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EXAMPLE-self

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EXAMPLE-self

EXAMPLE-self

EXAMPLE OF
PROBLEM SOLVING

SELF INDUCTANCE

Q1 Induced e.m.f of 5.0 V is developed
across a coil when the current
flowing through it changes at 25 A s-1.

Determine the self inductance of the
coil.

Example for Solution

1. List the information given;

= 5.0

= 25 −1

2. Choose a−s u i t a b le formula :

=

3. Apply the formula.

= During calculation, we only consider the
magnitude of the e.m.f induced.
So, = Minus sign that indicates the direction of e.m.f
( ) (Lenz’s Law ) can be ignored.

5
= 25
= 0.2

Q2 Calculate the magnetic flux through
the area enclosed by a 300 turn, 7.20
mH coil when the current in the coil
is 10 mA.

Example for Solution

1. List the information given;

= 300
= 7.2 × 10−3
= 10 × 10−3 A

2. Choose a suitable formula :

=

3. Apply the formula.

=

So, ∅ =

(7.2 × 10−3)(10 × 10−3)
= 300
= 2.4 × 10−7

Q3 (a) Calculate the self inductance of a solenoid
containing 300 turns if the length of the
solenoid is 25.0 cm and its cross sectional
area is 4 cm2.

(b) If the current through solenoid is
decreasing at the rate of 50.0 A s-1, what is
the self induced e.m.f?

Example for Solution

(a) 1. List the information given;

= 300
= 0.25
= 25 × 10−4 2

2. Choose a suitable formula :

=

3. Apply the formula.

= 0 2

× 10−7) 300 2(4 × 10−4)
(4
= 25 × 10−2
= 1.81 × 10−4

Example for Solution

(b) 1. List the information given;

= 50 −1 (From previous question)
= 1.81 × 10−4

2. Choose a suitable formula :

=

3. Apply the formula.

=

= (1.81 × 10−4)(50)

= 9.05 × 10−3 V

Q4 The inductance of a coil is 3.0 mH. The
current flow through the coil changes
from 0.26 A to 1.3 A in 0.18 s.
Calculate the magnitude of the average
induced e.m.f in the coil within the
period.

Example for Solution

(b) 1. List the information given;

= 3 × 10−3
= 1.04
--------- (0.26 A to 1.3 A)

= 0.18

2. Choose a s u i t able formula :

=

3. Apply the formula.

=

= (3 × 10−3)(10..1084)

= 17.33 × 10−3 V

2019 Minesing Booklet
Magic English 44 [Rus.] - Country life (Playtime)