EE 302 - HW Solution - 2 - 3 and the 10-volt source we get...

EE302 Homework #2
Chapter 2, Problem 12.
In the circuit in Fig. 2.76, obtain v
1,
v
2
, and v
3
.
Chapter 2, Solution 12.
For loop 1, -20 -25 +10 + v
1
= 0 v
1
= 35v
For loop 2, -10 +15 -v
2
= 0 v
2
= 5v
For loop 3, -v
1
+v
2
+v
3
= 0 v
3
= 30v
+
v
1
-
+ v
2
-
+
v
3
-
– 25v +
+ 10v -
+ 15v -
+
20v
-
loop 1
loop 2
loop 3

Chapter 2, Problem 16.
Determine V
o
in the circuit in Fig. 2.80.
Figure 2.80 For Prob. 2.16.
Chapter 2, Solution 16.
Apply KVL,
-9 + (6+2)I + 3 = 0, 8I = 9-3=6 , I = 6/8
Also,
-9 + 6I + V
o
= 0
V
o
= 9- 6I = 4.5 V
6 Ω
2 Ω
+
_
+
_
+
_
V
o
9 V
3 V
circle5
circle5

Chapter 2, Problem 17.
Obtain v
1
through v
3
in the circuit
Chapter 2, Solution 17.
Applying KVL around the entire outside loop we get,
–24 + v
1
+ 10 + 12 = 0 or v
1
= 2V
Applying KVL around the loop containing v
2
, the 10-volt source, and the
12-volt source we get,
v
2
+ 10 + 12 = 0 or v
2
= –22V