138

S
SERIES CIRCUITS
4
I
–
12.5 V
+
7 V
–
50 V
+
–
+
37.5 V
7 V
–
FIG. 5.22
Redrawing the circuit of Fig. 5.21.
+
R1
4
I
+
+
–
4
4
6 12 V
–
+
E
R2
20 V
R3
3 6V
–
5.6 VOLTAGE DIVIDER RULE
+
In a series circuit,
1 2V
–
the voltage across the resistive elements will divide as the magnitude
of the resistance levels.
FIG. 5.23
Revealing how the voltage will divide across
series resistive elements.
+
R1
6 M 12 V
R2
3 M
R3
1 M
–
+
E
20 V
6V
–
+
2V
–
FIG. 5.24
The ratio of the resistive values determines the
voltage division of a series dc circuit.
+
R1
1 M V1 >> V2 or V3
–
For example, the voltages across the resistive elements of Fig. 5.23
are provided. The largest resistor of 6 captures the bulk of the applied
voltage, while the smallest resistor R3 has the least. Note in addition
that, since the resistance level of R1 is 6 times that of R3, the voltage
across R1 is 6 times that of R3. The fact that the resistance level of R2 is
3 times that of R1 results in three times the voltage across R2. Finally,
since R1 is twice R2, the voltage across R1 is twice that of R2. In general, therefore, the voltage across series resistors will have the same
ratio as their resistance levels.
It is particularly interesting to note that, if the resistance levels of all
the resistors of Fig. 5.23 are increased by the same amount, as shown in
Fig. 5.24, the voltage levels will all remain the same. In other words,
even though the resistance levels were increased by a factor of 1 million, the voltage ratios remain the same. Clearly, therefore, it is the ratio
of resistor values that counts when it comes to voltage division and not
the relative magnitude of all the resistors. The current level of the network will be severely affected by the change in resistance level from
Fig. 5.23 to Fig. 5.24, but the voltage levels will remain the same.
Based on the above, a first glance at the series network of Fig. 5.25
should suggest that the major part of the applied voltage will appear
across the 1-M resistor and very little across the 100- resistor. In
fact, 1 M (1000)1 k (10,000)100 , revealing that V1 1000V2 10,000V3.
Solving for the current and then the three voltage levels will result in
100 V
E
I 99.89 mA
RT 1,001,100 +
E
100 V
R2
1 k V2 = 10V3
–
and
+
R3
100 V3
–
R1 >> R2 or R3
FIG. 5.25
The largest of the series resistive elements will
capture the major share of the applied
voltage.
V1 IR1 (99.89 mA)(1 M) 99.89 V
V2 IR2 (99.89 mA)(1 k) 99.89 mV 0.09989 V
V3 IR3 (99.89 mA)(100 ) 9.989 mV 0.009989 V
clearly substantiating the above conclusions. For the future, therefore,
use this approach to estimate the share of the input voltage across series
elements to act as a check against the actual calculations or to simply
obtain an estimate with a minimum of effort.
S

VOLTAGE DIVIDER RULE
In the above discussion the current was determined before the voltages of the network were determined. There is, however, a method
referred to as the voltage divider rule (VDR) that permits determining
the voltage levels without first finding the current. The rule can be
derived by analyzing the network of Fig. 5.26.
I
RT R1 R2
+
RT
E
I RT
and
R1
V1
–
E
+
Applying Ohm’s law:
R2
E
RE
V1 IR1 R1 1
RT
RT
E
RE
V2 IR2 R2 2
RT
RT
with
FIG. 5.26
Developing the voltage divider rule.
Note that the format for V1 and V2 is
Rx E
Vx RT
(voltage divider rule)
V2
–
(5.10)
where Vx is the voltage across Rx , E is the impressed voltage across the
series elements, and RT is the total resistance of the series circuit.
In words, the voltage divider rule states that
the voltage across a resistor in a series circuit is equal to the value of
that resistor times the total impressed voltage across the series
elements divided by the total resistance of the series elements.
+ V1 –
EXAMPLE 5.10 Determine the voltage V1 for the network of Fig.
5.27.
Solution:
20 60 R1
R2
Eq. (5.10):
R1E
1280 V
(20 )(64 V)
RE
V1 1 16 V
R1 R2
80
20 60 RT
64 V
E
FIG. 5.27
Example 5.10.
EXAMPLE 5.11 Using the voltage divider rule, determine the voltages
V1 and V3 for the series circuit of Fig. 5.28.
Solution:
(2 k)(45 V)
(2 k)(45 V)
RE
V1 1 2 k 5 k 8 k
15 k
RT
90 V
(2 103 )(45 V)
6V
15
15 103 (8 103 )(45 V)
(8 k)(45 V)
RE
V3 3 15 k
15 103 RT
360 V
24 V
15
The rule can be extended to the voltage across two or more series
elements if the resistance in the numerator of Eq. (5.10) is expanded to
+
R1
2 k V1
R2
5 k
R3
8 k V3
–
V'
+
E
–
+
45 V
–
+
FIG. 5.28
Example 5.11.
–
139
140

S
SERIES CIRCUITS
include the total resistance of the series elements that the voltage is to
be found across (R′); that is,
R′E
V′ RT
(volts)
(5.11)
EXAMPLE 5.12 Determine the voltage V′ in Fig. 5.28 across resistors
R1 and R2.
Solution:
R′E
(2 k 5 k)(45 V)
(7 k)(45 V)
V′ 21 V
RT
15 k
15 k
There is also no need for the voltage E in the equation to be the
source voltage of the network. For example, if V is the total voltage
across a number of series elements such as those shown in Fig. 5.29,
then
54 V
(2 )(27 V)
V2 6 V
9
423
+
4
V = 27 V
2
–
3
+ V2 –
FIG. 5.29
The total voltage across series elements need not be an independent voltage
source.
4 mA
E
+
R1
VR1
R2
VR2
20 V
FIG. 5.30
Example 5.13.
–
+
EXAMPLE 5.13 Design the voltage divider of Fig. 5.30 such that
VR1 4VR2.
Solution:
The total resistance is defined by
20 V
E
RT 5 k
4 mA
I
–
Since VR1 4VR2,
R1 4R2
Thus
RT R1 R2 4R2 R2 5R2
and
5R2 5 k
R2 1 k
R1 4R2 4 k
and
5.7
NOTATION
Notation will play an increasingly important role in the analysis to follow. It is important, therefore, that we begin to examine the notation
used throughout the industry.
S
INTERNAL RESISTANCE OF VOLTAGE SOURCES
E = +24 V
EXAMPLE 5.19 Using the voltage divider rule, determine the voltages V1 and V2 of Fig. 5.48.
+
Solution: Redrawing the network with the standard battery symbol
will result in the network of Fig. 5.49. Applying the voltage divider
rule,
V1 R1
E
V2
+
V2 R2
2
–
FIG. 5.48
Example 5.19.
EXAMPLE 5.20 For the network of Fig. 5.50:
a
4
–
R1E
(4 )(24 V)
V1 16 V
R1 R2
42
R2E
(2 )(24 V)
V2 8 V
R1 R2
42
+
–
Vab
R1
b
R2
2
+
3
10 V
Vb
–
5
R3
c
+
E
–
R1
+
4 V
1
–
R2
+
2 V
2
–
24 V
FIG. 5.50
Example 5.20.
a. Calculate Vab.
b. Determine Vb.
c. Calculate Vc.
Solutions:
a. Voltage divider rule:
(2 )(10 V)
RE
Vab 1 2 V
235
RT
b. Voltage divider rule:
(3 5 )(10 V)
(R2 R3)E
8 V
Vb VR2 VR3 10 RT
or

Vb Va Vab E Vab 10 V 2 V 8 V
c. Vc ground potential 0 V
5.8 INTERNAL RESISTANCE
OF VOLTAGE SOURCES
Every source of voltage, whether a generator, battery, or laboratory supply as shown in Fig. 5.51(a), will have some internal resistance. The
equivalent circuit of any source of voltage will therefore appear as
shown in Fig. 5.51(b). In this section, we will examine the effect of the
internal resistance on the output voltage so that any unexpected changes
in terminal characteristics can be explained.
In all the circuit analyses to this point, the ideal voltage source (no
internal resistance) was used [see Fig. 5.52(a)]. The ideal voltage
source has no internal resistance and an output voltage of E volts with
no load or full load. In the practical case [Fig. 5.52(b)], where we con-
FIG. 5.49
Circuit of Fig. 5.48 redrawn.
145