EXAMPLE-mutual

EXAMPLE OF PROBLEM
SOLVING

MUTUAL INDUCTANCE

Q1

Primary coil of a cylindrical former with the length
of 50 cm and diameter 3 cm has 1000 turns. If the
secondary coil has 50 turns, calculate ;
(a) mutual inductance
(b) the induced e.m.f in the secondary coil if the
current flowing in the primary coil is changing at
the rate of 4.8 A s-1.

Example for Solution

(a) 1. List the information given;

1 = 0.5  = 2
1 = 0.03
1 = 100 = 4 (0.03)2
2 = 50 =
4 10−4 2
7.07 ×

2. Choose a suitable formula:

= 0 1 2


3. Apply the formula.

= 0 1 2


= (4 × 10−7)(1000)(50)(7.07 × 10−4 2)
0.5

= 8.88 × 10−5

Example for Solution

(b) 1. List the information given;

1 = 4.8 −1

= 8.88 × 10−5
(From previous question)

2. Choose a suitable formula:
1
2 =

3. Apply the formula.

2 = 1


Q2

If the flux linkage in coil 2 is 5 ×
10−3 Wb and it has 300 turns and
the current in coil 1 is 7 A.

Calculate the mutual inductance.

Example for Solution

1. List the information given;

Φ2 = 5 × 10−3
2 = 300

1= 7

2. Choose a suitable formula:
2Φ2
= 1

3. Apply the formula.

= 2Φ2
1
(300)(5 × 10−3)
=7
= 2.14 × 10−5

Q3

The flux linkage in coil 1 is 3 Wb
and it has x turns and the current
in coil 2 is 2 A, calculate the
value of x if the mutual
inductance is 750 H.

Example for Solution

1. List the information given;

Φ1 = 3
2 = 2
= 750

2. Choose a suitable formula:
1Φ1
= 2

3. Apply the formula.

= 1Φ1
1 = 2 2
Φ1
(750)(2)
= (3)

= 500

Q4

Two coils are placed side by side and
fixed to their positions. When there is no
current in coil 1, the current in coil 2
increases at a rate of 12 As-1. The induced
e.m.f in coil 1 is 100 mV.
But if there is no current in coil 2 and
current in coil 1 changes at rate of 2.4 As-
1, what will the induced e.m.f in coil 2?

Example for Solution

1. List the information given;

Case 1, Case 2,
1 = 0
2 2 = 0
1
= 12 −1 = 2.4 −1
× 10−3
1 = 100

2. Choose a suitable formula:
1
2 =

3. Apply the formula. 1
2
=

100 × 10−3
= 12 −1
= 8.33 × 10−3 H

So, 2 = 1

= (8.33 × 10−3)(2.4)
= 20 × 10−3