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352 ⏐⏐⏐ NETWORK THEOREMS
I T
G T
FIG. 9.94
Reducing all the current sources of Fig. 9.93
to a single current source.
R eq
E eq
+
–
1
G T
I T
G T
FIG. 9.95
Converting the current source of Fig. 9.94 to
a voltage source.
R 1 5 � R 2 4 � R 3 2 �
E 1
E2 10 V
E3 16 V
8 V
FIG. 9.96
Example 9.18.
R L
R L
R L
I L
3 �
+
VL –
Step 2: Combine parallel current sources as described in Section 8.4.
The resulting network is shown in Fig. 9.94, where
IT � I1 � I2 � I3 and GT � G1 � G2 � G3 Step 3: Convert the resulting current source to a voltage source, and the
desired single-source network is obtained, as shown in Fig. 9.95.
In general, Millman’s theorem states that for any number of parallel
voltage sources,
IT �I1 � I2 � I3 � •••� IN Eeq ������ GT G1 � G2 � G3 � •••� GN �E1G1 � E2G2 � E3G3 � •••� ENGN or Eeq � ����� (9.10)
G1 � G2 � G3 � •••� GN The plus-and-minus signs appear in Eq. (9.10) to include those cases
where the sources may not be supplying energy in the same direction.
(Note Example 9.18.)
The equivalent resistance is
1
1
Req ��� ��� (9.11)
GT G1 � G2 � G3 � •••� GN In terms of the resistance values,
� �
Eeq � (9.12)
E1
� � �
R1
E2
� � �
R2
E3
� � •••� �
R3
EN
�
RN
————
1 1 1 1
�� � �� � �� � •••� �� R R R R
1
1
and Req � ———
1 1 1 1
(9.13)
�� � �� � �� � •••� �� R R R R
1
2
2
The relatively few direct steps required may result in the student’s
applying each step rather than memorizing and employing Eqs. (9.10)
through (9.13).
EXAMPLE 9.18 Using Millman’s theorem, find the current through
and voltage across the resistor RL of Fig. 9.96.
Solution: By Eq. (9.12),
� �
Eeq �
E1
� � �
R1
E2
� � �
R2
E3
�
R3
——
1 1 1
�� � �� � �� R R R
1
The minus sign is used for E 2/R 2 because that supply has the opposite
polarity of the other two. The chosen reference direction is therefore
3
3
2
3
N
N
Th