Basic Laws of Electric Circuits
Equivalent Resistance
Lesson 5
Basic Laws of Circuits
Equivalent Resistance:
We know the following for series resistors:
R1
R2
. . .
Req
RN
. . .
Figure 5.1: Resistors in series.
Req = R1 + R2 + . . . + RN
1
Basic Laws of Circuits
Equivalent Resistance:
We know the following for parallel resistors:
. . .
R eq
R1
RN
R2
. . .
Figure 5.2: Resistors in parallel.
1
1
1
1


 . . . 
Req
R1 R2
RN
2
Basic Laws of Circuits
Equivalent Resistance:
For the special case of two resistors in parallel:
R eq
R1
R2
Figure 5.3: Two resistors in parallel.
Req
3
R1 R2

R1  R2
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
By combination we mean we have a mix of series and
Parallel. This is illustrated below.
R1
Req
R3
R2
R4
R5
Figure 5.4: Resistors In Series – Parallel Combination
To find the equivalent resistance we usually start at
the output of the circuit and work back to the input.
4
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
R1
R3
R2
Req
Rx
R4 R5
Rx 
R4  R5
R1
R eq
5
R2
Ry
Figure 5.5: Resistance reduction.
R y  Rx  R3
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
R1
RZ
Req
Req
6
RZ 
R2 RY
R2  RY
Req  RZ  R1
Figure 5.6: Resistance reduction, final steps.
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
It is easier to work the previous problem using numbers than to
work out a general expression. This is illustrated below.
Example 5.1: Given the circuit below. Find Req.
10 
R eq
8
10 
3
Figure 5.7: Circuit for Example 5.1.
7
6
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.1: Continued . We start at the right hand side
of the circuit and work to the left.
10 
Req
10 
8
10 
2
Req
Figure 5.8: Reduction steps for Example 5.1.
Ans:
8
Req  15 
5
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Given the circuit shown below. Find Req.
6
12 
c
Req
10 
b
4
d
Figure 5.9: Diagram for Example 5.2.
9
a
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Continued.
6
12 
c
Req
10 
b
a
4
d
c
Fig 5.10: Reduction
steps.
10
Req
4
12 
b
6
10 
d, a
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Continued.
c
Req
4
12 
b
6
10  resistor
shorted out
10 
d, a
Req
Fig 5.11: Reduction
steps.
11
4
6
12 
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Continued.
4
Req
Fig 5.12: Reduction
steps.
12
Req
6
4
12 
4
Basic Electric Circuits
Wye to Delta Transformation:
You are given the following circuit. Determine Req.
I
9
10 
V
+
_
5
10 
Req
8
4
Figure 5.1: Diagram to start wye to delta.
13
Basic Electric Circuits
Wye to Delta Transformation:
You are given the following circuit. Determine Req.
I
9
10 
V
+
_
5
10 
Req
8
4
Figure 5.13: Diagram to start wye to delta.
14
Basic Electric Circuits
Wye to Delta Transformation:
I
9
10 
V
+
_
5
10 
Req
8
4
We cannot use resistors in parallel. We cannot use
resistors in series. If we knew V and I we could solve
Req =
15
V
I
There is another way to solve the problem without solving
for I (given, assume, V) and calculating Req for V/I.
Basic Electric Circuits
Wye to Delta Transformation:
Consider the following:
a

a

Ra
R1
Rc
c 
R2
Rb
 b
c 
 b
R3
(a) wye configuration
(b) delta configuration
Figure 5.14: Wye to delta circuits.
We equate the resistance of Rab, Rac and Rca of (a)
to Rab , Rac and Rca of (b) respectively.
16
Basic Electric Circuits
Wye to Delta Transformation:
Consider the following:
a

a

Ra
R1
Rc
c 
R2
Rb
 b
c 
 b
R3
(a) wye configuration
(b) delta configuration
R2(R1 + R3)
Eq 5.1
Rab = Ra + Rb =
R1 + R2 + R3
R1(R2 + R3)
Rac = Ra + Rc =
Eq 5.2
R1 + R2 + R3
R3(R1 + R2)
17
Rca = Rb + Rc =
R1 + R2 + R3
Eq 5.3
Basic Electric Circuits
Wye to Delta Transformation:
Consider the following:
a

a

Ra
R1
Rc
R2
Rb
c 
 b
c 
 b
R3
(a) wye configuration
18
(b) delta configuration
Ra 
R1 R2
R1  R2  R3
R1 
Ra Rb  Rb Rc  Rc Ra
Rb
Eq 5.4
Rb 
R2 R3
R1  R2  R3
R2 
Ra Rb  Rb Rc  Rc Ra
Rc
Eq 5.5
Rc 
R1 R3
R1  R2  R3
R3 
Ra Rb  Rb Rc  Rc Ra
Ra
Eq 5.6
Basic Electric Circuits
Wye to Delta Transformation:
Observe the following:
Go to wye
Go to delta
Ra Rb  Rb Rc  Rc Ra
Rb
Eq 5.4
R2 R3
Rb 
R1  R2  R3
R R  Rb Rc  Rc Ra
R2  a b
Rc
Eq 5.5
R1 R3
Rc 
R1  R2  R3
Ra Rb  Rb Rc  Rc Ra
R3 
Ra
Eq 5.6
Ra 
R1 R2
R1  R2  R3
R1 
We note that the denominator for Ra, Rb, Rc is the same.
We note that the numerator for R1, R2, R3 is the same.
We could say “Y” below: “D”
19
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.3: Return to the circuit of Figure 5.13 and find Req.
I
9
a
10 
V
+
_
Req
5
10 
c
8
b
4
Convert the delta around a – b – c to a wye.
20
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.3: continued
9
2
Req
4
2
8
4
Figure 5.15: Example 5.3 diagram.
It is easy to see that Req = 15 
21
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.4: Using wye to delta. The circuit of 5.13
may be redrawn as shown in 5.16.
9
a
10 
Req
c
5
10 
8
4
b
Figure 5.16: “Stretching” (rearranging) the circuit.
Convert the wye of a – b – c to a delta.
22
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.4: continued
9
9
a
10 
7.33 
27.5 
Req
c
8
a
11 
Req
c
22 
11 
5.87 
b
(a)
b
(b)
Figure 5.17: Circuit reduction of Example 5.4.
23
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.4: continued
9
Req
13.2 
11 
Figure 5.18: Reduction of Figure 5.17.
Req = 15 
This answer checks with the delta to wye solution earlier.
24
Basic Laws of Circuits
circuits
End of Lesson 5
Equivalent Resistance