Module2 Lesson4

Module 2
LESSON INDUCTANCE
4
Objectives:
At the end of the lesson, the students will be able to:
• explain the principles of the operation of a transformer
• define the mutual inductance of two coupled circuits such as the primary and
secondary circuits of a transformer
Introduction:
Hi there! This is a continuation of our study of electromagnetic induction. The mutual
inductance of two circuit, and the self-inductance of a single circuit, are quantities that are
introduced so that an induced e.m.f can be related directly to a changing current, rather
than to a changing flux. Also, we will investigate the relationship between the energy stored
in a current – carrying circuit and the circuit self-inductance, as well as the influence of self-
inductance on the transient behavior of resistive circuits.
Activity:
The image below is a typical set up of a transformer.
Observe the details of the image and write your observation in a paper.
Analysis:
After performing the activity, answer the following questions. Use the provided space below
for your answers.
1. How do the coils differ?
2. How does this difference affect the electric field?
Abstraction:
E.m.f. is associated with a changing magnetic flux are used to transform a varying voltage
in one circuit into a large or smaller voltage in another circuit. This is accomplished using
a transformer consisting of two coils electrically insulated from each other and wound on
the same ferromagnetic core.
Power is supplied to one coil named the primary coil. A varying current in this coil sets up
a varying magnetic flux, largely confined to the ferromagnetic core. The other coil, called
the secondary coil, thus encloses a varying flux. While this flux is changing at the ratet,
each of the Ns turns in the secondary coil experiences, according to Faraday’s flux rule,
an induced e.m.f. equal to 𝑑Φ /dt, and the e.m.f for the entire secondary coil is…
There is also an e.m.f. 𝑑Φ /dt induced in each of the NP turns of the primary coil and
the total e.m.f. for this coil is..
dividing these equations we get…
according to this result, when there are varying currents, the ratio of the e.m.f. induced in
the secondary and primary coils at any instant is equal to the “turn ratio”, Ns/Np. When Ns
is greater than Np, the secondary e.m.f exceeds the primary e.m.f and the transformer is
called a step-up transformer. In a step-down transformer, Ns is less than Np. It is desirable
to transmit electrical power at high voltages and small currents in order to minimize the I2R
poor dissipation in the transmission line. But this power must be generated and ultimately
delivered at relatively low voltages to avoid problems of insulation and safety. A very useful
feature of alternating current is the fact that the voltage and the current can be changed
with ease and efficiency by the use of transformers.
Energy Losses in a Transformer

Although transformers are very efficient devices, small energy losses do occur in them due
to four main causes.
(i) Resistance of windings. The copper wire used for the windings has resistance and
so ordinary (I2R) heat losses occur. In high-current, low p.d. windings these are
minimized by using thick wire.
(ii) Eddy currents. The alternating magnetic flux induces eddy currents in the iron core
and causes heating. The effect is reduced by having a laminated core.
(iii) Hysteresis. The magnetization of the core is repeatedly reversed by the alternating
magnetic field. The resulting expenditure of energy in the core appears as heat and is
kept to a minimum by using a magnetic material which has a low hysteresis loss.
(iv) Flux leakage. The flux due to the primary may not all link the secondary if the core
is badly designed or has air gaps in it.
Mutual Inductance
To faulitate the analysis of couple circuits such as the primary and secondary circuits of a
transformer, we should know the relationship between the changing current in one circuit
and the e.m.f that it induces in the other circuit.
Suppose a current Ip in the primary circuit produces a flux Φ through each turn of the
secondary coil. The mutual inductance M of the primary and secondary circuits is defined
by
If there are no ferromagnetic materials present, Φ is proportional to Ip and M is then a

constant, with a value that depends only on the geometry of the circuits. For a transformer
with a ferromagnetic core, M depends also on the permeability of the core which is a
function of the magnetic field B established within the core.
The SI unit of mutual inductance is the weber per ampere (Wb. A -1), which is called the
henry (H).
IH = Wb/A = 1V.s/A = 1J/A2
A current Is in the secondary will create a flux Φ enclosed by each turn of the primary. And
so we define M by…
EP = -M dIs/dt …similarly…. ES = -m dIS/dt
Then… M=
Self-Inductance
When the current through a coil is changing, the flux produced by this current and enclosed
by the turns of the coil is also changing,; consequently there is an e.m.f induced in the coil.
Such an e.m.f. is called a self-induced e.m.f. Suppose a coil current I produces a flux f
through each of N turns of a coil. The self-inductance L of the coil is defined by the equation
is…. 𝑁Φ = 𝐿𝐼
If there are no ferromagnetic materials present, f is proportional to I. Then L is a constant

with a value determined by the coil geometry. The self-Inductance of a given coil can be
increased greatly by providing a ferromagnetic core, but, because the core’s magnetic
permeability μ depends on B1 the self-inductance is then a complicated function of the coil
current.
𝑁𝑑Φ 𝐿𝑑𝐼
When L is constant = 𝑑𝑡 or E = L dI/dt
𝑑𝑡
Inductance in a Solenoid
Let us consider a long, air-cored solenoid of length l, cross-sectional area A having N turns
and carrying current I. the flux density B is almost constant over A and , neglecting the
ends, is given by
Energy Stored by an Inductor
And so the energy stored in a magnetic field of an inductor carrying a current is

Example1
A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second
coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance
of the combination of the two coils?
Example2
Two toroidal solenoids are wound around the same form so that the magnetic field of one
passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400
turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of
solenoid 2 is 0.0320 Wb.
(a) What is the mutual inductance of the pair of solenoids?
(b) When the current in solenoid 2 is 2.54 A, what is the average flux through each turn of
solenoid 1?
Example3
A toroidal solenoid has 500 turns, cross-sectional area of 6.25 cm2 and mean radius 4.00
cm.
(a) Calculate the coil’s self-inductance.
(b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-
induced emf in the coil.

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Module 2

dividing these equations we get…

Energy Losses in a Transformer

If there are no ferromagnetic materials present, Φ is proportional to Ip and M is then a

EP = -M dIs/dt …similarly…. ES = -m dIS/dt

If there are no ferromagnetic materials present, f is proportional to I. Then L is a constant

And so the energy stored in a magnetic field of an inductor carrying a current is

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