E M K 11203
E L E C T R I C CIRCUIT THE ORY 1
Chapter 2:
Resistive Circuit (cont.)
Delta (∆) - Wye (Y) Circuit
⦁ Circuit configuration are often encountered in which the
resistor do not appear to be in series or parallel.
⦁ Need to convert the circuit form one form to another to
solve.
⦁ Two circuit configuration that often account for these
difficulties configuration:
Wye (Y)
⦁ Delta (∆)
⦁
The bridge network
2
Delta (∆) - Wye (Y) Circuit (Cont…)
⦁ Delta (PI) configuration
3
Delta (∆) - Wye (Y) Circuit (Cont…)
⦁ Wye configuration
4
Delta (∆) - Wye (Y) Circuit (Cont…)
⦁ ∆ - Y transformation
5
Delta (∆) - Wye (Y) Circuit (Cont…)
⦁ Y - ∆ transformation
6
E xam ple
⦁ Convert the ∆ to a Y
𝑅𝐵𝑅𝐶
𝑅1 =
𝑅Æ + 𝑅𝐵 + 𝑅𝐶
(20Ω)(10Ω)
𝑅1 =
= 3.3333Ω
30Ω + 20Ω + 10Ω
𝑅Æ𝑅𝐶
𝑅2 =
𝑅Æ+ 𝑅𝐵 + 𝑅𝐶
(30Ω)(10Ω)
𝑅2 =
= 5Ω
30Ω + 20Ω + 10Ω
𝑅Æ𝑅𝐵
𝑅3 =
𝑅Æ + 𝑅𝐵 + 𝑅𝐶
𝑅3 =
7
(30Ω)(20Ω)
= 10Ω
30Ω + 20Ω + 10Ω
E xam ple
⦁ Convert the Y to a ∆
𝑅1 𝑅2 + 𝑅 2 𝑅 3 + 𝑅1 𝑅 3
𝑅Æ =
𝑅1
(10 Ω)(5Ω) + (5Ω)(10Ω) + (10 Ω)(10Ω)
3
𝑅Æ = 3
= 30Ω
10 Ω
3
𝑅𝐵 =
𝑅1 𝑅2 + 𝑅 2 𝑅 3 + 𝑅 1 𝑅3
𝑅2
(10 Ω)(5Ω) + (5Ω)(10Ω) + (10 Ω)(10Ω)
3
𝑅𝐵 = 3
= 20Ω
5Ω
𝑅𝐶 =
𝑅1 𝑅2 + 𝑅 2 𝑅 3 + 𝑅 1 𝑅3
𝑅3
(10 Ω)(5Ω) + (5Ω)(10Ω) + (10 Ω)(10Ω)
3
𝑅𝐶 = 3
= 10Ω
10Ω
8
E xercise
⦁ Find the total resistance (RT) of the network
Ans:
RT = 2.889Ω
9
E xam ple
⦁ Find the current and power supplied by the 40V sources in
the circuit shown below.
We can find this equivalent resistance easily after
replacing either the upper Δ (100, 125, 25Ω) or
the lower Δ (40, 25, 37.5Ω) with its equivalent Y.
We choose to replace the upper Δ. Thus,
100125
= 50
250
125 25
R2 =
= 12.5
250
100 25
R3 =
= 10
250
R1 =
10
Equivalent resistance,
Req = 55 +
50  50
= 80
100
Then, the current and power values
are,
40
= 0.5A
80
p = 40  0.5 = 20W
i=
11
E xercise
⦁ Find the equivalent resistance (Rab) & current i.
Ans:
Rab = 9.632Ω
i = 12.458A
12
Practice 1
⦁
Find i and Vab
Ans:
i = -1A
Vab = 56V
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Practice 2
⦁
Find Rab, Is, Vx and Power at 80 V.
Ans:
Rab = 40Ω
IS = 2A
Vx = -60V
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Practice 3
⦁
Find V1, V2 and Is using star-delta transformation technique
Ans:
V1 = 10V
V2 = 5V
IS = 1.0476A
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Practice 4
⦁
Find V1, V2 and power in 5kΩ and 20kΩ using voltage divider
rule or current divider rule
Ans:
V1 = 45V
V2 = 60V
P5 = 0.18W
P20 = 0.18W
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Practice 5
⦁
Find Vx, Vy and Ix using voltage divider rule or current divider
rule
Ans:
Vx = 10.8 V
Vy = 4.8V
Ix = 6A
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Practice 7
⦁
Find Req, IT and power developed by 30V.
Ans:
Req = 8.963Ω
IT= 3.347A
P30V= 100.41W
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Practice 8
⦁
Find Vo and power absorbed by 10Ω.
Ans:
Vo= 4.376V
P10= 748.03mW
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Practice 9
⦁
Find I1, I2, I3 and I4 using current divider rule and Kirchhoff
Current Law.
Ans:
I1= 5A
I2= 2.5A
I3= -2.5A
I4= 1.25A
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Practice 10
⦁
Find V0 using voltage divider rule.
Ans:
Vo= 14.084V
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Practice 11
⦁
Find RT and Vx
Ans:
RT=36.25Ω
Vx= 9.310V
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Practice 12
⦁
Find VX and VY using voltage divider rule and power at 2 Ω.
Ans:
Vx= 5V
VY= 3V
P2= 2W
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