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190 ⏐⏐⏐ PARALLEL CIRCUITS
+
E1 –
+
E1 –
10 V
10 V
R 1
10 Ω
a
+
V ab
–
b
FIG. 6.46
Example 6.22.
a
+
V ab
–
b
E 2
+ –
30 V
R 2
50 Ω
E 2
+ –
30 V
FIG. 6.47
Circuit of Fig. 6.46 redrawn.
I T = 12 mA
+
V
–
R 1
R 1
6 �
(a)
I = 0 A
6 �
R 2
R 2
(a)
c
+
V cd
–
d
c
+
V cd
–
d
12 �
EXAMPLE 6.22 Determine the voltages Vab and Vcd for the network
of Fig. 6.46.
Solution: The current through the system is zero amperes due to the
open circuit, resulting in a 0-V drop across each resistor. Both resistors
can therefore be replaced by short circuits, as shown in Fig. 6.47. The
voltage Vab is then directly across the 10-V battery, and
Vab � E1 � 10 V
The voltage Vcd requires an application of Kirchhoff’s voltage law:
�E1 � E2 � Vcd � 0
or Vcd � E1 � E2 � 10 V � 30 V � �20 V
The negative sign in the solution simply indicates that the actual voltage
Vcd has the opposite polarity of that appearing in Fig. 6.46.
EXAMPLE 6.23 Determine the unknown voltage and current for each
network of Fig. 6.48.
I = 0 A 12 mA
+
12 �
V = 0 V
I
E
FIG. 6.48
Example 6.23.
E
I
22 V
22 V
R 1
1.2 k�
R 1
(b)
(b)
R 2
8.2 k�
R 2
I = 0 A
+ V –
Solution: For the network of Fig. 6.48(a), the current I T will take the
path of least resistance, and, since the short-circuit condition at the end
of the network is the least-resistance path, all the current will pass
through the short circuit. This conclusion can be verified using Eq.
(6.9). The voltage across the network is the same as that across the
short circuit and will be zero volts, as shown in Fig. 6.49(a).
–
FIG. 6.49
Solutions to Example 6.23.
+ 22 V –
For the network of Fig. 6.48(b), the open-circuit condition requires
that the current be zero amperes. The voltage drops across the resistors
P