1E6 Electrical Engineering
DC Circuit Analysis
Lecture 4: Resistive Configurations
4.1 Introduction
Resistors are the fundamental passive components of electric circuits. In
high energy electric circuits they are used to represent equivalent loads for
power-dissipating operations or processes. In lower power electronic circuits
resistors are used to generate voltages and currents at particular locations in a
circuit to establish the correct operating conditions for semiconductor devices.
Hence, an understanding of the effect of varying configurations of resistor
networks is essential.
4.2 Resistors in Series
Consider the configuration shown in Fig. 1 where N resistors are
connected end-to-end and are referred to as being connected in series. A battery
of emf E Volts is used to drive the circuit so that a current I flows into the
terminals of the combination of resistors. Note that the same current flows
through each of the resistors as charge must be conserved. The dashed line in
the circuit is not intended to represent any electrical break in the connection but
merely to indicate that the series arrangement can be extended to any number of
resistors, N.
R1
current I
+
+
_
R2
+
V1
V2
_
R3
+
_
RN
+
V3
E
_
Fig. 1 A Configuration of Resistors Connected in Series
1
VN
_
The aim is to determine the value of a single equivalent resistance, REQ,
such that if the series combination of resistors were replaced with this single
value the circuit would behave the same way in all respects. We want then a
single resistance such that in accordance with Ohm’s Law gives:
REQ 
E
I
If the same current flows through each resistor then:
I  I1  I 2  I 3...........  I N
By Kirchhoff’s Voltage Law:
E1  V1  V2  V3...........  VN  0
or
E1  V1  V2  V3...........  VN
Then:
REQ 
REQ 
REQ 
E V1  V2  V3 ...........  VN

I
I
E V1 V2 V3
V
   ...........  N
I
I
I
I
I
E V1 V2 V3
V
   ...........  N
I
I1 I 2 I 3
IN
REQ  R1  R2  R3 ...........  RN
Therefore it can be seen that the equivalent resistance is simply the sum of the
individual resistors
2
4.3 Case Study 1
Find the equivalent resistance of the circuit shown in Fig. 2 below and the
current which it draws from the battery supply. In addition, determine the
potential difference developed across each of the individual resistors.
R1
I
E
4kΩ
I
2.5kΩ
10V
R2
≡
E
10V
REQ
1.5kΩ
R3
Fig. 2 A Circuit of Resistors in Series
The equivalent single resistance of the circuit is given as:
REQ  R1  R2  R3  4k  2.5k  1.5k  8k
I
E
10V

 1.25mA
R EQ 8k
The potential differences developed across the resistors are:
V1  I  R1  1.25 103  4 103  5V
V2  I  R 2  1.25 103  2.5 103  3.125V
V3  I  R 3  1.25 10 3 1.5 103  1.875V
3
4.4 Resistors in Parallel
Consider the configuration shown in Fig. 3 where N resistors are
connected side-by-side and are referred to as being connected in parallel. A
battery of emf E Volts is used to drive the circuit so that a current I flows into
the terminals of the combination of resistors. Note that this time the same
potential difference is developed across each of the resistors and is equal to the
battery emf, E. The dashed line in the circuit is again not intended to represent
any electrical break in the connection but merely to indicate that the parallel
arrangement can be extended to any number of resistors, N.
current I
+
R1
V1
_
+
+
+
E
I3
I2
I1
R2
V2
+
R3
V3
VN
_
_
IN
I1
RN
_
_
Fig. 3 A Configuration of Resistors Connected in Parallel
The aim again is to determine the value of a single equivalent resistance,
REQ, such that if the series parallel combination of resistors were replaced with
this single value the circuit would behave the same in all respects. We want
again then a single resistance such that in accordance with Ohm’s Law gives:
REQ 
E
I
If the same potential difference is developed across each resistor then:
E  V1  V2  V3...........  VN
By Kirchhoff’s Current Law as applied to the top node of the circuit:
I  I1  I 2  I3...........  I N  0
or
4
I  I1  I 2  I 3 ...........  I N
Then:
REQ 
Inverting gives:
E
E

I I1  I 2  I 3 ...........  I N
1
I  I  I ...........  I N
 1 2 3
REQ
E
1
I I
I
I
 1  2  3 ............  N
REQ E E E
E
1
I
I
I
I
 1  2  3 ............  N
REQ V1 V2 V3
VN
1
1
1
1
1
 
 ............ 
REQ R1 R2 R3
RN
This can also be written somewhat cumbersomely as:
REQ 
1
1
1
1
1

 ............ 
R1 R2 R3
RN
Therefore it can be seen that in the simplest form to express, the inverse of the
equivalent resistance is the sum of the inverses of the individual resistors.
Consider the particular case of only two resistors in parallel:
1
1
1 R1  R2
RR
 

or REQ  1 2
REQ R1 R2
R1R2
R1  R2
This is sometimes referred to as the ‘product over sum’
5
4.5 Case Study 2
Find the equivalent resistance of the circuit shown in Fig. 4 below and the
current which it draws from the battery supply. In addition, determine the
current which flows through each of the individual resistors.
I
I
E
R1
R2
R3
4kΩ
2.5kΩ
1.5kΩ
10V
≡
E
10V
REQ
Fig. 4 A Circuit of Resistors in Parallel
1
1
1
1



REQ 4 103 2.5 103 1.5 103
1
 0.25 103  0.4 103  0.67 103
REQ
1
 (0.25  0.4  0.67) 103  1.32 103
REQ
REQ 
1
 0.76k  760
3
1.32  10
Then:
I
E
10

 13.2mA
REQ 0.76 103
and
I1 
10
10
10

2
.
5
mA
,
I


4
mA
,
I

 6.7mA
2
3
3
3
3
4  10
2.5  10
1.5  10
Checking
I1  I 2  I 3  2.5  4  6.7  13.2mA
6
4.6 Applications:
Potential Divider:
In electronic circuits using semiconductor and other devices as active
elements it is often necessary to obtain a range of dc voltages other than the
battery or supply voltage for the purposes of biasing these devices for correct
operation or optimum performance. A simple potential divider network for this
purpose is shown in Fig. 5 where two resistors are connected is series across the
battery supply.
I
+
V1
R1
V2
R2
E
_
VO
Fig. 5 A Potential Divider Network
The current I flows through both resistors so that:
VO  V2  I 2 R2  IR2
and
I
E
E

REQ R1  R2
so that:
VO 
E
R2
R2 
E
R1  R2
R1  R2
The ratio of the resistors therefore sets the potential VO as a fraction of the
supply voltage E. The values of the resistors R1 and R2 can be chosen
accordingly to obtain any desired bias voltage from any supply voltage. It is of
course possible to obtain a range of bias voltages by using a chain of resistors
connected in series across the supply.
7
Current Splitting:
Sometimes in electronic circuits it necessary to divide current between
different paths or branches in a circuit. Of the total current drawn from the
supply, it may be required to deliver a fixed proportion or percentage to a
particular element or load. A current splitting network such as is shown in Fig. 6
accomplishes this purpose.
I
+
E
I2
I1
R1
V1
V2
R2
_
Fig. 6 A Current Splitting Network
The same potential drop is developed across both resistors so that:
V1  V2  E
and
I
E
E
E ( R1  R2 )


REQ R1 // R2
R1R2
with
I1 
V1 E

and
R1 R1
I2 
V2
E

R2 R2
Then:
I1 E
R1R2
R2


I
R1 E ( R1  R2 ) R1  R2
which gives:
I1 
R2
I
R1  R2
8
Similarly:
I2
E
R1R2
R1


I
R2 E ( R1  R2 ) R1  R2
which gives:
I2 
Consider then:
R1
I
R1  R2
I1
R2 I
R  R2 R2

 1

I 2 R1  R2
R1I
R1
It can be seen therefore that the current drawn from the supply is split between
the two branches of the circuit in the ratio of the resistances of the opposite
branches. This analysis can be carried out for a combination of more resistors in
parallel but gives a more complex expression.
4.7 Case Study 3
Find an equivalent single value of resistance as seen by the supply for the
circuit shown in Fig. 7 below and determine the value of the current, I, drawn
from the battery supply and the value of the output voltage, VO.
I
5kΩ
R1
2.5kΩ
R2
+
E
_
10kΩ
22kΩ
15kΩ
R3
R4
R5
Fig. 7 A Combination Resistor Configuration
9
VO
Then for the parallel section across the output terminals:
1
1
1
1



ROEQ R3 R4 R5
1
1
1
1



ROEQ 10  103 22  103 15  103
1
 (0.1  0.47  0.067)  10 3  0.637  10 3
ROEQ
ROEQ
103

 1.6k
0.637
The equivalent resistance of the entire circuit is then given as:
REQ  R1  R2  ROEQ
REQ  5k  2.5k  1.6k  9.1k
Then the current draw from the supply is given as:
I
E
12

 1.35mA
REQ 9.1k
Finally, the output voltage is given as:
VO  IROEQ  1.35 103 1.6 103  2.4V
10