1 Crore+ students have signed up on EduRev. Have you? 
A 5 H coil is coupled with a 20 H coil. What could be the maximum value of mutual inductance between them considering no flux linkage?
Concept:
Mutual inductance:
where K is the coefficient of coupling.
K ranges from 0 to 1.
When two coils are not in proximity to each other, then flux linkage between them will be zero, so, the value of K would be 1 in that case.
Given:
L_{1} = 5 H and L_{2} = 20 H.
Here K = 1.
From 1, mutual inductance (M) can be given as
= 10 H.
Identify the INCORRECT statement among the given options regarding mutual inductance.
Mutual Inductance:
Mutual inductance is given by
Where μ_{0} = Permeability of free space = 4π × 10^{7} H/m
μ_{r} = Relative permeability of the core
N_{1} = Number of turns of coil 1
N_{2} = Number of turns of coil 2
A = Crosssectional area in m^{2}
What is the basic operating principle of mutual induction?
Where:
µ_{o} is the permeability of free space (4π x 10^{7})
µ_{r} is the relative permeability of the soft iron core
N_{1}, N_{2} is in the number of coil turns
A is in the crosssectional area in m^{2}
L is the coil's length in meters.
Two identical coils X and Y of 500 turns each lie in parallel planes such that 80% of flux produced by one coil links with the other. If a current of 5 A flowing in X produces a flux of 10 mWb in it, find the mutual inductance between X and Y
Concept:
Consider two coils having selfinductance L_{1} and L_{2} placed very close to each other. Let the number of turns of the two coils be N_{1} and N_{2} respectively. Let coil A carries current I_{1} and coil B carries current I_{2}.
Due to current I_{1}, the flux produced is ϕ_{1} which links with both the coils. Then the mutual inductance between two coils can be written as
Here, ϕ_{12} is the part of the flux ϕ_{1} linking with the coil 2
Calculation:
Number of turns = N_{1} = N_{2} = 500
Current (I) = 5 A
Flux produced in coil X (ϕ_{1}) = 10 mwb
Flux linked with Y (ϕ_{12}) = 80% of flux produced in coil 1 = 8 mwb
Mutual inductance
Two coupled coils with L_{1} = 0.5 H and L_{2} = 4.0 H have a coefficient of coupling 0.8. Find maximum value of the inductance EMF in the coil 2 if a current of i_{1} = 20 sin 314t A is passed in coil 1.
Concept :
Selfinductance is the property of the currentcarrying coil that resists or opposes the change of current flowing through it. This occurs mainly due to the selfinduced emf produced in the coil itself. In simple terms, we can also say that selfinductance is a phenomenon where there is the induction of a voltage in a currentcarrying wire.
Selfinductance, usually just called inductance, L is the ratio between the induced voltage and the rate of change of the current
V_{1}(t) = L (di_{1}/dt)
Mutual Inductance between the two coils is defined as the property of the coil due to which it opposes the change of current in the other coil, or you can say in the neighboring coil. When the current in the neighboring coil changes, the flux sets up in the coil and, because of this, changing flux emf is induced in the coil called Mutually Induced emf and, the phenomenon is known as Mutual Inductance.
V_{2}(t) = M (di_{1}/dt)
Coefficient of Coupling :
The amount of coupling between the inductively coupled coils is expressed in terms of the coefficient of coupling, which is defined as
where M = mutual inductance between the coils
L_{1} = selfinductance of the first coil, and
L_{2} = selfinductance of the second coil
Calculations :
Given
L_{1} = 0.5 H
L_{2} = 4 H
K = 0.8
I = 20 sin 314t
The induced emf in coil 2 due to current in coil 1 is given by
V_{2} = M (di_{1} / dt)
M = 1.13 H
V_{2} = 1.13 d/dt (20 sin 314t)
V_{2} = 1.13 × 20 × 314 × cos 314t
The maximum value of V_{2} = 1.13 × 20 × 314 = 7.1 kV
Two inductors L_{1} = 20 mH and L_{2} = 40 mH are connected in series so that their equivalent inductance is 50 mH. The mutual inductance between the two coils is _______.
Concept:
The equivalent inductance of series aiding connection is
L_{eq} = L_{1} + L_{2} + 2M
The equivalent inductance of series opposing connection is
L_{eq }= L_{1} + L_{2} – 2M
Calculation:
Given:
L_{1} = 20 mH
L_{2} = 40 mH
L_{eq} = 50 mH
From the given data we can conclude that series opposing connection will be considered as the mutual index cannot be negative.
L_{eq} = L_{1} + L_{2} – 2M
50 = 40 + 20  2M
M = 5 mH
The input impedance, Zin (s), for the network shown is
Concept:
Mutual Inductance:
When two coils are placed close to each other, a change in current in the first coil produces a change in magnetic flux, which cuts not only the coil itself but also the second coil as well. The change in the flux induces a voltage in the second coil, this voltage is called induced voltage and the two coils are said to have a mutual inductance.
Consider a pair of coupled inductors with selfinductance L_{1} and L_{2}, magnetically coupled through coupling coefficient k.
Input and output voltage expressions are given as
V_{1} = jωL_{1}I_{1} + jωMI_{2} ...(1)
V_{2} = jωL_{2}I_{2} + jωMI_{1} ...(2)
Where,
ω = 2πf
M = Mutual inductance
L_{1} = Inductance of coil one
L_{2} = Inductance of coil two
Calculation:
The given circuit in the Laplace domain is
Apply KVL in the input loop,
V_{1}(s) = (4 + 6s)I_{1}(s) + sI_{2}(s) ...(1)
Apply KVL in the output loop
I_{2}(s)[5 +4s] + sI_{1}(s) = 0
Substitute I2(s) in equation(1)
10 μF capacitor is connected across the secondary winding of a highfrequency transformer having primary to secondary turns ratio 5: 2. What is the value of capacitance seen across primary?
Concept:
For a transformer, the input impedance (Z_{in}) for a given load impedance (Z_{L}) is given by:
Where Zin is also called the Reflected impedance since it appears as if the load impedance is reflected in the primary side.
Calculation:
Therefore the value of capacitance seen across the primary is:
A transformer has 350 primary turns and 1050 secondary turns. The primary winding is connected across a 230 V, 50 Hz supply. The induced EMF in the secondary will be
Concept:
In a transformer, the relation between the number of turns, current, and voltages is given by:
N_{1} and N_{2} = number of turns in the primary and secondary windings respectively
V_{1} and I_{1} = Voltage and current respectively at the primary end
V_{2} and I_{2} = Voltage and current respectively at the secondary end
Calculation:
Given V_{1} = 230 V, 50 Hz supply
N_{1} = 350
N_{2} = 1050
Putting on the respective values, we get:
V_{2} = 690 V
The frequency of the secondary generated voltage will be the same as the input frequency.
Two inductors whose selfinductance is 80 mH and 60 mH are connected in parallel aiding. The equivalent inductance of the combination is 48.75 mH. Calculate their mutual inductance.
Concept:
For Additive coupling (assists the selfinductance):
Calculation:
Given: L_{1} = 80 mH L_{2} = 60 mH
L_{eq} = 48.75 mH
Mutual inductance can be calculated from equation (1):
2M^{2}  195M + 4050 = 0
By using the following formula we can find the roots
On solving the above expression we get:
M = 30 and M = 135/2.
From the given options we can say that M = 30 mH is the correct answer.
11 videos46 docs62 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
11 videos46 docs62 tests









