Chapter 3, Problem 2. For the circuit in Fig. 3.51, obtain v1 and v2.

Figure 3.51

Chapter 3, Solution 2 At node 1,

2

vv125v

10v 2111 −

+=−− 120 = - 8v1 + 5v2 (1)

At node 2,

2

vv1264v 212 −

++= 72 = – 2v1 + 3v2 (2)

Solving (1) and (2), v1 = 0 V, v2 = 24 V Chapter 3, Problem 4. Given the circuit in Fig. 3.53, calculate the currents i1 through i4.

Figure 3.53 Chapter 3, Solution 4

12 A

6 A

At node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20 At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Problem 9. Determine Ib in the circuit in Fig. 3.58 using nodal analysis.

Figure 3.58 For Prob. 3.9. Chapter 3, Solution 9 Let V1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground reference be placed at the bottom of the 50-Ω resistor. This leads to the following nodal equation:

i1

10 Ω 5 Ω 5 A 5 Ω 4 A

i2

v2v1

i3 i4

10 Ω

2A

250 Ω

50 Ω

+ –

150 Ω

+ _ 12 V

60 Ib Ib

which leads to

0I300V5V1536V3

0150

0I60V50

0V250

12V

b111

b111

=−++−

=−−

+−

+−

But 250

V12I 1b

−= . Substituting this into the nodal equation leads to

04.50V2.24 1 =− or V1 = 2.083 V. Thus, Ib = (12 – 2.083)/250 = 39.67 mA.

Chapter 3, Problem 10. Find i0 in the circuit in Fig. 3.59.

Figure 3.59

At Node V1: V1/8 + 4 + (V1-V2)/1 = 0 Simplifying gives Eqn1: 9V1 – 8V2 = -32 At Node V2: V2/4 – 2io – (V1-V2) = 0 io = V1/8 Substitute for io: V2/4 – 2(V1/8) – (V1-V2) = 0 Simplifying gives Eqn 2: -5V1 + 5V2 = 0 Solving these two equations gives: io = -4A Chapter 3, Problem 12.

V1 V2

Using nodal analysis, determine Vo in the circuit in Fig. 3.61. There are two unknown nodes, as shown in the circuit below. At node 1,

30V10V16

01

VV2

0V10

30V

o1

o111

=−

=−

+−

+−

(1)

At node o,

0I20V6V5

05

0VI4

1VV

xo1

ox

1o

=−+−

=−

+−−

(2)

But Ix = V1/2. Substituting this in (2) leads to –15V1 + 6Vo = 0 or V1 = 0.4Vo (3) Substituting (3) into 1,

16(0.4Vo) – 10Vo = 30 or Vo = –8.333 V.

Chapter 3, Problem 13. Calculate v1 and v2 in the circuit of Fig. 3.62 using nodal analysis.

10 Ω

+ _ 2 Ω 5 Ω 30 V

4 Ix

1 Ω Vo V1

Ix

Figure 3.62

Chapter 3, Solution 13

At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts

Chapter 3, Problem 14. Using nodal analysis, find vo in the circuit of Fig. 3.63.

Figure 3.63

Chapter 3, Solution 14 –

+40 V

–+20 V

8 Ω

5 A

v1 v0

4 Ω

2 Ω 1 Ω

At node 1,

[(v1–40)/1] + [(v1–vo)/2] + 5 = 0 or 3v1 – vo = 70 (1) At node o,

[(vo–v1)/2] – 5 + [(vo–0)/4] + [(vo+20)/8] = 0 or 7vo – 4v1 = 20 (2) Solving (1) and (2),

4(1) + 3(2) = –4vo + 21vo = 280 + 60 or 17vo = 340 or

vo = 20 V.

Checking, we get v1 = (70+vo)/3 = 30 V. At node 1,

[(30–40)/1] + [(30–20)/2] +5 = –10 + 5 + 5 = 0!

At node o, [(20–30)/2] + [(20–0)/4] + [(20+20)/8] – 5 = –5 + 5 + 5 – 5 = 0!

Chapter 3, Problem 16. Determine voltages v1 through v3 in the circuit of Fig. 3.65 using nodal analysis.

Figure 3.65

Chapter 3, Solution 16 At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1) But

v1 = v2 + 2v0 and v0 = v2. Hence

v1 = 3v2 (2) v3 = 13V (3)

Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V Chapter 3, Problem 17. Using nodal analysis, find current io in the circuit of Fig. 3.66.

2 A

v3 v2v1

8 S

4 S 1 S

i0

–+13 V

2 S

+ v0 –

Figure 3.66

Chapter 3, Solution 17

At node 1, 2

vv8v

4v60 2111 −

+=− 120 = 7v1 - 4v2 (1)

At node 2, 3i0 + 02

vv10

v60 212 =−

+−

But i0 = .4

v60 1−

Hence

( )

02

vv10

v604

v603 2121 =−

+−

+− 1020 = 5v1 + 12v2 (2)

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = =−4

v60 1 1.73 A

Chapter 3, Problem 21. For the circuit in Fig. 3.70, find v1 and v2 using nodal analysis.

60 V

–+60 V

4 Ω

10 Ω

2 Ω

8 Ω

3i0

i0

v1

v2

Figure 3.70

Chapter 3, Solution 21 Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source. At node 1,

2000

vv4000

vv10x3 31213 −+

−=− 12 = 3v1 - v2 - 2v3 (1)

At node 2,

1v

2vv

4vv 23121 =

−+

− 3v1 - 5v2 - 2v3 = 0 (2)

Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 v3 = - 2v2 (3) From (1) to (3), v1 = 1 V, v2 = 3 V

(b)

+

+ v3 –

+ v2 –

3v0

3 mA

+

1 kΩ

v3 v2v1 3v0

+ v0 –

4 kΩ

2 kΩ

(a)

Chapter 3, Problem 27. Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig. 3.76.

Figure 3.76

Chapter 3, Solution 27 At node 1, 2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence, 2 = 7v1 + 11v2 – 4v3 (1) At node 2, v1 – v2 = 4v2 + v2 – v3 0 = – v1 + 6v2 – v3 (2) At node 3,

2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3) or – 4 = 4v1 + 13v2 – 7v3 (3) In matrix form,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

402

vvv

71341614117

3

2

1

,1767134

1614117=

−−

−=Δ 110

71341604112

1 =−−

−−

=Δ

,66744

101427

2 =−−

−=Δ 286

41340612117

3 =−

−=Δ

v1 = ,V625.01761101 ==

ΔΔ

v2 = V375.0176662 ==

ΔΔ

v3 = .V625.11762863 ==

ΔΔ

v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V.

Chapter 3, Problem 31. Find the node voltages for the circuit in Fig. 3.80.

Figure 3.80

Chapter 3, Solution 31 At the supernode,

1 + 2v0 = 1

vv1

v4v 3121 −

++ (1)

But vo = v1 – v3. Hence (1) becomes, 4 = -3v1 + 4v2 +4v3 (2) At node 3,

2vo + 2

v10vv

4v 3

313 −

+−=

or 20 = 4v1 + 0v2 – v3 (3)

At the supernode, v2 = v1 + 4io. But io = 4

v 3 . Hence,

v2 = v1 + v3 (4) Solving (2) to (4) leads to,

v1 = 4.97V, v2 = 4.85V, v3 = –0.12V.

i0

4 Ω

v2v1

1 Ω

1 A –+10 V 4 Ω

2 Ω

1 Ω

v32v0

+ v0 –