซ ๐๐ต
๐ (1 โ ๐โ
๐ ๐ฟ
๐ก)5
๐ฟ๐
0
๐๐ก
๐ธ๐ต = ๐๐ต
2
๐ โซ (1 โ ๐โ
๐ ๐ฟ
๐ก)5
๐ฟ๐
0
๐๐ก
๐ธ๐ต = ๐๐ต
2
๐ [(5
๐ฟ
๐ +
๐ฟ
๐ ๐โ
๐ ๐ฟ
โ5๐ฟ๐ ) โ (0 +
๐ฟ
๐ ๐โ 0)]
๐ธ๐ต = ๐๐ต
2
๐ [5
๐ฟ
๐ +
๐ฟ
๐ ๐โ 5 โ
๐ฟ
๐ ]
๐ธ๐ต = ๐ฟ๐๐ต
2
๐ 2[4 + ๐โ 5]
๐ธ๐ต = 3 โ 212
42[4 + ๐โ 5]
๐ธ๐ต = 331.31 ๐ฝ
Finally, the energy dissipated by the resistor is
๐ธ๐ = ๐ธ๐ต โ ๐๐ฟ
๐ธ๐ = 331.31 โ 40.79
๐ธ๐ = 290.5
By: ferrantetutoring