February 5, 2006
CHAPTER 1
P.P.1.1
A proton has 1.602 x 10-19 C. Hence, 2 million million protons have +1.602 x 10-19 x 2 x 106 = 3.204 x 10–13 C
P.P.1.2
i = dq/dt = -10(–2)e-2t mA At t = 0.5 sec, i = 20e-1 =
P.P.1.3
q=
1
∫ idt = ∫ 2dt + ∫ 0
= 2 + 14/3 =
P.P.1.4
(a)
2
1
7.358 mA
2
2t dt = 2 t
2
1
+ (2 / 3) t 0
3
2 1
6.667 C
Vab = w/q = -30/2 =
–15 V
The negative sign indicates that point a is at higher potential than point b.
P.P.1.5
(b)
Vab = w/q = -30/-6 =
(a)
v = 2 i = 10 cos (60 π t)
5V
p = v i = 50 cos2 (60 π t) At t = 5 ms, ms, p = 50 cos2 (60 π 5x10-3) = 50 cos2 (0.3 π) = (b)
v = 10 + 5
∫
t 0
17.27 watts
idt = 10 +
∫
t 0
25 cos 60 π t dt = 10+
25 60π
sin 60
p = vi = 5 cos (60 πt)[10 + (25/(60 π)) sin (60 π t)] At t = 5 ms, p = 5 cos (0.3π){10 + (25/(60 π)) sin (0.3 π)} =
29.7 watts
π
t
P.P.1.6
p = v i = 15 x 120 = 1800 watts; w = p x t therefore, t = w/p = (30x103)/1800 =
P.P.1.7
16.667 seconds
p1 = 5(-8) = –40w p2 = 2(8) = 16w p3 = 0.61(3) = 0.6(5)(3) = 9w p4 = 3(5) = 15w
P.P.1.8
i=
= e
dn = -1.6 x 10-19 x 1013 = -1.6 x 10-6 A dt
p = v0 i = 30 x 103 x (1.6 x 10-6) = 48mW
P.P.1.9
Minimum monthly charge
= $12.00
First 100 kWh @ $0.16/kWh
= $16.00
Next 200 kWh @ $0.10/kWh
= $20.00
Remaining 100 kWh @ $0.06/kWh = $6.00 Total Charge = $54.00 Average cost = $54/[100+200+100] = 13.5 P.P.1.10
cents/kWh
This assigned practice problem is to apply the detailed problem solving technique to some of the more difficult problems of Chapter 1.
P.P.1.6
p = v i = 15 x 120 = 1800 watts; w = p x t therefore, t = w/p = (30x103)/1800 =
P.P.1.7
16.667 seconds
p1 = 5(-8) = –40w p2 = 2(8) = 16w p3 = 0.61(3) = 0.6(5)(3) = 9w p4 = 3(5) = 15w
P.P.1.8
i=
= e
dn = -1.6 x 10-19 x 1013 = -1.6 x 10-6 A dt
p = v0 i = 30 x 103 x (1.6 x 10-6) = 48mW
P.P.1.9
Minimum monthly charge
= $12.00
First 100 kWh @ $0.16/kWh
= $16.00
Next 200 kWh @ $0.10/kWh
= $20.00
Remaining 100 kWh @ $0.06/kWh = $6.00 Total Charge = $54.00 Average cost = $54/[100+200+100] = 13.5 P.P.1.10
cents/kWh
This assigned practice problem is to apply the detailed problem solving technique to some of the more difficult problems of Chapter 1.
February 5, 2006
CHAPTER 2
P.P.2.1
i = V/R = 110/12 = 9.167 A
P.P.2.2
(a)
v = iR = 2 mA[10 kohms] = 20 V
(b)
G = 1/R = 1/10 kohms = 100 µS
(c)
p = vi = 20 volts[2 mA] = 40 mW
P.P.2.3
p = vi which leads to i = p/v = [20 cos2 (t) mW]/[10cos(t) mA] or i = 2cos(t) mA R = v/i = 10cos(t)V/2cos(t)mA 10cos(t)V/2cos(t)mA = 5 k
P.P.2.4
5 branches and 3 nodes. nodes. The 1 ohm and 2 ohm ohm resistors are in parallel. parallel. The 4 ohm resistor and the 10 volt source are also in parallel.
P.P.2.5
Applying KVL to the loop we get: -10 + 4i – 8 + 2i = 0 which leads to i = 3A v1 = 4i = 12V and v2 = -2i = –6V
P.P.2.6
Applying KVL to the loop we get: -35 + 10i + 2v x + 5i = 0 But, vx = 10i and v0 = -5i. Hence, -35 + 10i + 20i + 5i = 0 which leads to i = 1A. Thus, vx = 10V and v0 = –5V
Applying KCL, 6 = i0 + [i0 /4] + [v0 /8], but i0 = v0/2
P.P.2.7
Which leads to: 6 = (v0/2) + (v0/8) + (v0/8) thus, v 0 = 8V and i0 = 4A P.P.2.8
2
+ -
5V
i1
+ V1 -
4 + V3 -
i2 + V2 -
Loop 1
i3
+
Loop 2
8
3V
At the top node,
i1 = i2 + i3
(1)
For loop 1 or
-5 + V1 + V2 = 0 V1 = 5 - V2
(2)
For loop 2 or
- V2 + V3 -3 = 0 V3 = V2 + 3
(3)
Using (1) and Ohm’s law, we get (V1/2) = (V2/8) + (V3/4) and now using (2) and (3) in the above yields [(5- V2)/2] = (V2/8) + (V2+3)/4 or
V2 = 2 V
V1 = 5- V2 = 3V, V3 = 2+3 = 5V, i1 = (5-2)/2 = 1.5A, i2 = 250 mA, i3 = 1.25A 2
P.P.2.9
R eq
6
1
3
4
5
4
3
Combining the 4 ohm, 5 ohm, and 3ohm resistors in series gives 4+3+5 = 12. But, 4 in parallel with 12 produces [4x12]/[4+12] = 48/16 = 3ohm. So that the equivalent circuit is shown below. 2
Req
3
3
6
1
Thus, R eq = 1 + 2 + [6x6]/[6+6] = 6
20
P.P.2.10
8
Req
5
20
18
1 9 2
Combining the 9 ohm resistor and the 18 ohm resistor yields [9x18]/[9+18] = 6 ohms.
Combining the 5 ohm and the 20 ohm resistors in parallel produces [5x20/(5+20)] = 4 ohms We now have the following circuit: 8
4
6
1
20 2
The 4 ohm and 1 ohm resistors can be combined into a 5 ohm resistor in parallel with a 20 ohm resistor. This will result in [5x20/(5+20)] = 4 ohms and the circuit shown below: 8
4
6
2
The 4 ohm and 2 ohm resistors are in series and can be replaced by a 6 ohm resistor. This gives a 6 ohm resistor in parallel with a 6 ohm resistor, [6x6/(6+6)] = 3 ohms. We now have a 3 ohm resistor in series with an 8 ohm resistor or 3 + 8 = 11ohms. Therefore: R eq = 11 ohms
P.P. 2.11 8S
4S
8||4 = 8+4 = 12S 12 S
Geq
Geq 2S
4S
2||4 = 2+4 = 6S
12 S in series with 6 S = {12x6/(12+6)] = 4 or:
Geq = 4 S
6S
12
P.P.2.12
15V
-
+ v
i1
i2 +
6
+ -
+ v1
10
15V
40
v2
+ -
+
4 8
v2
-
-
6||12 = [6x12/(6+12)] = 4 ohm and 10||40 = [10x40/(10+40)] = 8 ohm. Using voltage division we get: v1 = [4/(4+8)] (15) = 5 volts, v2 = [8/12] (15) = 10 volts
i1 = v1/12 = 5/12 = 416.7 mA, i2 = v2/40 = 10/40 = 250 mA P1 = v1 i1 = 5x5/12 = 2.083 watts, P2 = v2 i2 = 10x0.25 = 2.5 watts
P.P.2.13 i1
1k
+ 3k
v1
-
i2
+ 10mA
4k 5k
20k
v2
-
4k
10mA
Using current division, i 1 = i2 = (10 mA)(4 kohm/(4 kohm + 4 kohm)) = 5mA (a)
v1 = (3 kohm)(5 mA) = 15 volts v2 = (4 kohm)(5 mA) = 20 volts
(b)
For the 3k ohm resistor, P1 = v1 x i1 = 15x5 = 75 mw For the 20k ohm resistor, P2 = (v2)2 /20k = 20 mw The total power supplied by the current source is equal to: P = v2 x 10 mA = 20x10 = 200 mw
(c)
P.P.2.14 R a = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 1 = [10x20 + 20x40 + 40x10]/10 = 140 ohms R b = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 2 = 1400/20 = 70 ohms R c = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 3 = 1400/40 = 35 ohms
P.P.2.15
We first find the equivalent resistance, R. We convert the delta sub-network to a wye connected form as shown below: 13
i
a
24
100V + -
a
13 20
30
10
24
10
a’
b’
6
50
10
n
b
15 b
c’
R a’n = 20x30/[20 + 30 + 50] = 6 ohms, R b’n = 20x50/100 = 10 ohms R c’n = 30x50/100 = 15 ohms. Thus, R ab = 13 + (24 + 6)||(10 + 10) + 15 = 28 + 30x20/(30 + 20) = 40 ohms. i = 100/ R ab = 100/40 = 2.5 amps
P.P.2.16
For the parallel case, v = v0 = 110volts. p = vi i = p/v = 40/110 = 364 mA For the series case, v = v0/N = 110/10 = 11 volts i = p/v = 40/11 = 3.64 amps (a)
We use equation (2.61) -3 -3 R 1 = 50x10 / (1-10 ) = 0.05/999 = 50 mΩ (shunt)
(b)
R 2 = 50x10 /(100x10 – 10 ) = 50/99 = 505 mΩ (shunt)
(c)
R 3 = 50x10 /(10x10 -10 ) = 50/9 = 5.556 Ω (shunt)
P.P.2.17
-3
-3
-3
-3
-3
-3
February 5, 2006
CHAPTER 3
P.P.3.1
1A
6
i1
1
i1
i2
1A
2
4A i3
2
4A
7
At node 1, 1 = i1 + i2
1=
or 6 = 4v1 - v2
(1)
v1
− v2
+
v1
= 4+
v2
6
−0 2
At node 2, i1
v1
= 4 + i3
− v2 6
−0 7
or 168 = 7v1 - 13v2 (2) Solving (1) and (2) gives v1 = –2 V, v2 = –14 V 2
i1
P.P.3.2
4ix v1
i2
i2 3
10 A
v2
v3 ix 4
i3 6
At node 1, 10 = i1 + i2 =
v1 − v 3
+
− v2
v1
2 or 60 = 5v1 - 2v2 - 3v3
3 (1)
At node 2,
i2
v1
+ 4i x = i x
− v2 3
+3
v2 4
=0
or 4v1 + 5v2 = 0
(2)
At node 3,
v1
i1 = i3 + 4ix
− v3 2
=
v3
−0 6
+4
v2 4
or -3v1 + 6v2 + 4v3 = 0
(3)
Solving (1) to (3) gives v1 = 80 V, v2 = –64 V, v3 = 156 V
P.P.3.3 4
v
7V
v1
- +
+
3
+
2
v
-
6
-
At the supernode in Fig. (a),
7−v 4
v
v1
3
2
= +
+
or 21 = 7v + 8v1
v1 6 (1)
Applying KVL to the loop in Fig. (b), - v - 3 + v1 = 0
+
+
v
v1
-
(b)
(a)
v1 = v + 3
(2)
3V
Solving (1) and (2), v = – 200 mV v1 = v + 3 = 2.8, i1 =
v1 2
= 1.4
i1 = 1.4 A P.P.3.4 v1
v2
3V
v3
+ -
-
+
+
+
v1
v2
v3
-
-
-
(a)
(b)
From Fig. (a),
v1
+
v2
+
v3
=0
6v1 + 3v2 + 4v3 = 0
(1)
- v1 + 10 + v2 = 0
v1 = v2 + 10
(2)
- v2 - 5i + v3 = 0
v3 = v2 + 5i
(3)
2
4
3
From Fig. (b),
Solving (1) to (3), we obtain v1 = 3.043V, v2 = –6.956 V, v3 = 652.2 mV P.P.3.5 We apply KVL to the two loops and obtain
- 12 + 18ii - 12i2 = 0 8 + 24i2 - 12i1 = 0 From (1) and (2) we get i1 = 666.7 mA, i2 = 0A
+
3ii - 2i2 = 2
(1)
- 3i1 + 6i2 = -2
(2)
P.P.3.6 For mesh 1,
- 20 + 6i1 – 2i2 - 4i3 = 0
3i1 - i2 - 2i3 = 10
(1)
For mesh 2, 10i2 - 2i1 - 8i3 - 10i0 = 0 = -i1 + 5i2 – 9i3
(2)
But i0 = i3, 18i3 - 4i1 - 8i2 = 0
- 2i1 - 4i2 + 9i3 = 0
From (1) to (3),
⎡ 3 − 1 − 2⎤ ⎡ i 1 ⎤ ⎢ − 1 5 − 9⎥ ⎢i ⎥ ⎥ ⎢ 2⎥ ⎢ ⎢⎣− 2 − 4 9 ⎥⎦ ⎢⎣i 3 ⎥⎦ 3
−1 Δ = −2 3
−1 10 0
Δ1 =
0 10 0
−1 − 2 5 −9 − 4 9 = 135 - 8 - 18 - 20 - 108 - 9 = - 28 −1 − 2 5 −9 −1 − 2 5 −9 − 4 9 = 450 − 360 = 90 −1 − 2 5 −9
0
−2 −9
0
9
3
10
−1
0
−2 −9
3
−1 Δ2 = − 2
=
⎡10⎤ ⎢0⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
10
= 180 + 90 = 270
(3)
−1
10
−1 5 Δ3 = − 2 − 4 3 −1 −1 5
0
3
i1 =
= 40 + 100 = 140
0 10 0
Δ Δ 270 140 Δ1 90 = −9.643 , i3 = 3 = = −5A = = −3.214, i2 = 2 = Δ − 28 Δ − 28 Δ − 28
i0 = i3 = –5A P.P.3.7 2
i3
2
i1 6V
i1
i3
2
2
4
+
-
4
+
3A
-
3A
i2
8
i2 8
1
i1
0 i2 (a)
(b)
For the supermesh, - 6 + 2i1 - 2i3 + 12i2 - 4i3 = 0
i1 + 6i2 - 3i3 = 3
(1)
For mesh 3, 8i3 - 2i1 - 4i2 = 0
- i1 - 2i2 + 4i3 = 0
At node 0 in Fig. (a), i1 = 3 + i2
i1 - i2 = 3
Solving (1) to (3) yields i1 = 3.474A, i2 = 473.7 mA, i3 = 1.1052A
(2)
P.P.3.8 G11 = 1/(1) + 1/(10) + 1/(5) = 1.3, G12 = -1/(5) = -0.2, G33 = 1/(4) + 1 = 1.25, G44 = 1/(2) + 1/(4) = 0.75, G12 = -1/(5) = - 0.2, G 13 = - 1, G14 = 0, G21 = -0.2, G23 = 0 = G26, G31 = -1, G32 = 0, G34 = - 1/4 = - 0.25, G41 = 0, G42 = 0, G43 = 0.25, i1 = 0, i2 = 2 - 1 = 1, i3 = - 1, i4 = 3.
Hence, 0 ⎤ ⎡ 1.3 − 0.2 − 1 ⎢− 0.2 0.2 ⎥ 0 0 ⎢ ⎥ ⎢ −1 0 1.25 − 0.25⎥ ⎢ ⎥ 0 − 0.25 0.75 ⎦ ⎣ 0 P.P.3.9
⎡ v1 ⎤ ⎡ 0 ⎤ ⎢v ⎥ ⎢ 3 ⎥ ⎢ 2⎥ = ⎢ ⎥ ⎢ v3 ⎥ ⎢− 1⎥ ⎢ ⎥ ⎢ ⎥ ⎣v4 ⎦ ⎣ 3 ⎦
R 11 = 50 + 40 + 80 = 170, R 22 = 40 + 30 + 10 = 80, R 33 = 30 + 20 = 50, R 44 = 10 + 80 = 90, R 55 = 20 + 60 = 80, R 12 = -40, R 13 = 0, R 14 = -80, R 15 = 0, R 21 = -40, R 23 = -30, R 24 = -10, R 25 = 0, R 31 = 0, R 32 = -30, R 34 = 0, R 35 = -20, R 41 = -80, R 42 = -10, R 43 = 0, R 45 = 0, R 51 = 0, R 52 = 0, R 53 = -20, R 54 = 0, v1 = 24, v2 = 0, v3 = -12, v4 = 10, v5 = -10
Hence the mesh-current equations are
⎡ 170 − 40 0 − 80 0 ⎤ ⎢− 40 80 − 30 − 10 0 ⎥ ⎢ ⎥ ⎢ 0 − 30 50 0 − 20⎥ ⎢ ⎥ − − 80 10 0 90 0 ⎢ ⎥ ⎢⎣ 0 − 20 0 0 80 ⎥⎦ P.P.3.10
⎡ i1 ⎤ ⎢i ⎥ ⎢ 2⎥ ⎢i 3 ⎥ ⎢ ⎥ ⎢i 4 ⎥ ⎢⎣i 5 ⎥⎦
=
⎡ 24 ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢− 12⎥ ⎢ ⎥ ⎢ 10 ⎥ ⎢⎣− 10⎥⎦
The schematic is shown below. It is saved and simulated by selecting Analysis/Simulate. The results are shown on the viewpoints:
v1 = –40 V, v2 = 57.14 V, v3 = 200 V
-40.0000
P.P.3.11
57.1430
200.0000
The schematic is shown below. After saving it, it is simulated by choosing Analysis/Simulate. The results are shown on the IPROBES. i 1 = –428.6 mA, i2 = 2.286 A, i3 = 2 A
-4.286E-01
2.286E+00 2.000E+00
P.P.3.12
For the input loop, 3
-5 + 10 x 10 IB + VBE + V0 = 0
(1)
For the outer loop, -V0 - VCE - 500 I0 + 12 = 0
(2)
But
V0 = 200 IE
(3)
Also
IC = βIB = 100 IB, α = β/(1 + β) = 100/(101) IC = αIE
IE = IC/(α) = βIB/(α)
IE = 100 (101/(100)) IR = 101 IB
(4)
From (1), (3) and (4), 10,000 IB + 200(101) IR = 5 - VBE IB =
5 − 0.7 10,000 + 20,000
= 142.38μA
V0 = 200 IE = 20,000 IB = 2.876 V From (2), -6
VCE = 12 - V0 - 500 IC = 9.124 - 500 x 100 x 142.38 x 10 VCE = 1.984 V {often, this is rounded to 2.0 volts}
P.P.3.13
20 k
i1 i0
iC 30 k
iB
+ 20 k
+
1V
+
-
v0
-
VBE -
+
-
22V
First of all, it should be noted that the circuit in the textbook should have a 22V source on the right hand side rather than the 10 V source.
iB =
1 − 0.7
B
30k
= 10μA, iC = βiB = 0.8 mA
i1 = iC + i0 Also,
-20ki0 – 20ki1 + 22 = 0
(1) i1 = 1.1 mA – i0
Equating (1) and (2), 1.1mA – i0 = 0.8 mA + i0 v0 = 20 ki0 = 20 x 0.15 = 3 V
i0 = 150 A
(2)
February 5, 2006
CHAPTER 4
P.P.4.1 6
i2 i1
+ 2
iS
By current division, i 2
2
=
is
2+6+4 2 v 0 = 4i 2 = i s 3 2 When is = 15A, v 0 = (15) = 10V 3 2 When is = 30A, v 0 = (30) = 20V 3
=
1 6
4
vo
is
P.P.4.2 12
v1
+
+
VS = 10 V
Let v0 = 1. Then i =
1 8
5
−
and v 1
giving vs = 2.5V. If vs = 10V, then v0 = 4V
=
1 8
(12
+
8)
=
2 .5
vo
8
P.P.4.3
Let v0 = v1 + v2, where v1 and v2 are contributions to the 20-V and 8-A sources respectively. 3
5
i
+ v1
+
2
−
(a)
3
i2
i1
5
+ v2
8A
2
(b)
To get v1, consider the curcuit in Fig. (a). (2 + 3 + 5)i = 20 v1 = 2i = 4V
i = 20/(10) = 2A
To get v2, consider the circuit in Fig. (b). i1 = i2 = 4A, v2 = 2i2 = 8V Thus, v = v1 + v2 = 4 + 8 = 12V
20 V
P.P.4.4 Let vx = v1 + v2, where v1 and v2 are due to the 10-V and 2-A sources
respectively. 20
v1
+
10 V
20
(a)
v2
2A
4
(b)
To obtain v1, consider Fig. (a).
0.1v1
+
10 − v1 20
=
v1
v1 = 2.5
4
For v2, consider Fig. (b). 2 + 0.1v2 +
0 − v2 20
=
vx = v1 + v2 = 12.5V
0.1v1
4
−
v2 4
v2 = 10
0.1v2
P.P.4.5
Let i = i1 + i2 + i3
where i1, i2, and i3 are contributions due to the 16-V, 4-A, and 12-V sources respectively.
2 6
2
8
6
8 4A
i1 16V
+
i2
−
(a)
(b) 6
2
8 i3 12V
+ −
(c)
For i1, consider Fig. (a), i 1
=
16 6+2+8
= 1A
For i2, consider Fig. (b). By current division, i 2
For i3, consider Fig. (c), i 3
=
2 2 + 14
( 4 ) = 0 .5
− 12
= −0.75A 16 Thus, i = i1 + i2 + i3 = 1 + 0.5 - 0.75 = 750mA
P.P.4.6
=
Combining the 6-Ω and 3-Ω resistors in parallel gives 6 3 =
6x3
= 2Ω . 9 Adding the 1-Ω and 4-Ω resistors in series gives 1 + 4 = 5 Ω. Transforming the left current source in parallel with the 2-Ω resistor gives the equivalent circuit as shown in Fig. (a).
2
5V −
+
io
+
10V
−
7
5
3A
7
5
3A
(a)
io 7.5A
2
(b) io 10.5A
(10/7)
7
(c)
Adding the 10-V and 5-V voltage sources gives a 15-V voltage source. Transforming the 15-V voltage source in series with the 2-Ω resistor gives the equivalent circuit in Fig. (b). Combining the two current sources and the 2-Ω and 5-Ω resistors leads to the circuit in Fig. (c). Using circuit division, 10 io
=
7 10 7
+7
(10.5) = 1.78 A
P.P.4.7
We transform the dependent voltage source as shown in Fig. (a). We combine the two current sources in Fig. (a) to obtain Fig. (b). By the current division principle, ix
=
5 15
(4 − 0.4i x )
ix = 1.176A
ix 4A
10
0.4ix
5
(a)
ix 4 – 0.4ix A
10
5
(b) P.P.4.8
To find R Th, consider the circuit in Fig. (a). 6
6
4
RTh
(a) 6 + 2A
6
2A
(b)
R Th
=
( 6 + 6) 4 =
12 x 4 18
=3
4
VTh
To find VTh, we use source transformations as shown in Fig. (b) and (c). 6
6 + 4
+
24 V
VTh
−
(c)
Using current division in Fig. (c), VTh
i=
=
4 4 + 12
VTh R Th
+1
( 24) = 6V
=
6 3 +1
=
1.5A
P.P.4.9 To find VTh, consider the circuit in Fig. (a).
5
6V
+ −
3
Ix
a
+
i2 i1
VTh
4 1.5Ix i2
i1 o
b (a)
0.5Ix
5
3
Ix
1.5Ix
i
a
+
4
(b)
−
b
1V
Ix = i2 i2 - i1 = 1.5Ix = 1.5i2
i2 = -2i1
(1)
For the supermesh, -6 + 5i1 + 7i2 = 0
(2)
From (1) and (2), i2 = 4/(3)A VTh = 4i2 = 5.333V To find R Th, consider the circuit in Fig. (b). Applying KVL around the outer loop, 5(0.5I x ) − 1 − 3I x = 0 1 i = − I x = 2.25 4 1 1 R Th = = = 444.4 m 2.25 i
P.P.4.10
Ix = -2
Since there are no independent sources, VTh = 0
4vx
10
+
−
+ vx
+ 5
15
vo
io
(a) 4vx
10
+
−
+ vx
15 +
5
vo
i
+ –
15io
(b)
To find R Th, consider Fig.(a). Using source transformation, the circuit is transformed to that in Fig. (b). Applying KVL, ).
But vx = -5i. Hence, 30i - 20i + 15io = 0 vo = (15i + 15io) = 15(-1.5io + io) = -7.5io R Th = vo/(io) = –7.5
10i = -15io
P.P.4.11 3
3
6
RN
(a)
3
5A
3
4A
(b)
From Fig. (a), R N = (3 + 3) 6 = 3
From Fig. (b), I N =
1 2
(5 + 4) = 4.5A
IN
P.P.4.12
2vx i
+
−
+
+ 6
vx
2
ix
vx
+
1V
−
(a)
2vx
+
−
+ 6
2
10 A
Isc
vx
(b)
To get R N consider the circuit in Fig. (a). Applying KVL, 6 i x But vx = 1, 6ix = 3 ix = 0.5 v i = i x + x = 0.5 + 0.5 = 1 2 1 R N = R Th = = 1 i
−
2v x
−
1
=
0
To find I N, consider the circuit in Fig. (b). Because the 2Ω resistor is shorted, v x = 0 and the dependent source is inactive. Hence, I N = isc = 10A. P.P.4.13
Fig. (a).
We first need to find R Th and VTh. To find R Th, we consider the circuit in
