Solving Problems using Maximum Power Transfer Delta-Wye Conversions

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Announcement!!!
BEC Quiz 3 on 28th Feb (Tue) at 3 PM
45 minutes
20 Marks
Topics (All Sessions 12, 13 and 14 Series)
Thevenin and Norton Theorems
Maximum Power Transfer
Delta-Wye Conversions

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Session 14B: Focus
 Solving Problems using
◦ Maximum Power Transfer
◦ Delta-Wye Conversions

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Maximum Power Transfer

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S14_Problem 3: Max Power Transfer
Find the value of RL when maximum power is delivered to it: RTH = RL =40/3 Ω
RTH = 20Ω || 40Ω = 40/3 Ω = RL
I1
When 50V = 0, I1 =
RTH = RL = 40/3 Ω
The dependent current source 0.5 I1 = 0
So, it is equivalent to an open circuit
The maximum power delivered will be: 1.875 W
Thevinin’s Equivalent circuit:
40/3 Ω
10 V
I1
RTH
I1
0

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Delta-Wye Conversions

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Star-Delta or Y-Delta or Y-
7
a
c
b
Shaped a
c
bY Shaped
Δ
Δ
to Y
Δ
Y to
Δ

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S14_Problem1: Y- Transformation
Note: This is Example 5.12 on page 156, Figure 5.46 in the Ref 1 book (by Hayt)
Find Rin using Y- Transformation: Rin = 159/71 Ω
c
b
a
a
c
b
a
c
b
RA
RB
RC
Δ
Δ
to Y
Δ
Rin

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S14_Problem 2: Y- Transformation
9
Note: This is Practice 5.11 on page 156, Figure 5.47 in the Ref 1 book (by Hayt)
Find Thévenin resistance of the circuit: RTH = 11.43 Ω
c
b
a
R1
R2
R3
RA = 300/10 = 30Ω
Each R is 10Ω
RB = 30Ω
RC = 30Ω
c
ba
c
ba
c
ba
ba
c
6
6
6
6 2
2
30 10
Δ

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S14_Problem 2: Y- Transformation
10
Note: This is Practice 5.11 on page 156, Figure 5.47 in the Ref 1 book (by Hayt)
Find Equivalent resistance of the circuit: Rin = 11.43 Ω
c
b
a
R1
R2
R3
Each R is 10Ω
c
ba
c
ba
6
6
6
6 2
2
30 10
Δ
Right most leg: 2+10+2 = 14Ω
Middle leg: 6 + 6 = 12Ω
Combine 12 Ω ||14Ω = 84/13 Ω
Combine 6+ 84/13+ 6 = 240/13 Ω
Combine 30 Ω || 240/13 Ω = 11.43Ω = Rin

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Session 14B: Summary
 Solving Problems using
◦ Maximum Power Transfer
◦ Delta-Wye Conversions

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References
Ref 1 Ref 2