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