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EXAMPLE 5.14 Find the voltage Vab for the conditions of Fig. 5.38.
Solution: Applying Eq. (5.12):
Vab � Va � Vb � 16 V � 20 V
� �4 V
Note the negative sign to reflect the fact that point b is at a higher
potential than point a.
EXAMPLE 5.15 Find the voltage Va for the configuration of Fig. 5.39.
Solution: Applying Eq. (5.12):
Vab � Va � Vb and V a � V ab � V b � 5 V � 4 V
� 9 V
EXAMPLE 5.16 Find the voltage Vab for the configuration of Fig.
5.40.
Solution: Applying Eq. (5.12):
Vab � Va � Vb � 20 V � (�15 V) � 20 V � 15 V
� 35 V
Note in Example 5.16 the care that must be taken with the signs
when applying the equation. The voltage is dropping from a high level
of �20 V to a negative voltage of �15 V. As shown in Fig. 5.41, this
represents a drop in voltage of 35 V. In some ways it’s like going from
a positive checking balance of $20 to owing $15; the total expenditure
is $35.
EXAMPLE 5.17 Find the voltages V b, V c, and V ac for the network of
Fig. 5.42.
+
E1 = 10 V
–
a
4 V
+ –
–
+
FIG. 5.42
Example 5.17.
b
E 2
+ –
20 V
Solution: Starting at ground potential (zero volts), we proceed
through a rise of 10 V to reach point a and then pass through a drop in
potential of 4 V to point b. The result is that the meter will read
Vb ��10 V � 4 V � 6 V
as clearly demonstrated by Fig. 5.43.
V b
c
a R b
NOTATION ⏐⏐⏐ 143
V a = +16 V V b = +20 V
V a
V a = 20 V
V b = –15 V
FIG. 5.38
Example 5.14.
V ab = +5 V V b = 4 V
a R b
FIG. 5.39
Example 5.15.
V a = +20 V
+
R 10 k� Vab –
V b = –15 V
FIG. 5.40
Example 5.16.
V
Gnd (0 V)
V ab = 35 V
FIG. 5.41
The impact of positive and negative voltages
on the total voltage drop.
10 V
V
Gnd (0 V)
4 V
6 V
FIG. 5.43
Determining V b using the defined
voltage levels.