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EEEB123 EEEB123 Circuit Analysis 2Circuit Analysis 2

Chapter 13Chapter 13

Magnetically Coupled Magnetically Coupled CircuitsCircuits

Materials from Fundamentals of Electric Circuits (4th Edition), Alexander & Sadiku, McGraw-Hill Companies, Inc.

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Magnetically Coupled Circuit Magnetically Coupled Circuit Chapter 13Chapter 13

13.1 Introduction -What is a transformer? 13.2 Mutual Inductance 13.3 Energy in a Coupled Circuit 13.5 Ideal Transformers

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13.1 What is a transformer? (1)13.1 What is a transformer? (1)

• It is an electrical device designed on the basis of the concept of magnetic coupling.

• It uses magnetically coupled coils to transfer energy from one circuit to another.

• It is the key circuit elements for stepping up or stepping down ac voltages or currents, impedance matching, isolation, etc.

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13.2 Mutual Inductance (1)13.2 Mutual Inductance (1)• It is the ability of one inductor to induce a voltage across a

neighboring inductor, measured in henrys (H).

dt

diMv 1

212 dt

diMv 2

121

M21 = Mutual inductance of coil 2 with respect to coil 1.

M12 = Mutual inductance of coil 1 with respect to coil 2.

Open circuit Mutual inductance (or induced voltage) across coil 2.

Open circuit Mutual inductance (or induced voltage) across coil 1.

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13.2 Mutual Inductance (2)13.2 Mutual Inductance (2)• If a current enters the dotted terminal of one coil,

the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil.

Illustration of the dot convention.

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13.2 Mutual Inductance (3)13.2 Mutual Inductance (3)Dot convention

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13.2 Mutual Inductance (4)13.2 Mutual Inductance (4)

)connection aiding-(series

221 MLLL

Dot convention for coils in series; the sign indicates the polarity of the mutual voltage; (a) series-aiding connection, (b) series-opposing connection.

)connection opposing-(series

MLLL 221

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13.2 Mutual Inductance (5)13.2 Mutual Inductance (5)

Time-domain analysis of a circuit containing coupled coils.

Frequency-domain analysis of a circuit containing coupled coils.

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13.2 Mutual Inductance (6)13.2 Mutual Inductance (6)Example 13.1

Calculate the phasor currents I1 and I2 in the circuit shown below.

A04.1491.2I A;39.4901.13I 21 Ans:

*Refer to textbook pp. 561-562

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13.3 Energy in a Coupled Circuit (1)13.3 Energy in a Coupled Circuit (1)

• The coupling coefficient, k, is a measure of the magnetic coupling between two coils; 0≤k≤1.

• The instantaneous energy stored in the circuit is given by

21LLkM

21222

211 2

1

2

1IMIiLiLw

Positive sign for mutual term if both currents enter or leave the dotted terminals. Negative sign if otherwise.

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13.3 Energy in a Coupled Circuit (2)13.3 Energy in a Coupled Circuit (2)

Example 13.3 Consider the circuit below. Determine the coupling

coefficient. Calculate the energy stored in the coupled inductors at time t = 1s if v=60cos(4t +30°) V.

Ans: k=0.56; w(1)=20.73J

Refer to textbook pp. 566-567

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13.5 Ideal Transformer (1)13.5 Ideal Transformer (1)• A transformer is said to be ideal if:

1. Coils have very large reactances (L1, L2, M )2. Coupling coefficient is equal to unity (k = 1)3. Primary and secondary coil are loss less (R1 = 0 = R2)

• An ideal transformer is a unity-coupled, lossless transformer in which the primary and secondary coils have infinite self-inductances.

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13.5 Ideal Transformer (2)13.5 Ideal Transformer (2)

(a) Ideal Transformer(b) Circuit symbol

nN

Nn

N

N 1

2

1

1

2 1

2

1

2

I

I

V

V

V2>V1→ step-up transformerV2<V1→ step-down transformer

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13.5 Ideal Transformer (3)13.5 Ideal Transformer (3)• Polarity of VV and direction of II:

1. If VV11 and VV22 are both positive or both negative at the dotted terminals, use +n in . Otherwise use –n .

2. If II11 and II22 both enter or both leave the dotted terminals, use –n in . Otherwise use +n.

nN

N

1

2

1

2

V

V

nN

N 1

2

1 1

2

I

I

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13.5 Ideal Transformer (4)13.5 Ideal Transformer (4)

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13.5 Ideal Transformer (5)13.5 Ideal Transformer (5)• VV11 can be expressed in terms of VV2 2 and II11 in terms of II22, or vice versa:

or

or

• Complex power in primary winding is:

showing SS supplied to primary is delivered to secondary without loss.

2*22

*2

2*111 SIVI

VIVS n

n

n2

1

VV 12 VV n

21 II nn1

2

II

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13.5 Ideal Transformer (6)13.5 Ideal Transformer (6)

Input impedance as seen by the source figure above is:

Since , we get

2nL

in

ZZ

2

2

1

1in I

V

I

VZ

2

1

n

2

2L I

VZ

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13.5 Ideal Transformer (7)13.5 Ideal Transformer (7)

•Input impedance is also known as reflected impedance, since it appears as if load impedance is reflected to primary side.

•Common practice in analyzing circuit with ideal transformer is to eliminate transformer by reflecting impedances and sources from one side of the transformer to the other.

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13.5 Ideal Transformer (8)13.5 Ideal Transformer (8)•Example: Want to reflect secondary side of circuit below to primary side:

1. Find Thevenin equivalent of circuit to the right of terminals a-b. (i.e. Obtain VTh as open-circuit voltage at terminals a-b).

2. Get ZTh by removing voltage source at secondary winding and insert a unit source at terminals a-b.

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13.5 Ideal Transformer (8)13.5 Ideal Transformer (8)

1.Obtain VTh

nns22

1Th

VVVV

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13.5 Ideal Transformer (8)13.5 Ideal Transformer (8)

2.Obtain ZTh

where

2nn

n 2

2

2

1

1Th

Z

I

V

I

VZ

222 IZV

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13.5 Ideal Transformer (11)13.5 Ideal Transformer (11) The general rule for eliminating the transformer and reflecting the secondary circuit to the primary side is: divide the secondary impedance by n2, divide the secondary voltage by n, and multiply the secondary current by n.

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13.5 Ideal Transformer (12)13.5 Ideal Transformer (12) The general rule for eliminating the transformer and reflecting the primary circuit to the secondary side is: multiply the primary impedance by n2, multiply the primary voltage by n, and divide the primary current by n.

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13.5 Ideal Transformer (13)13.5 Ideal Transformer (13)

Example 13.7

An ideal transformer is rated at 2400/120V, 9.6 kVA, and has 50 turns on the secondary side.

Calculate: (a) the turns ratio, (b) the number of turns on the primary side, and (c) the current ratings for the primary and secondary

windings.

Ans:(a) This is a step-down transformer, n=0.05(b) N1 = 1000 turns(c) I1 = 4A and I2 = 80A

Refer to textbook pp. 578