**Table of contents**

**How is the mutual inductance used in a transformer?**

**Mutual Inductance Transformer**

A transformer consists of 2-types of winding’s.

- Primary winding.
- Secondary winding.

By the principle of mutual inductance, whenever the current changes in the primary coil, it changes the current in the secondary coil. The variable current in the primary coil creates variable magnetic flux in the core. This magnetic flux in the core induces a varying voltage in secondary winding; thus, mutual inductance in transformer is applied.

**Mutual inductance Formula**

Mutual inductance formula for any two inductor coils is [Latex]M = \phi i[/Latex] where phi is the magnetic flux produced in one coil and i is the current through another coil due to which the flux generates.

**What is self and mutual inductance?**

Self-inductance is the property of an inductor for which it opposes any change in current, if there are two or more coils, then any change in current passing through one coil induces EMF in other coils. This is mutual induction. Mutual inductance is the effect of mutual induction.

**What is the effect of mutual inductance?**

The main impacts of mutual inductance is the consequence variation in current at one coil will result in the generation of EMF in other coil.

**What is the mutual inductance formula of the two coils in between ?**

**Mutual Inductance Of Two Solenoids**

Mutual inductance of two solenoids, [Latex]M = \frac{\mu {0}\mu {r}N_{1}N_{2}A}{l}[/Latex]

Where,

µ_{o} = permeability of free space (4π x 10^{-7}).

µ_{r} = relative permeability of the iron core.

N_{1} and N_{2} = numbers of coil turns in two coils.

A = cross-sectional area.

ℓ = length of the coil.

**What are inductance and mutual inductance?**

Induction is the property of inductors coil due to which these opposes any current change in it and mutual inductance is the reason why EMF is induced in one coil for the change in current in another closely placed coil.

**What are the reciprocity properties of mutual inductance?**

The reciprocity property of mutual inductance says that M_{12} = M_{21}, i.e. there is no individual mutual inductance of two coils and mutual inductance will be same for the two.

**To know more about mutual inductance click here**

**What are the electrical characteristics of capacitance inductance and mutual inductance of 40 meters of uncoiled 3 core extension lead 1.5 sq mm copper flex?**

Generally, 3 core extensions have the inductance value of 1 mH/meter. So we can make the following conclusions-

- The mutual inductance can be as high as 0.8 microhenry/meter because the wires are right next to each other.
- It can be around 0.7 mm in diameter, and the separation is about 0.5mm.
- The dielectric constant is of an approximate value of 2 (some air, some plastic). Therefore the capacitance is nearly 20 pF.

**What do you mean by self and mutual inductance? Find a relationship between them by defining the coefficient of coupling?**

The current passing thru a coil generated from its own magnetic field is known as self-inductance and Contrarily, the current flowing in one coil due to the influence of the magnetic field in another coil is called mutual inductance.

The fractional part of magnetic flux generated by the current in one coil that linked with other coil is known as the co-efficient of coupling and generally denoted by (k).

[Latex]k = \frac{M}{\sqrt{L_{1}L_{2}}}[/Latex]

Where, k=co-efficient of coupling.

M=mutual inductance between the 2-coils.

L_{1}, L_{2} = self-inductance of 2-coils.

**When is the mutual inductance between two coils zero?**

Let us assume one coil is placed on one arm of the core. The other coil can be wound in such a way that half of the turns are in the clockwise direction, and the other half is in the counterclockwise direction. The magnetic flux due to the primary with one half of the coil gets canceled by that of the second half of the coil. Therefore, the overall effect of the primary side on the secondary side is zero, and mutual inductance is also zero.

**How to insulate two coils to prevent mutual inductance?**

In two ways, insulation can be done to prevent mutual inductance-

- By winding the coils in opposite directions, the first coil left-handed or counterclockwise, the second coil right-handed or clockwise
- By placing one cool on top of the PWB (Printed Wiring Board)
- By placing them 90 degrees to each other

**Why are the self-inductance and mutual inductance of an ideal transformer infinite?**

An ideal transformer is said to have infinite magnetic permeability, Thus the self-inductance and mutual inductance subsequently become infinite.

**How to achieve zero inductance?**

Zero inductance can be achieved through a process called non-inductive binding. The resistors in the resistance box are made by using wire of ‘manganin’. The wire of the required length is folded in the middle and then it is wound on a bobbin. The two ends of the wire are soldered to the two ends of the gap, if a wire is doubled up and wound like this the current is clockwise in one set of turns, but it is counter-clockwise in the other set of turns. So inductance effects cancel out. So, this is called non- inductive winding.

**Find mutual inductance of the two concentric coplanar coils?**

Let us assume two concentric co planar coils with radius R and r where R>r, current= i. Therefore magnetic field at the center = μ_{0}i/2R

Flux through the inner coil = μ_{0}i/2R x πr^{2}

Therefore mutual inductance M = flux/current = μ_{0}πr^{2}/2R

**Can mutual inductance be negative?**

The order of magnitude of the mutual inductance can never be negative, However, its sign can be negative or positive depending on the polarity of the induced EMF and direction of the induced current.

**What is magnetizing current?**

Transformers draw constant current from the supply for producing magnetic flux. It is known as magnetizing current. It does not depend upon the nature of load.

**What can happen if a transformer fails?**

Transformer failure can cause black-out in the total area where power is supplied. The oil used in the transformer core can increase the risk of fire.

**Define auto-transformer.**

An auto-transformer is a device that has the same winding for the primary and the secondary coils, unlike isolation transformers.

**What is a single-phase and a three-phase transformer?**

If a transformer works on single-phase supply then it is known as a single-phase transformer. Similarly, the transformers working on three-phase supply are known as three-phase transformers.

**Mutual inductance circuit problem – Mutual inductance transformer related | Mutual inductance mesh analysis**

Find the input impedance and current passing through the coil connected to the supply voltage in the circuit below. Z_{1} = 60 – j100 ohm, Z_{2} = 30 + j40 ohm and load impedance Z_{L} = 80 + j60 ohm. Supply voltage = 50∠60, mutual inductance = j5 ohm, primary coil impedance = j20 ohm, and secondary coil impedance = j40 ohm.

Let us assume the input impedance current is i_{1} and the reflected impedance current is i_{2}. Both are flowing in the clockwise direction.

We know, input impedance, [Latex]Z_{in} = Z_{P}+Z_{R} = R_{1} + j\omega L_{1} + \frac{\omega ^{2}M^{2}}{R^{2}+j\omega L_{2}+Z_{L}}[/Latex]

Putting all the given values we get,

[Latex]Z_{in} = 60-j80 + j20 +\frac{5^{2}}{30 + j40 + j40 + 80 + j60} = 60 – j60 + \frac{25}{110 + j140} = j59.53 – 59.63 ohm[/Latex]

Input impedance current i_{1} = V/Z_{in} = 50∠60 / 84∠-45 = 0.6∠105

**Mutual inductance transformer formula-**

There’s zero power loss in an ideal transformer. So, the input power = output power

[Latex]W_{1}i_{1}cos\phi = W_{2}i_{2}cos\phi[/Latex] or [Latex]W_{1}i_{1} = W_{2}i_{2}[/Latex]

Therefore, [Latex]\frac{i_{1}}{i_{2}} = \frac{W_{2}}{W_{1}}[/Latex]

Since voltage is directly proportional to the no. of turns in the coil.,

we can write,

[Latex]\frac{V_{2}}{V_{1}} = \frac{W_{2}}{W_{1}} = \frac{N_{2}}{N_{1}} = \frac{i_{1}}{i_{2}}[/Latex]

If V_{2}>V_{1}, then the transformer is called a **step-up transformer**.

If V2<V1, then the transformer is called a step-down **transformer **.

**Numerical problems | Mutual inductance example problem**

**If 2 co-axial solenoids are constructed by winding’s utilized by thin insulated wire over a pipe of cross-sec-area A = 10 cm**^{2} and L = 20 cm, and If one solenoid have 300 turns and the other has 400 turns, calculate the mutual-inductances in between.

^{2}and L = 20 cm, and If one solenoid have 300 turns and the other has 400 turns, calculate the mutual-inductances in between.

**Detailed Solution :**

We know, mutual inductance of two coaxial solenoids = [Latex]\frac{\mu {0}N{1}N_{2}A_{1}}{l} = \frac{4\pi \times 400 \times 300 \times 10 \times 10^{-4}}{20 \times 10^{-2}} = 0.75 \; mH[/Latex]

**Self-inductance and mutual inductance problem**

**Two solenoids (of the same length) s**_{1} and s_{2} have the areas in ratio 3:4 and number of turns in ratio 5:6. If the self-inductance of s_{1} is 10 mH, find the mutual inductance of the solenoids.

_{1}and s

_{2}have the areas in ratio 3:4 and number of turns in ratio 5:6. If the self-inductance of s

_{1}is 10 mH, find the mutual inductance of the solenoids.

**Detailed Solution :**

Self-inductance of s_{1}, [Latex]L_{1} =\frac{\mu {0}N{1}^{2}A_{1}}{l}[/Latex]

Mutual inductance, [Latex]M =\frac{\mu {0}N{1}N_{2}A_{2}}{l}[/Latex]

[Latex]\frac{M}{L_{1}} = \frac{\frac{\mu {0}N{1}N_{2}A_{2}}{l}}{\frac{\mu {0}N{1}^{2}A_{1}}{l}} = \frac{N_{2}}{N_{1}}\times \frac{A_{2}}{A_{1}}= \frac{6\times 4}{5\times 3} = \frac{8}{5}[/Latex]

So, M = 8/5 x L = 16 mH

**Combination of inductors with mutual inductance | Three inductors in series with mutual inductance**

**Q. Find the total inductance of three series connected mutually coupled coils with L**_{1} = 2 H, L_{2} = 4 H, L_{3} = 6 H and M_{12} = 1 H, M_{23 }= 2 H, M_{13} = 1 H

_{1}= 2 H, L

_{2}= 4 H, L

_{3}= 6 H and M

_{12}= 1 H, M

_{23 }= 2 H, M

_{13}= 1 H

**Detailed Solution :**

Total inductance of coil1 = L_{1} + M_{12} – M_{13} = 2 H

Total inductance of coil2 = L_{2 }+ M_{12} – M_{23} = 3 H

Total inductance of coil3 = L_{3} – M_{13} – M_{23} = 3 H

Therefore total = 2 + 3 + 3 = 8 H

**MCQ on Inductor**

**1. If in an iron core inductor the iron core is removed to make it air-core, the inductance will be**

- More
**b. Less**c. Same d. Insufficient data

**Detailed Solution :**

The inductance of the iron-core inductor = μ_{0}μ_{r}N_{2}A/l where μ_{r} is the relative permeability of the iron-core.

If the iron core is removed, the inductance of the air-core inductor = μ_{0}N_{2}A/l

μ_{r}>1, so the inductance decreases when the iron core is removed.

**2. If the current in one coil is steadied, what happen to the mutual inductance?**

**0**b. ∞ c. Twice d. half.

**Detailed Solution :**

A current is induced when the magnetic flux is changing. The induced current in the other coil is ‘0’ if current get steadied in one coil, So, answer is 0.

**3. Calculate the value of x if the Mutual inductance is 20 Henry, the inductance of coil-1 is x Henry and the inductance of coil-2 is 8Henry, assume coupling co-efficient is 5.**

**2****H****enry.**b) 4 Henry. c) 6 Henry. d) 8 Henry.

**Detailed Solution :**

We know , M=k√L_{1}L_{2 }

20 = 5√8x so x = 2 H

**4. There are two long coaxial solenoids of the same length l. The inner and outer coils have radius r**_{1},r_{2} and the no. of turn/ unit-length are n_{1}, n_{2}. Then calculate the ratio of mutual inductance/ self-inductance of the inner coil.

_{1},r

_{2}and the no. of turn/ unit-length are n

_{1}, n

_{2}. Then calculate the ratio of mutual inductance/ self-inductance of the inner coil.

**n**b. (n_{2}/n_{1}_{2}/n_{1})(r_{2}^{2}/r_{1}^{2}) c. (n_{2}/n_{1})(r_{1}/r_{2}) d. n_{1}/n_{2}

**Detailed Solution :**

Mutual inductance M = μ_{0}N_{p}N_{s}A_{s}/l_{p} where p denotes parameters of primary coil and s denotes those of secondary coil.

Therefore, M = μ_{0} n_{1}l x n_{2}l x A_{2}/l = μ_{0}n_{1}n_{2}A_{2}l

Self-inductance L_{2} of inner coil = μ_{0}n_{2}^{2}A_{2}/l

So, ratio M/L_{2} = n_{2}/n_{1}

**5. Two circular coils are arranged in three situations shown below, Their mutual inductance will be maximum at which of the arrangement.**

**In (i)**b. In (ii) c. In (iii) d. Same in all

**Detailed Solution :**

Mutual inductance M=ϕi where ϕ is the flux thru one coil due to the current i in other coil and the flux ϕ = B.A where B is the magnetic field vector and A is the area vector and B and A are parallel in (i) but perpendicular in (ii) and (iii). so, the flux and mutual inductance are maximum in (i).

**MCQ on mutual inductance transformer related**

**1. The transformer ratings are measured in _____________**

a) kW

b) kVAR

c) HP

**d) kVA**

**Detailed Solution :**

There are two kinds of losses in a transformer- copper Losses and core Losses. Copper losses depend on the current passing through the winding and core losses depend on voltage. So the rating of the transformer is given in kVA.

**2. What does a transformer transform?**

a) frequency

b) current

c) **power**

d) voltage

**Detailed Solution :**

The voltage and the current are changed in the transformer. So we can say power is transformed.

**3. We add ___________ to convert an ideal transformer to a real transformer**

a) Primary winding resistance and secondary winding’s resistance.

b) Primary winding leakage reactance and secondary winding’s leakage reactance.

c) **Primary winding****’s resistance, leakage****-reactance, and ****2 ^{nd} winding**

**’s leakage reactance**

**.**

d) Can’t solve.

**Detailed Solution :**

The primary and the secondary resistances along with leakage reactance are connected in the circuit as series parameters.

**4. A single-phase transformer ratings 250 KVA, 11000 V/415 V, 50 Hz, Find the primary current.**

a) 602.4Amp.

b) 602.4Amp.

**c) 22.7A****mp.**

d) 11.35Amp.

**Detailed Solution :**

The primary current is the ratio of the power of the transformer to the primary voltage.Thus, primary current = power/voltage= 250000/11000= 22.7 A.

**5. A 100 kVA transformer with R=700Ω and L=1.2 H can be operated on both 60 & 50 Hz frequencies. For the same rating, the output will be higher in**

a) 60 Hz

**b) 50 Hz**

c) same in both

d) data insufficient

**Detailed Solution :**

On 60 Hz frequency, [Latex]cos\theta =\frac{R}{Z} = \frac{R}{\sqrt{R^{2}+X_{L}^{2}}} = \frac{100}{\sqrt{100^{2}+(2\pi \times 1.2\times 60)^{2}}} = 0.84[/Latex]

real power of the transformer = [Latex]kVAcos\theta = kVA \times \frac{R}{Z} = 84\; kW[/Latex]

On 50 Hz frequency, [Latex]cos\theta =\frac{R}{Z} = \frac{R}{\sqrt{R^{2}+X_{L}^{2}}} = \frac{100}{\sqrt{100^{2}+(2\pi \times 1.2\times 50)^{2}}} = 0.88[/Latex]

real power of the transformer = [Latex]kVAcos\theta = kVA \times \frac{R}{Z} = 88\; kW[/Latex]

Therefore, for 50 Hz frequency, output is higher.

**6. Two single-phase transformers are connected in parallel. Which of the options are correct ?**

a) They must have the same efficiency.

b) They must have the power rating.

**c) They must have the same polarity.**

d) They must have the same number of turns in the secondary coil.

**Detailed Solution :**

Varied efficiency, different power rating or unequal number of turns in the coils doesn’t affect the kind of the connection in transformers. Only requirement for the parallel connection is that the polarity of the windings must be the same.

**7. Which factors impact the efficiency of a transformer ?**

a) Loading current.

b) Supply freq.

c) Powerfactor of the load.

d)** Both A and C**** option.**

**Detailed Solution :**

The efficiency of a transformer is the ratio of the o/p power and the I/p power and For both the calculations, we need to know the values of the power factor and the load current.

**8. Which one would have the maximum number of turn?**

a) Primary winding.

b) Secondary winding.

**c) High voltage winding****.**

d) Low voltage winding.

**Detailed Solution :**

We know that the voltage is directly proportional to the no. of turn in the coil. Therefore, high voltage winding carries the most number of turns.

**9. Which of the following is the correct relation between the voltage applied to the primary coil of the transformer (V) and EMF induced in that (E) ?**

a) V = E

**b) ****E = √2Vcos ωt**

c) V = √2Ecos ωt

d) E = Vcos ωt

**Detailed Solution :**

An ideal transformer has a primary coil with N_{1} turns and a secondary coil with N_{2} turns on a common core. The voltage of the source of the primary is E = √2 V cos ωt, while the secondary coil is initially assumed to be an open circuit.

**10. The ratio of the number of turns in the primary coil and the secondary coil of a transformer is n then What will be the ratio of their impedance ?**

**a) Z _{p} = Z_{s}/n^{2}**

b) Z_{p} = n^{2}Z_{s}

c) Z_{p} = Z_{s}/n

d) Z_{p} = nZ_{s}

**Detailed Solution :**

The ratio of impedances of the primary coil to the secondary coil is directly proportional to the reciprocal of the square of turns ratio of the transformer. Hence the primary impedance to the secondary impedance ratio will be Z_{p} = Z_{s}/n^{2}.