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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
articles in our series to help you ace your exam and have concepts made easy –
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Control Systems - Revision Test 1
Linear Algebra – Revision Test 1
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