Geoscience Reference
In-Depth Information
-
+
+
-
E
2
E
2
+
+
E
1
R
1
R
1
-
-
R
2
Series Aiding
R
2
Series Opposing
FIGURE 11.36
Series aiding and opposing sources.
E
b2
40 v
R
1
40 ohms
+
-
+-
A
+
+
E
b1
140 v
R
2
20 ohms
-
-
I
-
+
E
b3
20 v
FIGURE 11.37
Solving for circuit current in a multiple-source circuit.
Solution:
Start at point A.
Basic equation:
E
a
+
E
b
+
E
c
+ … +
E
n
= 0
From the circuit:
E
b
2
+
E
1
-
E
b
1
+
E
b
3
+
E
2
= 0
40 + 40
I
- 140 + 20 + 20
I
= 0
Combining like terms, we obtain
60
I
- 80 = 0
60
I
= 80
I
= 1.33 amps
11.7.7 p
arallel
dC C
irCuits
The principles we applied to solving simple series circuit calculations for determining the reactions of
such quantities as voltage, current, and resistance can be used in parallel and series-parallel circuits.
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