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Magnetically Coupled Circuit
Magnetically Coupled Circuit  When two inductors are placed physically closed to each other then due flow in the first inductor, some flux is produced in the first inductor. part ot this flux will link with second inductor. therefore, using faraday’s law of electromagnetic induction this rate of change of flux will induce. some voltage across the second inductor. this induced emf will be additive or subtractive in nature.
All the KVL equation in the given network are then modified.
. Effect of M is positive.
. Induced voltage is additive.
e1 = L1 dI1/dt + M dI2
e2 = L2 dI2/dt + M dI1/dt
Assume zero initial conditions,
e1(s) = sL1I1 (s) + s MI2 (s)
e2 (s) = sL2 I2 (s) + MsI1 (s)
|e1 (s)| = |sL1  sM||I1 (s)|
|e2 (s)| = |sM   sL2||I2 (s)|
[V1/V2] = joL1/JoM    joM/joL2] [I1/I2]
. Effect of M is negative.
. Induced voltage is subtractive.
e1 = L1 dI1/dt – M dI2/dt
e2 = – M dI1/dt + L2 dI2/dt (in time domain)
Assume zero initial conditions
|e1 (s)/e2 (s)] = sL1 – sM/-sM  sL2] [I1 (s)/I2 (s)
[e1/e2] = [joL1 – joM/-jM  JoL2] [I1/I2]
Representation of M
(i)  in terms of M
(ii) in terms of K
K = coupling coefficient
K = M/L1L2
K = 1, Critical coupling (critically coupled networks)
K < 1, Loose coupling (under coupled networks)
K > 1, Tight coupling (over coupled networks)
(iii) in terms of turns ratio
e1/e2 = n1/n2 = I2/I1
Example 1. (a) LEQ = L1 + L2
(b)  LEQ = L1 (effective) + L2 (effective)
LEQ = L1 + M + L2 + M
LEQ = L1 + L2 + 2M
(c) LEQ = L2 (effective) + L2 (effective)
LEQ = L1 + L2 – 2M
(d) LEQ = L1 (effective) L2 (effective)
LEQ = L1 + L2 – 2M
Example 2.  Draw the KVL equation of the circuit shown in fig. (a), after drawing the dotted equivalent.
Sol. The dotted equivalent and the voltage equivalent is shown in fig. (b) and fig. (c) respectively. Applying KVL to the circuit of fig. (c), we get
RI + L1 dI/dt – M dI/dt + 1/C |I dt + L2 di/dt – M dI/dt = V
RI = L dI/dt + 1/C |I dt = V
L’ = L1 + L2 – 2M
Example 3. Obtain the dotted equivalent of the circuit shown in fig. (a). write the KVL equation and obtain the voltage across the capacitor C, where R1 = R2 = 5. 0L1 = 0L2 = 5, 0C = 0.1 and 0M = 2 V1 = 10 V, V2 = J 10 V.
Sol. To place the dots, consider only the coils and their winding sense. drive the current into the top of the left coil and place a dot at this terminal. the corresponding flux is upward. by lenz’s law, the flux at the right coil must be upward, directed to oppose the first flux. then the natural current leaves this winding by the upper terminal which is marked with a dot. the dotted equivalent circuit is shown in figure. the KVL equations can be written from fig. (c) by putting the values of the components as
|5 – j5   5 + j3/5 + j3   10 + j6|[I1/I2] = [10/10 – J10]
Solving for I,
|10    5 + j3/10 – j10  10 + j6/5 – j5   5 + j3/ 5 + j3  10 + j6| = 1.015 < 113.960 A
The voltage across the capacitor C is
V = I1 (-j10) = 10.15 V
Intro Exercise – 7

  1. Effective inductance of the circuit across the terminal AB in the figure as shown below is

(a) 9 H
(b) 21 H
(c) 11 H
(d) 6 H

  1. Two coils in differential connection have self inductance of 2 mH and 4 mH and a mutual inductance of 0.15 mH. the equivalent inductance of the combination is

(a) 5.7 mH
(b) 5.85 mH
(c) 6 mH
(d) 6.15 mH

  1. When two coupled coils of equal self inductance are connected in series in one way the net inductance is 12 mh and when they are connected in the other way, the net inductance is 4 mh. the maximum value of net inductance when they are connected in parallel in a suitable way is

(a) 2 mH
(b) 3 mH
(c) 4 mH
(d) 6 mH

  1. Given two coupled inductors L1 and L2 their mutual inductance M satisfies

(a) M = L12 + L22
(b) M > (L1 + L2)/2
(c) M = L1L2
(d) M < 2/L1L2

  1. The impedance seen by the source in the circuit figure is given by

(a) (0.54 + j 0.313)
(b) (4 – j2)
(c) (4.54 – j 1.69)
(d) (4 + j2)

  1. In the transformer shown in the given figure, the inductance measured across the terminals 1 and 2 was 4 h with open terminals 3 and 4. it was 3 h when terminal 3 and 4 were short-circuited. the coefficient of coupling would be

(a) 1
(b) 0.707
(c) 0.5
(d) interminate due to insufficient data

  1. The resonant frequency of the given series circuit is

(a) 1/2  3 Hz
(b) 1/4  3 Hz
(c) 1/4  2 Hz
(d) 1/2  Hz

  1. The maximum value of mutual inductance of two inductively coupled coils with self inductance L1 = 49 mH and L2 = 81 mH is

(a) 130 mH
(b) 63 mH
(c) 32 mH
(d) 3969 mH

  1. Find the value of V1.

(a) – 16 cos 2t V
(b) 16 cos 2t V
(c) 4 cos 2t V
(d) -4 cos 2t V

  1. Find the value of V2.

(a) 2 cos 2t V
(b) – 2 cos 2t V
(c) 8 cos 2t V
(d) – 8 cos 2t V
Answers with Solutions

  1. (c)

Leffective (across)
AB = L1 + L2 + L3 – 2M12 – 2M13 + 2M23
= (4 + 5 + 6 – 2 x 1 – 2 x 3 + 2 x 2) H
= (15 – 2 – 6 + 4) H
= (15 – 8 + 4) H
= (15 – 4) H
= 11 H

  1. (a)

when two coils are connected in series, the effective inductance
Leffective = L1 + L2 + 2 M
For this case, Leffective = L1 + L2 – 2 M
= 2 + 4 – 2 x 0.15
= 5.7 mH

  1. (b)

L1 = L2 = L
Leffective = L1 + L2 + 2 M (when two coils are connected in series)
Leffective = L1 + L2 + 2 M
= 2L + 2M
2L + 2M = 12……………(1)
2L + 2M = 2
L = 4, M = 2
To get maximum value in parallel connection
LMAX (L + M) (L + M)/2L + 2M = (4 + 2) (4 + 2)/(8 + 4)
LMAX = 36/12 3 mH
LMAX = 3 mH

  1. (d)

M = K L1L2
K = M/L1L2
K < 1
M/L1L2 < 1
M < L1L2

  1. (c)

Z1 = 10 < 300 x (1/4)2
Z1 = (0.54 + 1 J 0.31)
Total impedance = (4 – j2) + (0.56 + j0.31)
= (4.54 – j1.69)

  1. (d)

data is insufficient to calculate the value of coupling coefficient (k).

  1. (b)

LEffective or LEQ = L1 + L2 + 2M
L1 = 2 + 2 + 2 x 1
LEQ = 6 H
At resonance,
0L = 1/0C
0 = 1/LEQC = 1/12
I = 1/4  3 Hz

  1. (b)

M > K L1L2
Maximum value of (m)
MMAX = K L1L2
MMAX = L1L2
MMAX = 40 x 81
MMAX = 63 mh

  1. (b)

V1 = 2dI1/dt + 1. di2/dt
V1 = 2di1/dt = 16 cos 2t V

  1. (c)

V2 (1)  di2/dt + (1) di1/dt
V2 = di1/dt
V3 = 8 cos 2t V