The document contains 34 problems involving circuit analysis techniques like Kirchhoff's laws, voltage and current division. The problems involve finding unknown currents, voltages, powers, and components in circuits. Suggested solutions are provided that apply circuit analysis methods to solve for the unknown quantities in 3 steps or less.

1. Problem 2.1
Find the current I and the power supplied by the source in the network shown.
I
6 V 20 kΩ
Suggested Solution
3
6
0.3 mA
20 10
I = =
×
3
(6)(0.3 10 ) 1.8 mWP VI −
= = × =

2. Problem 2.2
In the circuit shown, find the voltage across the current source and the power absorbed by the resistor.
3 mA 3 kΩ
+
VCS
_
Suggested Solution
( )( )3 3
3 10 3 10 9 VCSV −
= × × =
( )( )3
3 10 9 27 mWP VI −
= = × =

3. Problem 2.3
If the 10- resistor in the network shown absorbs 2.5 mW, find V .kΩ S
VS 10 kΩ
Suggested Solution
2
10 k
SV
P =
Ω
or ( )( )3 3
2.5 10 10 10 5 VS R −
= × = × × =V P

4. Problem 2.4
In the network shown, the power absorbed by xR is 5 mW. Find xR .
Rx
1
mA
2
Suggested Solution
( )
3
2 23
5 10
20 k
0.5 10
x
P
R
I
−
−
×
= = = Ω
×

5. Problem 2.5
In the network shown, the power absorbed by xR is 20 mW. Find xR .
2 mA Rx
Suggested Solution
( )
3
2 23
20 10
5 k
2 10
P
R
I
−
−
×
= = = Ω
×

6. Problem 2.6
A model for a standard two D-cell flashlight is shown. Find the power dissipated in the lamp.
1.5 V
1.5 V
1-Ω lamp
Suggested Solution
1.5 1.5 3 VlampV = + =
( )
22
3
9 W
1
lamp
lamp
lamp
V
P
R
= = =

7. Problem 2.7
An automobile uses two halogen headlights connected as shown. Determine the power supplied by the
battery if each headlight draws 3 A of current.
12 V
Suggested Solution
( )( )12 3 3 72 WSP VI= = + =

8. Problem 2.8
Many years ago a string of Christmas tree lights was manufactured in the form shown in (a). Today the
lights are manufactured as shown in (b). Is there a good reason for this change?
(a)
(b)
Suggested Solution
First of all, one must recognize the fact that a failed lamp becomes an open circuit. Then:
In (a), one faulty lamp would take out the whole string. There would be no way to identify the faulty lamp.
In (b), if one lamp fails, the others stay lit. It is easy to identify the faulty lamp.

9. Problem 2.9
Find 1I in the network shown.
I1
12 mA
4 mA
2 mA
Suggested Solution
I1
12 mA
4 mA
2 mA
I2
2 0.004 0.002 0.006 AI = + =
1 20.012 I I= +
1 6 mAI⇒ =

10. Problem 2.10
Find 1I and 2I in the circuit shown.
I3 = 5 mA
IS
I4 = 6 mA
I1
I6 = 3 mA I2
Suggested Solution
By KCL, 1 4 6 mAI I= =
Also, 6 2 4I I I+ =
or 20.003 0.006I+ =
2 3 mAI⇒ =

11. Problem 2.11
Find xI and yI in the network shown.
4 mA
12 mA 8 mA
Ix Iy
Suggested Solution
0.012 0.004 xI= +
8 mAxI⇒ =
0.004 0.008yI= +
4 mAyI⇒ = −

12. Problem 2.12
Find xI , yI and zI in the circuit shown.
Ix
Iy
Iz2 mA
12 mA
4 mA
Suggested Solution
Ix
Iy
Iz2 mA
12 mA
4 mA
0.012 0.004xI + =
8 mAxI⇒ = −
0.004 0.002 0.012yI + = +
10 mAyI⇒ =
0.004 0.002zI + =
2 mAzI⇒ = −

13. Problem 2.13
Find xI in the circuit shown.
1 mA
4 mA
2 Ix
Ix
Suggested Solution
1 mA
4 mA
2 Ix
Ix
0.001 0.004 2x xI I+ = +
3 mAxI⇒ = −

14. Problem 2.14
Find xI , yI and zI in the network shown.
12 mA
3 mA
2 mA
4 mA
Ix
Iy
Iz
Suggested Solution
12 mA
3 mA
2 mA
4 mA
Ix
Iy
Iz
0.003 0.012xI + =
9 mAxI⇒ =
0.012 0.002 0.004yI + + =
10 mAyI⇒ = −
0.004 0.002zI + =
2 mAzI⇒ = −

15. Problem 2.15
Find xI in the circuit shown.
2Ix
Ix
2 mA
6 mA
Suggested Solution
2Ix
Ix
2 mA
6 mA
2 0.002 0.006x xI I+ = +
4 mAxI⇒ =

16. Problem 2.16
Find xI in the network shown.
1 mA
4 Ix
Ix
6 mA
Suggested Solution
1 mA
4 Ix
Ix
6 mA
4 0.001 0.006x xI I+ = +
5
mA
3
xI⇒ =

17. Problem 2.17
Find xV in the circuit shown.
9 V
+ Vx -
2 Vx
Suggested Solution
KVL: 9 2xV V− + + = 0
3 VxV⇒ =

18. Problem 2.18
Find V in the circuit shown.bd
a
b
c
d
12 V
6 V
+
2 V
_
+ 4 V -
Suggested Solution
KVL: 12 4 0bdV− + + =
8 VbdV⇒ =

19. Problem 2.19
Find V in the network shown.ad
a
b
c
de
4 V 12 V
- 3 V + - 2 V +
+ 3 V -
Suggested Solution
KVL: 4 3adV− + − = 0
7 VadV⇒ =

20. Problem 2.20
Find V and V in the circuit shown.af ec
a b c d
efg
_
2 V
+
- 2 V +
+ 1 V -
12 V 3 V
+
3 V
_
- 1 V +
Suggested Solution
KVL: 2 2afV+ + = 0
4 VafV⇒ = −
KVL: 3 3 0ecV + + =
6 VecV⇒ = −

21. Problem 2.21
Find V in the circuit shown.ac
a
b
c
d
12 V
+ 4 V -
3 Vx
+
Vx = 2 V
_
Suggested Solution
KVL: 2 12 0acV + − =
10 VacV⇒ =

22. Problem 2.22
Find V and V in the circuit shown.ad ce
a b c
de
4 Vx 12 V
- 1 V +
- +
Vx = 2 V
+ 1 V -
Suggested Solution
KVL: 12 2 1 0adV − + + =
9 VadV⇒ =
KVL: 1 12 0ceV + − =
11 VceV⇒ =

23. Problem 2.23
Find V in the network shown.ab
a b
c
12 V
3 Ω
6 Ω
Suggested Solution
Using voltage division,
3
12 4 V
3 6
abV
 
= = 
+ 

24. Problem 2.24
Find V in the network shown.bd
12 V
3 kΩ 1 kΩ
4 V
a
b
c
d
Suggested Solution
12 V
3 kΩ 1 kΩ
4 V
a
b
c
d
I
12 4
2 mA
3000 1000
I
−
= =
+
KVL: 4 1000(0.002) 0bdV − − =
6 VbdV⇒ =

25. Problem 2.25
Find xV in the circuit shown.
24 V
6 V
2 kΩ
5 kΩ
2 kΩ
+
Vx
_
Suggested Solution
Using voltage division,
( )
5000
24 6 10 V
2000 5000 2000
xV
 
= − 
+ + 
=

26. Problem 2.26
Find xV in the circuit shown.
_
Vx
+
6 V
4 kΩ 6 kΩ
24 V
8 V
Suggested Solution
Using voltage division,
( )
4000
6 8 24 4 V
4000 6000
xV
 
= + − 
+ 
= −

27. Problem 2.27
Find xV in the network shown.
+
Vx
_
5 Ω
3 Vx
15 Ω
5 V
Suggested Solution
+
Vx
_
5 Ω
3 Vx
15 Ω
5 V
I
KVL: , where V I5 15 3 0x xI V V− + − + = 5x =
Therefore, ( )5 15 3 5 5 0I I I− + − + =
1 AI⇒ =
5 VxV∴ =

28. Problem 2.28
Find V in the network shown.1
+
V1
_
+ Vx -
5 V
20 kΩ 40 kΩ
2
xV
Suggested Solution
+
V1
_
+ Vx -
5 V
20 kΩ 40 kΩ
2
xV
I
KVL: 20000 40000 5 0
2
xV
I I− + + + = , where V I40000x =
Therefore,
40000
20000 40000 5 0
2
I
I I− + + + =
0.125 mAI⇒ = −
1 5 40000 5 5 5 0 VxV V I= + = + = − + =

29. Problem 2.29
Find the power absorbed by the resistor in the circuit shown.30-kΩ
30 kΩ
12 V
2 Vx
10 kΩ
+
Vx
_
Suggested Solution
30 kΩ
12 V
2 Vx
10 kΩ
+
Vx
_
I
KVL: where V I12 30000 2 0x xI V V− + + + = 10000x =
Therefore, ( )12 30000 2 10000 10000 0I I− + + + =I
200 AI µ⇒ =
( ) ( ) ( )
22 6
30k 30000) 200 10 30000 1.2 mWP I −
Ω = = × =

30. Problem 2.30
Find oI in the network shown.
12 mA 2 kΩ 6 kΩ 3 kΩ
Io
Suggested Solution
( )
1
2000 12 mA 6 mA
1 1 1
2000 6000 3000
oI
 
 
= = 
 + + 
 

31. Problem 2.31
Find oI in the circuit shown.
120 mA 4 kΩ 4 kΩ
8 kΩ
Io
Suggested Solution
Using current division,
( )
1
4000 0.120 90 mA
1 1
4000 8000 4000
oI
 
 
= = 
 +
 + 

32. Problem 2.32
Find oI in the network shown.
6 kΩ
12 mA
12 kΩ 12 kΩ
Io
Suggested Solution
Using current division,
( )
1
12000 0.012 3 mA
1 1 1
6000 12000 12000
oI
 
 
= − = − 
 + + 
 

33. Problem 2.33
Find V in the circuit shown.o
24 mA 8 kΩ
6 kΩ
6 kΩ
8 kΩ
+
Vo
_
Suggested Solution
Combining the 8- resistors in parallel yields the following circuit.kΩ
24 mA 4 kΩ
6 kΩ
6 kΩ
+
Vo
_
I
Using current division,
( )
1
6000 6000 0.024 6 mA
1 1
4000 6000 6000
I
 
 += = 
 + 
+ 
Then,
6000 36 VoV I= =

34. Problem 2.34
Find oI in the network shown.
12 mA
3 kΩ 6 kΩ
4 kΩ
2 kΩ
Io
Suggested Solution
Combining the 3- and resistors in parallel yields the following equivalent circuit.kΩ 6-kΩ
12 mA
2 kΩ
4 kΩ
2 kΩ
I1
( )
( )1
4000
0.012 6 mA
4000 2000 2000
I
 
= = 
+ +  
Then,
1
6000
4 mA
6000 3000
oI I
 
= = + 

35. Problem 2.35
Determine LI in the circuit shown.
Ix IL
2 kΩ
6 mA
4 kΩ
2
xI
Suggested Solution
Ix IL
2 kΩ
6 mA
4 kΩ
2
xI
+
V1
_
KCL: 0.006 0
2
x
x L
I
I I− + + = , where 1
2000
x
V
I = and 1
4000
L
V
I =
Therefore,
1 1
0.006 0
2000 4000 4000
V V V
− + + =1
1 6 VV⇒ =
Then,
6
1.5 mA
4000
LI = =

36. Problem 2.36
Determine LI in the circuit shown.
6 kΩ
6 mA
3 Ix
3 mA
2 kΩ 3 kΩ
Ix IL
Suggested Solution
6 kΩ
6 mA
3 Ix
3 mA
2 kΩ 3 kΩ
Ix IL
+
V1
_
KCL: 1
0.006 3 0.003 0
6000
x x
V
I I I− + + + + =L , where 1
2000
x
V
I = and 1
3000
L
V
I = .
Therefore,
1 1 1
0.006 3 0.003 0
6000 2000 2000 3000
V V V 
− + + + + = 
 
1V
1
6
1.2 V
5
V⇒ = =
and
1.2
0.4 mA
3000
LI = =

37. Problem 2.37
Find in the circuit shown.ABR
A
B
RAB
2 kΩ
2 kΩ
3 kΩ
6 kΩ
12 kΩ
Suggested Solution
The network can be redrawn as shown below.
A
B
RAB
2 kΩ
2 kΩ
3 kΩ 6 kΩ 12 kΩ
C B A
C B A
Then,
At A-A: 6000 12000 4 k= Ω
At B-B: 2000 4000 6 k+ = Ω
At C-C: 3000 6000 2 k= Ω
and
2000 2000 4 kABR = + = Ω

38. Problem 2.38
Find in the circuit shown.ABR
RAB
A
B
5 kΩ
2 kΩ 2 kΩ
2 kΩ2 kΩ2 kΩ
Suggested Solution
RAB
A
B
5 kΩ
2 kΩ 2 kΩ
2 kΩ2 kΩ2 kΩ
D C B A
D C B A
At A-A: 2000 2000 4 k+ = Ω
At B-B:
4
2000 4000 k
3
= Ω
At C-C:
4000 10
2000 k
3 3
+ = Ω
At D-D:
10000
2000 1250
3
= Ω
Then,
5000 1250 6250ABR = + = Ω

39. Problem 2.39
Find in the network shown.ABR
RAB
A
B
6 kΩ
3 kΩ 3 kΩ
8 kΩ
4 kΩ
5 kΩ 4 kΩ
Suggested Solution
The network can be redrawn as shown below.
RAB
A
B
6 kΩ 3 kΩ 3 kΩ 8 kΩ4 kΩ
5 kΩ 4 kΩ
C B A
C B A
At A-A: 4000 8000 12 k+ = Ω
At B-B: 3000 3000 4000 12000 1 k= Ω
At C-C: 5000 1000 6 k+ = Ω
Therefore,
6000 6000 3 kABR = = Ω

40. Problem 2.40
Find in the circuit shown.ABR
RAB
A
B
2 kΩ
4 kΩ
4 kΩ
3 kΩ
6 kΩ
12 kΩ
Suggested Solution
The circuit can be redrawn as shown below.
RAB
A
B
2 kΩ
4 kΩ
4 kΩ
6 kΩ
12 kΩ
3 kΩ
A
B
C
C
At A: 6000 3000 2 k= Ω
At B: 4000 2000 6 k+ = Ω
At C-C: 6000 12000 4 k= Ω
Then,
2000 4000 4000 10 kABR = + + = Ω

41. Problem 2.41
Find in the network shown.ABR
A
B
RAB 6 kΩ
6 kΩ
6 kΩ
2 kΩ
Suggested Solution
The 2- resistor is shorted. Therefore, the network reduces to:kΩ
A
B
RAB 6 kΩ 6 kΩ 6 kΩ
Then,
6000 6000 6000 2 kABR = = Ω

42. Problem 2.42
Find in the circuit shown.ABR
A
B
RAB
12 kΩ
6 kΩ2 kΩ
4 kΩ 12 kΩ
Suggested Solution
The circuit maybe redrawn as follows:
RAB
A
B
4 kΩ
2 kΩ
12 kΩ
12 kΩ
6 kΩ
A
B
At A: 12000 6000 4 k= Ω
At B: 2000 4000 6 k+ = Ω
Then,
4000 6000 12000 2 kABR = = Ω

43. Problem 2.43
Find in the circuit shown.ABR
2 kΩ 2 kΩ
2 kΩ 2 kΩ
2 kΩ2 kΩ
2 kΩ 2 kΩ
4 kΩ 4 kΩ
A
B
RAB
Suggested Solution
Combining each series pair of resistors, the circuit can be redrawn as follows:2-kΩ
⇒
4 kΩ 4 kΩ
4 kΩ4 kΩ
4 kΩ 4 kΩ
A
B
RAB
4 kΩ4 kΩ
2 kΩ 2 kΩ
A
B
RAB
or
hen,
6 kΩ 6 kΩ
A
B
RAB
T
R 6000 6000 3 kAB = = Ω

44. Problem 2.44
Find the range of resistance for the following resistors:
a) with a tolerance of 5%.1 kΩ
b) with a tolerance of 2%.470 Ω
c) with a tolerance of 10%.22 kΩ
Suggested Solution
a) Minimum value = ( )( )1 0.05 1 k 950− Ω = Ω
Ω
Ω
Ω
Ω
Ω
Maximum value = ( )( )1 0.05 1 k 1050+ Ω =
b) Minimum value = ( )( )1 0.02 470 460.6− Ω =
Maximum value = ( )( )1 0.02 470 479.4+ Ω =
c) Minimum value = ( )( )1 0.1 22 k 19.8 k− Ω =
Maximum value = ( )( )1 0.1 22 k 24.2 k+ Ω =

45. Problem 2.45
Given the network shown, find the possible range of values for the current and power dissipated by the
following resistors:
a) with a tolerance of 1%.390 Ω
b) with a tolerance of 2%.560 Ω
I
10 V R
Suggested Solution
Note that
10 V
I
R
= and
( )
2
10 V 100
P
R R
= = .
a) Minimum resistor value = ( )( )1 0.01 390 386.1− Ω = Ω
Maximum resistor value = ( )( )1 0.01 390 393.9+ Ω = Ω
Minimum current value =
10
25.39 mA
393.9
=
Maximum current value =
10
25.90 mA
386.1
=
Minimum power value =
100
253.9 mW
393.9
=
Maximum power value =
100
259 mW
386.1
=
b) Minimum resistor value = ( )( )1 0.02 560 548.8− Ω = Ω
Maximum resistor value = ( )( )1 0.02 560 571.2+ Ω = Ω
Minimum current value =
10
17.51 mA
571.2
=
Maximum current value =
10
18.22 mA
548.8
=
Minimum power value =
100
175.1 mW
571.2
=
Maximum power value =
100
182.2 mW
548.8
=

46. Problem 2.46
Given the circuit shown:
a) Find the required value of .R
b) Use Table 2.1 to select a standard 10% tolerance resistor for .R
c) Calculate the actual value of I .
d) Determine the percent error between the actual value of I and that shown in the circuit.
e) Determine the power rating for the resistor .R
I = 40 mA
510 Ω
100 mA
R
Suggested Solution
a) From KCL, the current in the 510-Ω resistor is 100 mA 40 mA 60 mA− = . From Ohm’s Law,
the voltage across the circuit is ( )( ) 30.6 V=510 60 mAΩ . Therefore, the required value of R is
( ) ( )30.6 V 40 mA 765= Ω .
b) From Table 2.1, the nearest 10% resistor value is 820 Ω .
c) To find the actual value of I, first find the actual voltage across the circuit, which is
( )( )100 mA 510 820 31.44 VΩ Ω = . Then,
31.44 V
38.35 mA
820
I = =
Ω
.
d) The percent error is
( )
( )
38.35 40
100 4.14 %
40
−
= − .
e) Since the power consumption in R is
( )
2
31.44 V
1.206 W
820
=
Ω
, a resistor should be used.2-W

47. Problem 2.47
The resistors and shown in the circuit are 11R 2R Ω with a tolerance of 5% and with a tolerance of
10%, respectively.
2 Ω
a) What is the nominal value of the equivalent resistance?
b) Determine the positive and negative tolerance for the equivalent resistance.
Req
R1
R2
Suggested Solution
a) The nominal value is 1 2 3eqR R R= + = Ω
b) The minimum and maximum values of are1R 0.95 Ω and 1.05 Ω , and the minimum and
maximum values of are 1.82R Ω and 2.2 Ω . Thus, the minimum and maximum values of
are and 1.0
eqR
0.95 1.8 2.75+ = Ω 5 2.2 3.25+ = Ω . The positive and negative tolerances are
Positive Tolerance =
( )
( )
3.25 3
100 8.33 %
3
−
=
Negative Tolerance =
( )
( )
2.75 3
100 8.33 %
3
−
= −

48. Problem 2.48
Find 1I and V in the circuit shown.o
6 V
1 kΩ
12 kΩ 3 kΩ
I1
+
Vo
_
Suggested Solution
From Ohm’s Law: 1
6 V
0.5 mA
12 k
I
−
= = −
Ω
By application of voltage division: ( )
3 k
6 V 4.5 V
3 k 1 k
oV
Ω 
= − = 
Ω + Ω 
−

49. Problem 2.49
Find 1I and V in the circuit shown.o
12 V
2 kΩ
6 kΩ
8 kΩ
4 kΩ
I1
+
Vo
_
Suggested Solution
Combining resistors ( )6 k 8 k +4 k 4 k Ω Ω Ω = Ω  reduces the network to the following:
12 V
2 kΩ
4 kΩ
+
V1
_
Using voltage division, then
( )1
4000
12 V 8 V
2000 4000
V
 
= = 
+ 
.
Looking back at the original circuit,
12 V
2 kΩ
6 kΩ
8 kΩ
4 kΩ
I1
+
Vo
_
+
8 V
_
Ohm’s Law: 1
8 V 4
mA
6 k 3
I = =
Ω
Voltage division: ( )
4000 8
8 V
8000 4000 3
oV
 
= = 
+ 

50. Problem 2.50
Find 1I in the circuit shown.
12 mA 10 kΩ
2 kΩ
2 kΩ 2 kΩ
2 kΩ
I1
Suggested Solution
The circuit can be simplified as follows:
12 mA 10 kΩ
2 kΩ
I2
( )
4
2 k 2 k +2 k k
3
Ω Ω Ω = Ω
Then, ( )2
10000
0.012 9 mA
4000
10000 2000
3
I
 
 
 = =
  + +    
and
( )
( )1
2000
9 mA 3 mA
2000 2000 2000
I
 
= = 
+ +  

51. Problem 2.51
Determine V in the network shown.o
14 kΩ
4 kΩ
9 kΩ
12 kΩ
2 kΩ
2 mA
+
Vo
_
Suggested Solution
The network can be redrawn as:
14 kΩ
4 kΩ
12 kΩ
2 kΩ
2 mA
+
Vo
_
9 kΩ
A
A
The equivalent resistance to the left of A-A is ( )14 k 4 k 9 k 12 k 4 kΩ + Ω Ω Ω = Ω .
4 kΩ
2 kΩ
2 mA
+
Vo
_
Then, V .( )( )2 k 4 k 2 mA 12 Vo = − Ω + Ω = −

52. Problem 2.52
Find V in the network shown.o
2 Ω 8 Ω
2 Ω
4 Ω 4 Ω
24 V
+
Vo
_
Suggested Solution
The network can be redrawn as:
Combining the four resistors on the right-hand side ( ) ( )4 2 8 4 4 Ω + Ω Ω + Ω = Ω  yields:
2 Ω
8 Ω
2 Ω
4 Ω
4 Ω
24 V
+
Vo
_
2 Ω
4 Ω
24 V
I1
and
( )1
24 V
4 A
2 4
I = =
Ω + Ω
.
Then, reconsidering the original circuit,

53. 2 Ω 8 Ω
2 Ω
4 Ω 4 Ω
24 V
+
Vo
_
4 A
I2
( )
( ) ( )
( )2
4 2 4
4 A A
4 2 8 4 3
I
 +
= = 
+ + +  
and
( )
4 16
4 A A V
3 3
oV
 
= − = − 
 

54. Problem 2.53
Find oI in the circuit shown.
9 kΩ 4 kΩ
6 kΩ
6 kΩ
3 kΩ24 V
Io
Suggested Solution
The circuit can be redrawn as:
9 kΩ 4 kΩ
6 kΩ 6 kΩ 3 kΩ24 V
I2
Io
I1
B A
B A
At A-A: 6000 3000 2 k= Ω
At B-B: ( )6000 4000 2000 3 k+ = Ω
The circuit simplifies to:
9 kΩ
3 kΩ24 V
I1
Then,
( )1 2
24 6000
2 mA 1 mA
9000 3000 6000 4000 2000
I I
 
= = ⇒ = = 
+ + +  
1I
and 2
3000 1
mA
3000 6000 3
oI I
 
= = 
+ 
.

55. Problem 2.54
Find V in the circuit shown.o
2 kΩ
2 kΩ4 kΩ
12 kΩ 3 kΩ
6 mA +
Vo
_
Suggested Solution
2 kΩ
2 kΩ4 kΩ
12 kΩ 3 kΩ
6 mA +
Vo
_
A B
A B
At A-A: ( )2000 4000 12000 4 k+ = Ω
At B-B: ( )4000 2000 3000 2 k+ = Ω
Then, V .( )( )2 k 6 mA 12 Vo = Ω =

56. Problem 2.55
Find oI in the network shown.
1 kΩ 10 kΩ 2 kΩ
6 kΩ
3 kΩ
4 kΩ
4 kΩ
6 kΩ
12 V
Io
Suggested Solution
Redrawing the network:
1 kΩ 10 kΩ 2 kΩ
6 kΩ3 kΩ4 kΩ4 kΩ 6 kΩ12 V
Io
D C B A
I1
D C B A
At A-A: 3000 6000 2 k= Ω
At B-B: ( )4000 2000 2000 2 k+ = Ω
At C-C: ( )6000 10000 2000 4 k+ = Ω
At D-D: 4000 4000 2 k= Ω
1
12
4 mA
1000 2000
I = =
+
Using current division, 1
4000
2 mA
4000 4000
oI I
 
= = + 

57. Problem 2.56
Find V in the network shown.o
6 kΩ
9 kΩ 8 kΩ
4 kΩ4 kΩ
12 mA
+
Vo
_
Suggested Solution
6 kΩ
9 kΩ 8 kΩ
4 kΩ4 kΩ
12 mA
+
Vo
_
I1 I2
A
A
At A-A: ( )4000 8000 4000 3 k+ = Ω
Using current division,
( )1
6000
0.012 4 mA
6000 9000 3000
I
 
= = 
+ +  
Again using current division,
( )2 1
4000
1 mA
4000 8000 4000
I I
 
= = 
+ +  
Then, V .24000 4 Vo I= =

58. Problem 2.57
Find oI in the circuit shown.
6 kΩ
6 kΩ
2 kΩ 4 kΩ
12 V
Io
3 kΩ
Suggested Solution
The circuit can be redrawn as:
⇒12 V
At A-A: ( )2000 2000 4000 2 k+ = Ω
6 kΩ
6 kΩ
2 kΩ 4 kΩ
Io
3 kΩ
6 kΩ
2 kΩ
2 kΩ 4 kΩ
12 V
Io
I1
A
A
1
12
1.5 mA
6000 2000
I = =
+
Using current division,
( )
( ) 1
2000 2000
0.75 mA
2000 2000 4000
oI I
 +
= = 
+ +  

59. Problem 2.58
If V in the network shown, find V .4 Vo = S
VS
8 kΩ
4 kΩ
+
Vo = 4 V
_
Suggested Solution
VS
8 kΩ
4 kΩ
+
Vo = 4 V
_
I
4
1 mA
4000 4000
oV
I = = =
Then, V I .8000 8 4 12 VS oV= + = + =

60. Problem 2.59
If the power absorbed by the resistor in the network shown is 36 mW, find4-kΩ oI .
9 kΩ
12 kΩ 4 kΩ
Io
Suggested Solution
9 kΩ
12 kΩ 4 kΩ
Io
+
V1
_
2
1
4 k 136 mW 12 V
4 k
V
P VΩ = = ⇒ =
Ω
12 V
1 mA
12 k
oI = =
Ω

61. Problem 2.60
If the power absorbed by the resistor in the circuit shown is 36 mW, find V .4-kΩ S
VS
9 kΩ
12 kΩ 4 kΩ
Suggested Solution
VS
9 kΩ
12 kΩ 4 kΩ
+
V1
_
I2 I1
Io
2
1
4 k 136 mW 12 V
4000
V
P VΩ = = ⇒ =
1
1 mA
12 k
o
V
I = =
Ω
1
1 3 mA
4 k
V
I = =
Ω
2 1 1 mA 3 mA 4 mAoI I I= + = + =
( ) 2 19 k 48 VSV I V= Ω + =

62. Problem 2.61
In the network shown, the power absorbed by the 4-Ω resistor is 100 W. Find V .S
VS
3 Ω
4 Ω
7 Ω
3 Ω
Suggested Solution
VS
3 Ω
4 Ω
7 Ω
3 Ω
+
V1
_
I1
I2I3
( )2
4 1 14 100 W 5 AP I IΩ = Ω = ⇒ =
1 14 20V I= = V
1
2 2 A
7 3
V
I = =
Ω + Ω
3 1 2 7 AI I I= + =
13 21 20 4S SV I V= + = + = 1 V

63. Problem 2.62
In the network shown, V . Find6 Vo = SI .
4 kΩ
8 kΩ 4 kΩ
2 kΩ
IS
+
Vo
_
Suggested Solution
4 kΩ
8 kΩ 4 kΩ
2 kΩ
IS
+
Vo
_
+
VS
_
I1I2
1
6
3 mA
2 k 2000
oV
I = = =
Ω
( ) 14 k 12 6 18 VS oV I V= Ω + = + =
2
18
1.5 mA
8 k 4 k 12000
SV
I = = =
Ω + Ω
1 2 3 mA 1.5 mA 4.5 mASI I I= + = + =

64. Problem 2.63
In the circuit shown, 4 mAI = . Find V .S
2 kΩ
6 kΩ
5 kΩ
3 kΩVS
I
Suggested Solution
2 kΩ
6 kΩ
5 kΩ
3 kΩVS
I
+
V1
_
I2 I1
( )1 6 k 24 VV I= Ω =
1
1
24
3 mA
5 k 3 k 8000
V
I = = =
Ω + Ω
2 1 0.004 0.003 7 mAI I I= + = + =
( ) 2 12 k 14 24 38 VSV I V= Ω + = + =

65. Problem 2.64
In the circuit shown, . Find2 mAoI = SI .
4 kΩ
4 kΩ 2 kΩ
6 kΩ
1 kΩ
2 kΩ
IS
Io
Suggested Solution
4 kΩ
4 kΩ 2 kΩ
6 kΩ
1 kΩ
2 kΩIS
Io
+
V1
_
+ V2 -
+
V3
_
I3
I2
I1
( )1 6 k 12 VoV I= Ω =
1
1
12
4 mA
1 k 2 k 3000
V
I = = =
Ω + Ω
2 1 0.002 0.004 6 mAoI I I= + = + =
( )2 22 k 12 VV I= Ω =
3 2 1 12 12 24 VV V V= + = + =
3
3
24
3 mA
4 k 4 k 8000
V
I = = =
Ω + Ω
3 2 0.003 0.006 9 mASI I I= + = + =

66. Problem 2.65
In the network shown, V . Find V .1 12 V= S
2 kΩ
6 kΩ
4 kΩ
4 kΩ
1 kΩ
3 kΩVS
+ V1 -
Suggested Solution
2 kΩ
6 kΩ
4 kΩ
4 kΩ
1 kΩ
3 kΩVS
+ V1 -+
V2
_
A
A
I2
I1I3
At A-A: ( )4000 1000 3000 2 k+ = Ω
1
1
12
3 mA
4 k 4000
V
I = = =
Ω
( )2 1 12 k 12 6 18 VV V I= + Ω = + =
2
2
18
3 mA
6 k 6000
V
I = = =
Ω
3 1 2 0.003 0.003 6 mAI I I= + = + =
( ) 3 22 k 12 18 30 VSV I V= Ω + = + =

67. Problem 2.66
In the circuit shown, V . Find2 Vo = SI .
IS
10 Ω
12 Ω 2 Ω
3 Ω
8 Ω
4 Ω
+
Vo
_
Suggested Solution
IS
10 Ω
12 Ω 2 Ω
3 Ω
8 Ω
4 Ω
+
Vo
_
+
V1
_
+
V2
_
I1
I2
I3I4
1
2
0.5 A
4 4
oV
I = = =
Ω
( )1 18 6 VoV I V= Ω + =
1
2 2 A
3
V
I = =
Ω
3 2 1 2.5 AI I I= + =
( )2 3 12 11 VV I V= Ω + =
2
4 0.5 A
10 12
V
I = =
Ω + Ω
3 4 2.5 0.5 3 ASI I I= + = + =

68. Problem 2.67
In the network shown, V . Find6 Vo = SI .
IS 9 kΩ
3 kΩ
2 kΩ
1 kΩ
5 kΩ
+
Vo
_
Suggested Solution
IS 9 kΩ
3 kΩ
2 kΩ
1 kΩ
5 kΩ
+
Vo
_
+ V1 -
+
V2
_
I1
I2
I3
I4
1
6
3 mA
2 k 2000
oV
I = = =
Ω
2
6
1 mA
1 k 5 k 6000
oV
I = = =
Ω + Ω
3 1 2 0.003 0.001 4 mAI I I= + = + =
( )1 33 k 12 VV I= Ω =
2 1 12 6 18 VoV V V= + = + =
2
4
18
2 mA
9 k 9000
V
I = = =
Ω
3 4 0.004 0.003 6 mASI I I= + = + =

69. Problem 2.68
In the circuit shown, . Find2 AoI = SI .
8 Ω
4 Ω
4 Ω3 ΩIS
Io
6 V
Suggested Solution
8 Ω
4 Ω
4 Ω3 ΩIS
Io
6 V
+
V1
_
+
V2
_
I1I2I3
( )1 3 6 VoV I= Ω =
1
1
6
1.5 A
4 4
V
I = = =
Ω
2 1 2 1.5 3.5 AoI I I= + = + =
2 16 V 6 6 12 VV V= + = + =
2
3
12
1 A
8 4 12
V
I = =
Ω + Ω
=
2 3 3.5 1 4.5 ASI I I= + = + =

70. Problem 2.69
If in the circuit shown, find4 mAoI = SI .
10 kΩ
1 kΩ
4 kΩ
2 kΩ
12 V
IS
Io
Suggested Solution
10 kΩ
1 kΩ
4 kΩ
2 kΩ
12 V
IS
Io
+
V1
_
+ V2 -
+
V3
_
I1
I2
I3
( )1 4 k 12 V 16 12 4 VoV I= Ω − = − =
1
1
4
2 mA
2 k 2000
V
I = = =
Ω
2 1 0.004 0.002 6 mAoI I I= + = + =
( )2 21 k 6 VV I= Ω =
3 2 1 6 4 10 VV V V= + = + =
3
3
10
1 mA
10 k 10000
V
I = = =
Ω
2 3 0.006 0.001 7 mASI I I= + = + =

71. Problem 2.70
Find oI in the circuit shown.
12 kΩ
2 kΩ
12 kΩ
12 kΩ
8 kΩ
I1 = 5 mA
Io
9 mA
Suggested Solution
12 kΩ
2 kΩ
12 kΩ
12 kΩ
8 kΩ
I1 = 5 mA
Io
9 mA
_
V1
+
+ V3 -
_
V2
+
I2
I3
1 2 29 mA 4 mAI I I+ = ⇒ =
( )1 112 k 60 VV I= Ω =
( )2 212 k 48 VV I= Ω =
3 1 2 60 48 12 VV V V= − = − =
3
3
12
1 mA
12 k 12000
V
I = = =
Ω
1 3 0.005 0.001 6 mAoI I I= + = + =

72. Problem 2.71
Find V in the circuit shown.o
12 kΩ
4 kΩ 6 kΩ 18 kΩ
6 kΩ
42 VI1 = 4 mA
+
Vo
_
Suggested Solution
12 kΩ
4 kΩ 6 kΩ 18 kΩ
6 kΩ
42 VI1 = 4 mA
+
Vo
_
+ V3 -
_
V2
+
+
V1
_
I3
I2
( )1 14 k 16 VV I= Ω =
2 142 V 42 16 26 VV V= − = − =
2
2
26 13
mA
6 k 6000 3
V
I = = =
Ω
3 2 1
13 1
mA 4 mA mA
3 3
I I I= − = − =
( )3 312 k 4 VV I= Ω =
1 3 16 4 12 VoV V V= − = − =

73. Problem 2.72
Find the power absorbed by the network shown.
12 kΩ
6 kΩ 6 kΩ
2 kΩ 18 kΩ
21 V
Suggested Solution
Redrawing the network:
Y∆ →
⇒
12 kΩ
2 kΩ 6 kΩ
18 kΩ 6 kΩ
21 V
3 kΩ
2 kΩ 6 kΩ
6 kΩ 2 kΩ
21 V
The equivalent resistance seen by the source is:
( ) ( )2 k 6 k 6 k 2 k 3 k
4 k 3 k
7 k
eqR  = Ω + Ω Ω + Ω + 
= Ω + Ω
= Ω
Ω
Then,
( )
2
21 V 441
63 mA
7000eq
P
R
= = =

74. Problem 2.73
Find oI in the circuit shown.
2 Ω
3 Ω
4 Ω
5 Ω
18 Ω
12 Ω
9 Ω
36 V
Io
12 Ω
Suggested Solution
Note that the four right-most resistors can be combined as ( )3 4 5 12 6Ω + Ω + Ω Ω = Ω . Then the circuit
can be redrawn as:
Y∆ →
⇒
Io
36 V
9 Ω
3 Ω 2 Ω
2 Ω
6 Ω
I
Io
36 V
9 Ω
18 Ω 12 Ω
2 Ω
6 Ω
( ) ( )
36 V
4 A
9 3 2 2 6
I = =
 Ω + Ω Ω + Ω + Ω 
( )
( ) ( )
9 3
3 A
9 3 2 2
oI I
 +
= = 
+ + +  

75. Problem 2.74
Find oI in the circuit shown.
Suggested Solution
Applying the transformation to the three 12-Y∆ → Ω resistors yields:
12 Ω
5 Ω
12 Ω
12 Ω
14 Ω
12 V
Io
5 Ω 14 Ω
12 V
Io
4 Ω 4 Ω
4 Ω
Ix
( )
( ) ( )
4 14
4 5 4 14
o xI I
 +
= −  
+ + +  
where
( ) ( )
12 V 12
1.2 A
104 4 5 4 14
xI = =
 Ω + Ω + Ω Ω + Ω 
=
18 12 4
0.80 A
27 10 5
oI∴ = − = − ≈ −i

76. Problem 2.75
Find oI in the circuit shown.
Io
12 Ω
6 Ω
4 Ω
18 Ω
6 Ω
12 A
Suggested Solution
Converting the to a Y yields the circuit:∆
Io
4 Ω 6 Ω
12 A
3 Ω
2 Ω 6 Ω
( )
( ) ( )
( )
2 4
12 A 4 A
2 4 6 6
oI
 +
= = 
+ + +  

77. Problem 2.76
Find oI in the circuit shown.
4 Ω
4 Ω
4 Ω
12 Ω
12 Ω
Io
12 A
Suggested Solution
Applying the transformation:Y → ∆
12 A
⇒
12 Ω
12 Ω
12 Ω
Io
12 Ω
12 Ω
I Io
12 A 12 Ω
12 Ω
12 Ω
12 Ω
12 Ω
The circuit can be further simplified to:
I
12 A
12 Ω
12 Ω
12 A
6 A
2
I = =
3 A
2
o
I
I = =

78. Problem 2.77
Find oI in the circuit shown.
12 Ω 12 Ω
6 Ω
6 Ω 6 Ω
6 Ω
12 V
Io
Suggested Solution
Combining the two 12-Ω resistors yields:
6
Y∆ →
⇒
6 Ω
Ω
Ω 6 Ω
6 Ω
12 V
Io
6 Ω
6 Ω 2 Ω
2 Ω
12 V
Io
2 Ω
6
The circuit can be further simplified to the following:
⇒
I
12 V
o
2 Ω
8 Ω
8 Ω
12 V
Io
2 Ω
4 Ω
12 V
2 A
6
oI = =
Ω

79. Problem 2.78
Find V in the circuit shown.o
12 V
+
Vo
_
IS
2000 IS
3 kΩ
5 kΩ
Suggested Solution
KVL: 12 3000 2000 5000 0S S SI I I− + − + =
2 mASI⇒ =
5000 10 Vo SV I= =

80. Problem 2.79
Find V in the network shown.o
12 V
2 kΩ
2 Vo
2 kΩ
+
Vo
_
Suggested Solution
12 V
2 kΩ
2 Vo
2 kΩ
+
Vo
_
I
KVL: where V I12 2000 2 0o oI V V− + + + = 2000o =
Therefore,
12 2000 4000 2000 0I I I− + + + =
1.5 mAI⇒ =
( )3
2000 1.5 10 3 VoV −
= × =

81. Problem 2.80
Find V in the network shown.o
5 mA 2 kΩ 1 kΩ2 Ix
Ix
+
Vo
_
Suggested Solution
KCL: 0.005 2 0
2000 1000
o o
x
V V
I− + + + = where
2 k
o
x
V
I =
Ω
Therefore,
0.005 2 0 2 V
2000 2000 1000
o o o
o
V V V
V− + + + = ⇒ =

82. Problem 2.81
Find oI in the network shown.
Io
+
Vx
_
5 A 3 Vx6 Ω
4 Ω
8 Ω
Suggested Solution
Io
+
Vx
_
5 A 3 Vx6 Ω
4 Ω
8 Ω
+
V1
_
KCL: 1 1
5 3 0
6 8 4
x
V V
V+ − + =
+
where 1
1
4
8 4 3
x
V 
V V= = 
+ 
Therefore,
1 1 1
5 3 0
6 12 3
V V V
+ − + =
1 4 VV⇒ =
1 4 2
A
6 6 3
o
V
I = = =
Ω

83. Problem 2.82
Find V in the circuit shown.o
6 kΩ
2 kΩ
1 kΩ
+
Vo
_
4 mA
2000
oV
Suggested Solution
6 kΩ
2 kΩ
1 kΩ
+
Vo
_
4 mA
2000
oV
+
V1
_
KCL: 1 1
0.004 0
6000 2000 2000 1000
oVV V
− − + =
+
where 1
1
1000
2000 1000 3
o
V 
V V= = 
+ 
Therefore,
3 3
0.004 0
6000 2000 3000
o o oV V V
− − + =
4 VoV⇒ =

84. Problem 2.83
A single-stage transistor amplifier is modeled as shown. Find the current in the load LR .
+
VS = 250 mV
_
RS = 1 kΩ Rb = 250 Ω
Ib
110 Ib
Ro = 4 kΩ
RL = 400 Ω
Io
Suggested Solution
0.25
0.2 mA
1000 250
S
b
S b
V
I
R R
= = =
+ +
( )
4000
110 100 20 mA
4000 400
o b bI I I
 
= − = − = − 
+ 

85. Problem 2.84
A typical transistor amplifier is shown. Find the amplifier gain G (i.e., the ratio of the output voltage to
the input voltage).
250 mVSV =
5
3.5 10 bI×
100 Ω
5 kΩ
500 Ω
Ib
5 kΩ
300 Ω
+
Vo
_
Suggested Solution
250 mVSV =
5
3.5 10 bI×
100 Ω
5 kΩ
500 Ω
Ib
5 kΩ
300 Ω
+
Vo
_
I1
( )1
0.25
451 A
554.5100 5 k 500
SV
I µ= =
Ω + Ω Ω
=
1
5000
409.8 A
5000 500
bI I µ
 
= = 
+ 
( )5300
3.5 10 8.12 V
300 5000
o bV I
  = − × = −  + 
8.12
32.5
0.250
o
S
V
G
V
−
= ≈ −

86. Problem 2.85
For the network shown, choose the values of and such that V is maximized. What is the resulting
ratio,
inR oR o
o SV V ?
VS
RS
Rin
Ro
RLµ Vin
+
Vin
_
+
Vo
_
Suggested Solution
in
in S
S in
R
V V
R R
 
=  
+ 
( ) inL L
o in
o L S in o L
RR R
V V
R R R R R R
µ µ
   
= =   
+ +   
SV


+ 
Therefore,
o in L
S S in o
V R R
V R R R R
µ
 
=  
+ +  L



Clearly, to maximize o
S
V
V
, we must maximize . This is becauseinR in
S i
R
R R+ n
is always less than 1, but gets
closer and closer to 1 as is made larger and larger.inR
On the other hand, to maximize o
S
V
V
, we must minimize . This is becauseoR L
o L
R
R R+
is always less than
1, but gets closer and closer to 1 as is made smaller and smaller.oR
In the limit,
0 0
lim lim lim lim 1 1
in o in o
o in L
R R R R
S S in o L
V R R
V R R R R
µ µ µ
→∞ → →∞ →
    
= =    
+ +    
× × =

87. Problem 2.86
In many amplifier applications we are concerned not only with voltage gain, but also with power gain.
( ) ( )Power gain = = power delivered to the load power delivered by the inputpA
Find the power gain for the circuit shown, where 50 kLR = Ω .
VS
+
V
_
8 kinR = Ω
2 kSR = Ω
V/100
100 koR = Ω
50 kLR = Ω
+
Vo
_
Suggested Solution
VS
+
V
_
8 kinR = Ω
2 kSR = Ω
V/100
100 koR = Ω
50 kLR = Ω
+
Vo
_
IS Io
8000
0.8
8000 2000
S SV V
 
= = 
+ 
V
3100,000
5.33 10
100,000 50,000 100
o S
V
I V−  
= − = − ×  
+   
( ) ( )
22 3
5.33 10 50,000) 1.422load o L S SP I R V V−
= = × = 2
2
8000 2000 10,000
S S
in S S S
V V
P V I V
 
= = = + 
2
3
2
1.422
14.22 10
10,000
load S
p
in S
P V
A
P V
= = = ×
 
 
 

88. Problem 2.87
Find the power absorbed by the 10- resistor in the circuit shown.kΩ
11 mA 2 kΩ
4 kΩ 4 kΩ
3 kΩ
10 kΩ
+
Vx
_
+
Vo
_
2000
xV
Suggested Solution
Combining the two resistors in parallel yields the following simplified circuit:4-kΩ
11 mA 2 kΩ
2 kΩ
3 kΩ
10 kΩ
+
Vx
_
+
Vo
_
2000
xV
0.011 0
2 k 2000 2 k 3 k 10 k
o x o oV V V V
− + + + + =
Ω Ω + Ω Ω
where
3000
0.6
2000 3000
x o o
 
= = 
+ 
V V V
Therefore,
3
0.011 0 10 V
2000 10,000 5000 10,000
o o o o
o
V V V V
V− + + + + = ⇒ =
( )
22
10 k
10
10 mW
10 k 10,000
oV
P Ω = = =
Ω

89. Problem 2.88
Find the power absorbed by the 12- resistor in the circuit shown.kΩ
6 mA 6 kΩ
3 Io
6 kΩ 3 kΩ
4 kΩ
12 kΩ
Io
+
Vo
_
Suggested Solution
6 mA 6 kΩ
3 Io
6 kΩ 3 kΩ
4 kΩ
12 kΩ
Io
+
Vo
_
I1
( )
0.006 3 0
6 k 12 k4 k 6 k 3 k
o o
o
V V
I− + + + + =
Ω ΩΩ + Ω Ω
oV
where 1 1
6000 2
6000 3000 3
oI I I
 
= = 
+ 
( )1
2
6000 3 6000 90004 k 6 k 3 k
o o o
o
V V V
I I
 
= = ⇒ =  
Ω + Ω Ω  
oV
=
Then, substituting:
0.006 3 0 8 V
6000 9000 6000 12,000
o o o o
o
V V V V
V
 
− + + + + = ⇒ = 
 
2
0
12 k
64
5.33 mW
12 k 12,000
V
P Ω = = =
Ω

90. Problem 2FE-1
Find the power generated by the source in the network shown.
5 kΩ
6 kΩ 18 kΩ
12 kΩ
4 kΩ 6 kΩ
120 V
Suggested Solution
Applying the transformation yields:Y∆ →
⇒
5 kΩ
4 kΩ 6 kΩ
120 V
3 kΩ
2 kΩ 6 kΩ
5 kΩ
120 V
3 kΩ
4 kΩ
I
120 V
10 mA
12 k
I = =
Ω
( )120 V 120 0.010 1.2 WP I= = × =

91. Problem 2FE-2
Find the equivalent resistance of the circuit shown at the terminals -A B .
A
B
RAB
12 kΩ
6 kΩ
6 kΩ
4 kΩ
12 kΩ
12 kΩ
12 kΩ
Suggested Solution
Combining resistors,
12 k 6 k 4 kΩ Ω = Ω
( )12 k 12 kΩ 12 kΩ 8 kΩ + = Ω
The circuit is now reduced to:
A
B
RAB
4 kΩ
6 kΩ
4 kΩ
8 kΩ
and further to:
A
B
RAB
4 kΩ
( )6 k 8 k 4 k 4 kΩ Ω+ Ω = Ω
Therefore,
4 k 4 k 8 kABR = Ω + Ω = Ω

92. Problem 2FE-3
Find the voltage V in the network shown.o
24 mA
1 kΩ 2 kΩ
6 kΩ3 kΩ
6 kΩ
12 kΩ
+
Vo
_
Suggested Solution
Combining resistors:
1 k 3 k 4 kΩ + Ω = Ω
6 k 12 k 4 kΩ Ω = Ω
The network is now reduced to:
24 mA
2 kΩ
6 kΩ4 kΩ
4 kΩ
+
Vo
_
I
Using current division,
( )
( )
4000
24 mA 6 mA
4000 2000 6000 4000
I
 
= = 
+ + +  
( )6 k 36 VoV I= Ω =

93. Problem 2FE-4
Find the current oI in the circuit shown.
Io
3 kΩ
3 kΩ 6 kΩ6 kΩ
6 kΩ
12 kΩ
4 kΩ
12 V
Suggested Solution
Combining resistors:
3 k 6 k 9 kΩ + Ω = Ω
3 k 6 k 2 kΩ Ω = Ω
Thus, the circuit simplifies to:
⇒ ⇒9 k
2 kΩ
6 kΩ
9 kΩ
12 kΩ
4 kΩ
12 V
Ix
I
18 kΩ
6 kΩ
Ω
12 V
Ix
I
6 kΩ
12 V
6 kΩ
I
12V
1 mA
6 k 6 k
I = − = −
Ω + Ω
9000 1
mA
9000 18,000 3
xI I
 
= = 
+ 
−
6000 2
mA
3000 6000 9
o xI I
 
= = 
+ 
−