3.1 Resistors in Series
When two elements connected at a single node, they are said to be in series
Series-connected circuit elements carry the same current
The resistors in the circuit shown are connected on series
a
is
vs
+
-
h
R1
b
R3
c
i2
i1
R7
R2
g
R6
R4
f
R5
e
The resistors carry the same current as can be shown
KCL @ node a
KCL @ node b
i s - i1
0
i1 - i 2
0
i s  i1
i1  i 2
i s  i1
 i2
We can apply KCL t o the other nodes b,c,e,f,g,h to conclude that the current is the same
namely is
Applying KVL around the loop we obtain
R1
a
b
R2
R3
c
is
R4
+
-
vs
h
R7
g
R6
f
R5
e
-v s  i s R 1  i s R 2  i s R 3  i s R 4  i s R 5  i s R 6  i s R 7  0
v s  i s (R1  R 2  R 3  R 4  R 5  R 6  R 7 )
The significant of the equation is that the seven resistors can be replaced by a single
Resistor whose value is the sum of the individual resistor
a
R1
b
R2
a
R3
c
is
vs
R4
+
-

is
vs
+
-
h
h
R7
g
R6
f
R5
e
R eq  R1  R 2  R 3  R 4  R 5  R 6  R 7
Req
is
a
R1
b
R2
R3
c
is
+
R4
vs
-

a
7
+
Req
vs
R eq 
R
k
i 1
h
R7
g
R6
f
R5
e
h
In general, if k resistors are connected in series, the equivalent single resistor has a resistance
equal to the sum of the k resistances
k
R eq 
R
i 1
k
3.2 Resistors in Parallel
When two elements connected at a single node pair, they are said to be in parallel
Parallel-connected circuit elements have the same voltage across their terminals
The resistors in the circuit shown are connected on parallel
a
vs
+
-
R1
R2
R3
R4
h
Do not make the mistake of assuming that two elements are parallel connected merely
Because they are lined up in parallel
R
2
R1
R3
We can show that the parallel connection have the same voltage by applying KVL
to the each loop in the circuit
Applying KCL at node a
a
is
i s  i1  i 2  i 3  i 4
vs
R1
+
-
i1
R2
i2
R3
R4
i3
i3
h
From the parallel connection the elements have the same voltage and polarity namely vs
i1 
is 
vs
R1
i2 
vs
R2
vs vs vs vs



R1 R 2 R 3 R 4
From which
is
1

vs
R eq

i3 
vs
R3
v s

i4 
vs
R4
1
1
1
1



R1 R 2 R 3 R 4
1
1
1
1




R1 R 2 R 3 R 4


The four resistors
on the circuit can be replaced by a single equivalent resistor
i
a
vs
s
R1
+
-
i1
R2
i2
R3
R4
i3
i3
h
a
is
is
+
vs
+
R1
R2
R3
R4
Req
vs
-

1
1
1
1
1




Req
R1 R 2 R 3 R 4

-
h
In general, if k resistors are connected in parallel, the equivalent single resistor has a resistance
equal to
k
1
1

Req
R
i 1 i
Using conductance when dealing with resistor in parallel is more convenient

k
G eq 
G
i 1
i
 G1
 G2 
 Gk
3.3 The voltage-Divider and Current-Divider Circuits
In some applications, we need to develop more than one voltage level from a single voltage
supply
One way of doing this is by using a voltage-divider-circuit
+
R1
vs
+
-
v1
-
+
R2 v 2
-
We analyze this circuit as follows
i
vs
+
-
+
R1
v1
-
+
R2 v 2
-
i 
v s  iR 1  iR 2
Using Ohm’s law, we have
v 1  iR 1  v s
R1
R1  R 2
vs
R1  R 2
v 2  iR 2
vs
R2
R1  R 2
+
R1
vs
+
-
v1
v1 
-
+
R1
R1  R 2
v2 
vs
R2
R1  R 2
vs
R2 v 2
-
We can see that v1 and v2 are each fraction of vs .
Each fraction is the ratio of the resistance across which the divided voltage is defined to the
Sum of the two resistances.
Because this ratio is always less than 1.0 , the divided voltages v1 and v2 are always less than
the source voltage vs
If vs is specified , say 15 V and a particular value of v2 is desire say 5 V
v2 vs 
1
R2
3
R1  R 2
There are an infinite combinations of R1 and R2

1
3
R2

that will satisfy this ratio
1
2
R1
Consider connecting a resistor RL in parallel with R2 as shown
R1
vs
+
-
The resistor RL acts as a load on the voltage-divider circuit
+
R2
vo
RL
-
vo 
The output voltage vo becomes
vo 
Note as RL
→∞
v2 
R2
R 1 1 R 2 R L   R 2
R eq
R 1  R eq
vs
, vo reduce to
R2
R1  R 2
vs
as it should be
vs
were
R eq 
R2R L
R2  RL
The Current-Divider
The current divider consist of two resistors connected in parallel as shown
+
R1
is
i1 v i 2
R2
-
We will find the relationship between is and i1 and i2 as follows
v  i 1R 1  i 2 R 2

R1R 2
i
R1  R 2 s
i1 
R2
i
R1  R 2 s
i2 
R1
i
R1  R 2 s
R1||R 2
The equation shows that the current is divided between two resistors in parallel such that
The current in one resistor is controlled by the other resistor
This is similar to a main water pipe that split into two pipes with different dirt and sand and
obstacle ( ‫)معيقات‬. The pipe with the less obstacle will have the current flow in it more than the
Other pipe with more obstacle
3.4 The Voltage and Current Division
We can generalize the voltage and current division as follows
Voltage division
R1
R2
+
+
Circuit
Rj
v
i
i 
R1  R 2 
vj
-
-
v
were
R n -1
Rn
 Rn
R eq
R eq  R1  R 2 
v j  iR j 
Current division

Rj
v
 Rn
v
R eq
i
+
Circuit
R1
Rj
R2
i j R n -1
Rn
ij 
v
Rj
R eq

i
Rj
|| R n )  iR eq
v
were
-
using Ohm's law
v  i ( R 1 || R 2 ||
R eq  ( R 1 || R 2 ||
|| R n )
Example 3.4 Use current division to find the current io and use voltage division to find
the voltage vo
R eq  (36 44) || 10 || ( 401030) || 24  80 || 10 || 80 || 24  6
80
io 
80
6
(8A)  2A
24
vo 
v
 ( 24)(8A )  48 V
30
(48 V)
401030

30
( 48 V)
80
 18 V
3.7 Delta to Wye Equivalent Circuit
Consider the following circuit
R1
v
R2
Rm

-
R3
R4
Rc
a
b
Rc
Ra
Ra
Rb
Ra
c
b
a
R1
R2
R2
R1
R3
R3
c
c
b
a
b
b
a
Rc
Ra
Rb
R2
R1
c
R3
c
R ab 
Rc ( R a Rb )
R a Rb Rc
 R1  R 2
R bc 
R a ( Rb  Rc )
R a Rb Rc
 R2  R3
R ca 
Rb ( Rc R a )
R a Rb Rc
 R1  R 3
a
b
b
a
Rc
Rb
R2
R1
Ra
c
R ab 
R bc 
R ca 
Rc ( R a Rb )
R a Rb Rc
R a ( Rb  Rc )
R a Rb Rc
Rb ( Rc R a )
R a Rb Rc
Solving for
R3
c
 R1  R 2
 R2  R3
 R1  R 3
R 1 , R 2 , R 3 we have,
Rb Rc
R1 
R a Rb Rc
Rc R a
R2 
R a Rb Rc
R3 
R a Rb
R a Rb Rc
b
Rc
Rb
R2
R1
Ra
R3
c
R1 
Rb Rc
R a Rb Rc
Solving for
R2 
Rc R a
R a Rb Rc
R3 
R a Rb
R a Rb Rc
R a , R b , R c we have,
R R R R R R
Ra  1 2 2 3 3 1
R1
Rb 
R1R 2  R 2 R 3  R 3R1
R2
Rc 
R1R 2  R 2 R 3  R 3R1
R3
R ab 
Rc ( R a Rb )
R a Rb Rc
 R1  R 2
R bc 
R a ( Rb  Rc )
R a Rb Rc
 R2  R3
R ca 
Rb ( Rc R a )
R a Rb Rc
 R1  R 3
Ra 
R1R 2  R 2 R 3  R 3R1
R1
Rb 
R1R 2  R 2 R 3  R 3R1
R2
Rc 
R1R 2  R 2 R 3  R 3R1
R3
Rb Rc
R a Rb Rc
Rc R a
R2 
R a Rb Rc
R1 
R3 
R a Rb
R a Rb Rc
Example 3.7 : Find the current and power supplied by the 40 V source ?
5
i
100 
40 V
125 
25 

-
40 
37.5 
5
100 X 125
 50 Ω
100 + 125+ 25
125 X 25
 12.5 Ω
R2 
100 + 125+ 25
R1 
100 
i
R1
R3
40 V

-
125 
R2
25 
40 
R3 
37.5 
100 X 25
 10 Ω
100 + 125+ 25
5
i
50 
5
10 
40 V
12.5 

40 V
40 

-
i
37.5 
R eq  50  (10 40)||(12.5 37.5)  75 Ω
i 
40 V

-
i
80 
P
40V
40 V
1
 A
80 
2
1
 - (40)( )  -20 W
2
Req