4
Magnetic Circuits
OBJECTIVE
OBJECTIVE--TYPE QUESTIONS
1. Mutual inductance between two magnetically coupled coils depends on
(a) the number of their turns (b) permeability of core
(c) cross-sectional area of their common core
(d) all of the above
2. Hysterisis is the phenomenon of - - - - in a magnetic circuit.
(a) leading of B behind H
(b) lagging of H behind B
(c) setting up constant flux
(d) none
3. Two coils are placed closely to each other having mutual inductance of
1 H. If the primary coil current varies linearly from 0 to 10 A in 0.01 s,
then the emf induced is
(a) 0 V
(b) 10 V
(c) 100 V
(d) 1000 V
4. The coefficient of coupling between the primary and secondary of a practical iron core transformer is about
(a) 0.9
(b) 0.1
(c) 1.5
(d) 0.2
5. A 10-turn coil has a second layer of 10 turns wound over the first. Then
the total inductance will be about - - - - the original inductance.
(a) 2 times
(b) 4 times
(c) 6 times
(d) 3 times
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4.2
Basic Electrical Engineering
6. When two coils are connected in series, their total inductance is 80 mH.
When the connection of one coil is reversed, the total inductance becomes
40 mH. Then the mutual inductance between the coils is
(a) 10 mH
(b) 20 mH
(c) 60 mH
(d) 120 mH
7. The total inductance of the circuit is
(a) 13 H
(b) 10 H
(c) 12 H
(d) 19 H
M=3H
L1 = 9 H
L2 = 4 H
A
B
Fig. 4.1
8. The magnetic field intensity at the centre of a circular current loop is
(a) proportional to the radius of the loop
(b) inversely proportional to the radius of the loop
(c) inversely proportional to the square of the radius of the loop
(d) none
9. The magnetic field intensity (in A/m) at the centre of a circular coil of
diameter 1 m when carrying current of 2A is
(a) 8
(b) 4
(c) 3
(d) 2
10. The permeance in a magnetic circuit corresponds to - - - - in an electric
circuit.
(a) conductivity
(b) resistivity
(c) conductance
(d) resistance
PROBLEMS
4.1 Two coupled coils of L1 = 0.8 H and L2 = 0.2 H have a coupling coefficient
N
K = 0.9. Find the mutual inductance M and turns ratio 1 .
N2
Solution:
M=K
M = 0.9
L1 L2
0.8 × 0.2
= 0.36 H
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Magnetic Circuits
4.3
M=
N2 φ 12
I1
(1)
K=
φ 12
I1
(2)
From equation (2) f12 = Kf1
(3)
We know that
Substitute Eq. (3) in Eq. (1) and multiply Eq. (1) by
M=
N1
N2
N2
Kf1 N1
I1
N2
M=K
N2
N1
FG N φ IJ = K N
H I K N
1
1
2
L1
1
1
KL1
0.9 × 0.8
N1
=
=
=2
M
0.36
N1
4.2 A coil of 100 turns is wound on a magnetic core having a reluctance of 106
AT/wb when the coil current is 5 A.
Solution:
L=
L=
N 2 µ 0 aµ r
l
N2
N2
=
l
S
a µoµ r
N2
(1000 ) 2
=
=1H
S
106
1
1
W = LI2 = ´ 1 ´ 5 ´ 5 = 12.5 J
2
2
4.3 Coupled coils with L1 = 0.03 H, L2 = 0.02 H and K = 0.6 are connected in
four different ways: series aiding, series opposing and parallel with both
arrangements of the winding sense. What are the four equivalent inductances?
Solution: (i) Series aiding
L=
Leq = L1 + L2 + 2 M
M=K
L1 L2 = 0.6
0.03 × 0.02
= 0.0146 H
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4.4
Basic Electrical Engineering
Leq = 0.03 + 0.02 + 2 (0.0146)
= 0.0792 H
M
I
Fig. 4.2
(ii) Series opposing
Leq = L1 + L2 – 2M
= 0.03 + 0.02 – 2 (0.0146)
Leq = 0.0208 H
M
Fig. 4.3
(iii) Parallel aiding
Leq =
=
L1 L2 − M 2
L1 + L2 − 2 M
( 0.03) (0.02) − (0.0146) 2
0.03 + 0.02 − 2 ( 0.0146 )
= 185.98 ´ 10 – 4 H
M
L1
L2
Fig. 4.4
(iv) Parallel opposing
Leq =
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L1 L2 − M 2
L1 + L2 − 2 M
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Magnetic Circuits
=
4.5
( 0.03) (0.02) − ( 0.0146) 2
( 0.03) + (0.02) + 2 ( 0.0146)
= 48.84 ´ 10–4 H
M
L1
L2
Fig. 4.5
4.4 An iron ring has a cross-section of 80 millisquare metres and a mean
diameter of 38.2 cm. It is uniformly wound with a coil of 480 turns. For a
current of 2 A, the flux set up is 1.2 mwb. Find the relative permeability of
iron at the flux density.
Solution: Area of cross-section
(a) = 80 ´ 10–3 sq.m
Mean diameter = 38.2 cm
Current through the coil = 2A
Number of turns = 480
MMF = NI
= 480 ´ 2 = 960 AT
Flux (f) = 1.2 ´ 10–3 wb
Flux density =
φ
1.2 × 10 −3
=
= 0.015 wb/m2
A
80 × 10 −3
Mean length = pD
= p ´ 0.382 m = p 0.382
H=
MMF
960
=
AT/m
Length 0.382 π
B
0.015 × 0.382π
=
H
960
0.015 × 0.382π
mr =
= 14.92
4π × 10 −7 × 960
Permeability (m0 mr) =
4.5 An iron ring has a mean diameter of 25 cm and a cross-sectional area of
4 cm2. It is wound with a coil of 1200 turns. An airgap of 1.5 mm width
is cut in the ring. Determine the current required in the coil to produce a
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4.6
Basic Electrical Engineering
flux of 0.48 mwb in the airgap. The rative permeability of iron under this
condition is 800. Neglect leakage.
Solution:
Flux = 0.48 ´ 10–3 wb
Area of cross section = 4 cm2 = 4 ´ 10–4 m2
B=
AT for air gap, Hair =
0.48 × 10 −3
φ
=
= 1.2 wb/m2
4 × 10 −4
A
B
1.2
=
AT/m
4π × 10 −7
µ0
lair = 1.5 ´ 10–3 m
AT required for air gap =
1.2
´ 1.5 ´ 10–3
4π × 10 −7
= 1430 AT
B
1.2
=
4π × 10 −7 × 800
µ0 µr
Length (lIron) = p ´ 0.25 m
HIron =
AT for iron portion =
1.2 × 10 7
´ p ´ 0.25 = 937 AT
4π × 800
Total AT required = 1430 + 937 = 2367 AT
Current required =
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2367
= 1.975 Amp
1200
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