S10HW4
Transcript of S10HW4
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 –