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Maximum Power Transfer Theorem - GATE

Study Material in PDF

In the previous article we discussed about Thevenin’s theorem, Norton’s theorem,

Reciprocity theorem and Tellegen’s theorem. These lead us to learn about Maximum

Power Transfer Theorem and other theorems that help us further learn Network

Theorems to simplify and analyze complex circuits.

In this article we mainly concentrate on three articles. They are

(i) Maximum power transfer theorem

(ii) Superposition Theorem

(iii) Millman’s Theorem

In these GATE 2018 Notes we will discuss Maximum Power Transfer Theorem and the

other two theorems. You can download these GATE Study Material in PDF. These notes

are useful for GATE EC, EE, BARC, IES, DRDO, BSNL, ECIL and other exams! Before

you get started, however, it is important to get through some basics!

Recommended Reading –

Basic Network Theory Concepts

Source Transformation & Reciprocity Theorem

Kirchhoff’s Laws – KCL & KVL

Nodal & Mesh Analysis

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Thevenin’s, Norton's & Tellegen’s Theorems

Maximum Power Transfer Theorem

According to this theorem, the load will receive maximum power from a source when its

resistance (RL) is equal to the internal resistance (or) Thevinin’s (or) Norton Resistance.

Maximum Power Transfer Power in DC Circuits

For the given circuit RL = RTh for maximum power transfer

∴ IL =VTh

RTh+RL=

VTh

RTh+RTh=

VTh

2RTh

Pmax = IL2. RL = (

VTh

2RTh)

2. RTh

∴ Pmax =VTh

2

4RTh

But Ptotal = IL2(RL + RTh) =

VTh2

4RTh2 . (RL + RTh) =

VTh2

2RTh (∵ RL = RTh)

Efficiency, η =Pmax

Ptotal× 100

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=(

VTh2

4RTh)

(VTh

2

2RTh)

× 100 =1

2× 100 = 50%

The efficiency of Maximum power can be shown in the below wave form.

Example 1:

Find the maximum power delivered to the load R in the given circuit

Solution:

Step 1: Find VTh

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VTh+50

5+

VTh

10+ 0 = 0

2VTh + 100 + VTh = 0

VTh = −33.3V

Step 2: Find RTh

∴ RTh = 5 + (10||5) = 5 +50

15= 5 +

10

3= 8.33Ω

Maximum power delivered =VTh

2

4RTh=

(−33.3)2

4×8.33= 33.28 watt

Maximum Power Transfer Power in AC Circuits.

For the given network the maximum power transfer theorem can be analyzed in various

cases.

Case i: only RL is variable

Then for maximum power transfer we should have

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RL = √RS2 + (XS + XL)2

Case ii: only XL is variable

Then for maximum power transfer we should have

XS + XL = 0

Case iii: Both RL and XL are variable

Then for maximum power transfer we should have

XS + XL = 0 ⇒ XS = -XL

RL = √RS2 + (XS + XL)2

∴ RL = √RS2 + 0 = RS

RL = RS

∴ For Pmax , ZL = RL + jXL

= Rs = jXS = ZS∗

∴ ZL = ZS∗

Example 2:

Find the maximum power delivered

to the load

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Solution:

Step 1: Find VTh

VTh = Voc = 10∠0°.j8

6−j8+j8

VTh = 10∠0°.j8

6=

80

6∠90° = 13.33∠90°

Step 2: Find ZTh

ZTh =(6−j8)(j8)

6−j8+j8=

j16(3−j4)

6= (10.67 + j8)Ω

For Pmax , ZL = ZTh = (10.67 − j8)

So given network becomes

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IL =13.33∠90°

10.67+10.67= 0.624j

Pmax = IL2. RL = |0.624|2 × 10.67 = 4.163 watt

Note:

While calculating Pmax, consider only real part of load resistance i.e. RL.

Superposition Theorem

If a number of voltage or current sources are acting simultaneously in a linear network,

the resultant current in any branch is the algebraic sum of the currents that would be

produced in it when each source acts alone replacing all other independent sources by

their internal resistances.

i.e.

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The current i’ in figure (1) is calculated by adding the currents of i1 and i2 from the figure

(2) and figure (3)

∴i = i1 + i2

Example 3:

Find I in the circuit shown below using superposition theorem

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Solution:

Case i:

When only current source is active and voltage source is replaced by short circuit then,

∴ i1 = 1 ×1

1+2=

1

3= 0.33A

Case ii:

When only voltage source is active and current source is replaced by open circuit. Then

2i2 + 1V + i2 = 0

3i2 = −1

i2 = −1

3A

From superposition theorem, I = i1 + i2

∴ i =1

3−

1

3= 0A

∴ I = 0A

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Millman’s Theorem

When a number of voltage source V1, V2, V3 --- VN are in parallel having internal

resistances R1, R2, R3 --- Rn respectively, the arrangement can be replaced by a single

equivalent voltage source V in series with an equivalent series resistance R as given

below.

Where V1 =V1G1+V2G2+−−−VnGn

G1+G2+−−−Gn

R =1

G=

1

G1+G2+G3+−−−Gn

Example 4:

Find the value of current through RL using Millman’s theorem.

Solution:

Given R1 = R2 = R3 = 4

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G = G1 + G2 + G3

=1

4+

1

4+

1

4=

3

4

∴ R =1

G=

4

3Ω

V =V1G1+V2G2+V3G3

G1+G2+G3

=(−4)(

1

4)+(−2)(

1

4)+10(

1

4)

3

4

=−4−2+10

3=

4

3

So given circuit becomes

∴ IL =V

R+RL=

4

34

3+10

=4

34= 117.6 mA

Liked this article on Maximum Power Transfer Theorem? Enjoyed reading about

Superposition Theorem? Let us know in the comments. You may also like some more

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