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7/25/2019 prob2-21s
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Problem 2.21 Find I1
to I3
in the circuit of Fig. P2.21.
3 k
4 k2 k16 V
+
_
12 V
8 V
+
_
+_
I1 I3
I2 +
_
+
_
+_
Figure P2.21: Circuit for Problem 2.21.
Solution: Application of KVL to the outer-perimeter loop gives
16+3103
I18+12 = 0,
which leads to
I1 = 4 mA.
For the left loop,
16+3103I1+4103I2 = 0,
I2 =163103I1
4103 =
1631034103
4103 = 1 mA.
Consequently,
I3= I
1I
2=
41=
3 mA.
All rights reserved. Do not reproduce or distribute. c2013 National Technology and Science Press
