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with the bottom of the determinant determined by:
Finally,
BRIDGE NETWORKS ⏐⏐⏐ 293
det[[(1/3�1/4�1/2),�1/4,�1/2][�1/4,(1/4�1/2�1/5),�1/5][�1/2,�1/5,(1/5�1/2�1/1)]] ENTER 1.312
and V 1 � 8 V
CALC. 8.5
10.5/1.312 ENTER 8
CALC. 8.6
Similarly, V2 � 2.667 V and V3 � 2.667 V
and the voltage across the 5-� resistor is
V5� � V2 � V3 � 2.667 V � 2.667 V � 0 V
Since V5� � 0V,wecan insert a short in place of the bridge arm without
affecting the network behavior. (Certainly V � IR � I·(0) �
0V.) In Fig. 8.67, a short circuit has replaced the resistor R5, and the voltage
across R4 is to be determined. The network is redrawn in Fig. 8.68, and
V1� � (voltage divider rule)
�
� �
2(20 V) 40 V
��� ���2.667 V
2 � 4 � 9 15
as obtained earlier.
We found through mesh analysis that I5� � 0A,which has as its
equivalent an open circuit as shown in Fig. 8.69(a). (Certainly I �
V/R � 0/(∞ �) � 0A.) The voltage across the resistor R4 will again
be determined and compared with the result above.
The network is redrawn after combining series elements, as shown in
Fig. 8.69(b), and
(6 � � 3 �)(20 V) 2 �(20 V)
V3� ��� ��� �8 V
6 � � 3 ��3 � 2 ��3 �
2
�(20 V)
3
��
� 2
� � �4 � � �9
3 3 3 �
� 2
�(20 V)
3
��
� 2
(2 � � 1 �)20 V
����
(2 � � 1 �) � (4 � � 2 �) � 3 �
� � �8 � � 3
3 6
1 �(8 V) 8 V
and V1� ��� ���2.667 V
1 ��2 � 3
as above.
The condition V5� � 0 V or I5� � 0 A exists only for a particular
relationship between the resistors of the network. Let us now derive this
relationship using the network of Fig. 8.70, in which it is indicated that
I � 0 A and V � 0 V. Note that resistor Rs of the network of Fig. 8.69
will not appear in the following analysis.
The bridge network is said to be balanced when the condition of
I � 0AorV � 0Vexists.
If V � 0 V (short circuit between a and b), then
V1 � V2 R s
E
3 �
20 V
R 1
R 3
4 � R2 V = 0
2 �
R 4
2 �
1 �
–
+
V 1�
FIG. 8.67
Substituting the short-circuit equivalent for
the balance arm of a balanced bridge.
R s
3 �
R 1
4 � R 2
2 �
+
+
E 20 V R3 2 � R4 1 � V1� –
–
FIG. 8.68
Redrawing the network of Fig. 8.67.