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February 5, 2006
CHAPTER 10
P.P.10.1 10 sin( 2 t) ⎯⎯→ 10 ∠ 0 °, ω= 2
2 H ⎯⎯→ jωL=j 4
0. 2 F =
ω
Hence, the circuit in the frequency domain is as shown below.
At node 1, 2 - j2.
V 1 V 1 − V 2
100 = ( 5 +j 4 ) V 1 (^) −j 4 V 2 (1)
At node 2, 4
j 4 - j2.
V 2 (^) V 1 V 2 V x− V 2
= where V x (^) = V 1
- j2.5 j 4 ( ) 2. 5 ( 3 ) 2 1 2 1 2
V = V − V + V − V
0 = - ( 7. 5 +j 4 ) V 1 (^) +( 2. 5 +j 1. 5 ) V 2 (2)
Put (1) and (2) in matrix form.
5 j 4 - j
2
1
V
V
where Δ =( 5 +j 4 )( 2. 5 +j. 15 )−(-j4)(-(7.5+j4))= 22. 5 −j 12. 5 = 25. 74 ∠-29.05°
- 5 j 12. 5
7.5 j4 5 j 4
- 5 j 1. 5 j
2
1
V
V
5 j 12. 5
5 j 1. 5 V 1
5 j 12. 5
5 j 4 V 2
- j2.5 Ω
Vx
10 ∠ 0 ° A (^) j 4 Ω
+ − 2 Ω 3Vx
V 1 V 2
In the time domain,
v 1 (t) = 11.32 sin(2t + 60.01 ° ) V
v 2 (t) = 33.02 sin(2t + 57.12 ° ) V
P.P.10.2 The only non-reference node is a supernode.
4 j 4 - j 2
15 V 1 V 1 V 2 V 2
15 − V 1 (^) =-j V 1 +j 4 V 2 + 2 V 2
15 = ( 1 −j) V 1 (^) +( 2 +j 4 ) V 2 (1)
The supernode gives the constraint of
V 1 = V 2 + 20 ∠ 60 ° (2)
Substituting (2) into (1) gives
15 =( 1 −j)( 20 ∠ 60 °)+( 3 +j 3 ) V 2
3 j 3
15 ( 1 j)( 20 60 ) V 2
V 1 = V 2 + 20 ∠ 60 °=(- 3. 272 +j 0. 8327 )+( 10 +j 17. 32 )
V 1 = 6. 728 +j 18. 154
Therefore, V 1 = 19.36 ∠ 69.67 ° V , V 2 = 3.376 ∠ 165.7 ° V
P.P.10.3 Consider the circuit below.
For mesh 1, ( 8 − j 2 +j 4 ) I 1 −j 4 I 2 = 0
( 8 +j 2 ) I 1 (^) =j 4 I 2 (1)
8 Ω j 4 Ω
- j 2 Ω
I 1 I 2
−
10 ∠ 30 ° V
I 3
2 ∠ 0 ° A
Eliminating I 3 from (1) and (2)
( 15 − j 4 ) I 1 (^) +(- 5 +j 4 ) I 2 = 60 (4)
(- 5 + j 4 ) I 1 (^) +( 5 −j 2 ) I 2 =- 10 +j (^12) (5)
From (4) and (5),
15 j 4 - 5 j
2
1
I
I
Δ = 58 j 10 58. 86 - 9.
15 j 4 - 5 j
Δ = 298 j 20 298. 67 - 3.
60 - 5 j
1
Thus, = Δ
1 I (^) o I 1 5.074 ∠ 5.94 ° A
P.P.10.5 Let , where and are due to the voltage source and
current source respectively. For consider the circuit in Fig. (a).
" o
' I (^) o = I o+ I
' I (^) o
" I o
' I o
For mesh 1, ( 8 + j 2 ) I 1 −j 4 I 2 = 0
I (^) 2 =( 0. 5 −j 2 ) I 1 (1)
For mesh 2, ( 6 + j 4 ) I (^) 2 −j 4 I 1 − 10 ∠ 30 °= 0 (2)
Substituting (1) into (2),
( 6 + j 4 )( 0. 5 −j 2 ) I 1 −j 4 I 1 = 10 ∠ 30 °
- 08 j 0. 556 11 j 14
1
' o = + −
I = I =
- j 2 Ω 6 Ω
I 1 10 ∠ 30 ° V
−
j 4 Ω I^2
I (^) o
'
(a)
For consider the circuit in Fig. (b).
" I o
Let Z 1^ =^8 −j^2 Ω, = + Ω
= = 1. 846 j 2. 769 6 j 4
j 24 Z 2 6 ||j 4
- 4164 j 0. 53
- 846 j 0. 77
( 2 )( 1. 846 j 2. 769 ) ( 2 ) 1 2
" 2 o = +
Z Z
Z
I
Therefore, 0. 4961 j 1. 086
" o
' I o = I o+ I = +
I (^) o = 1.1939 ∠ 65.45 ° A
P.P.10.6 Let , where is due to the voltage source and is due to
the current source. For , we remove the current source.
" o
' v (^) o = vo+v
' v (^) o
" vo
' vo
30 sin( 5 t) ⎯⎯→ 30 ∠ 0 °, ω= 5
j C
0. 2 F = =
ω
1 H ⎯⎯→ jωL=j( 5 )( 1 )=j 5
The circuit in the frequency domain is shown in Fig. (a).
8 Ω j 4 Ω
- j 2 Ω
2 ∠ 0 ° A
I (^) o
"
(b)
Vo
'
−
30 ∠ 0 ° V
- j Ω j 5 Ω
(a)
P.P.10.7 If we transform the current source to a voltage source, we obtain the
circuit shown in Fig. (a).
(a)
VS
−
4 Ω - j 3 Ω 2 Ω j Ω
- j 2 Ω
Io
j 5 Ω
V s = I s Z s=(j 4 )( 4 −j 3 )= 12 +j 16
We transform the voltage source to a current source as shown in Fig. (b).
Let Z = 4 −j 3 + 2 +j= 6 −j 2. Then, 1. 5 j 3
6 j 2
s^12 j^16 s = + −
Z
V
I.
Io
IS j 5 Ω
- j 2 Ω - j 2 Ω
(b)
Note that ( 1 j) 3
6 j 3
( 6 j 2 )(j 5 ) || j 5 = +
Z =.
By current division,
( 1. 5 j 3 )
( 1 j) ( 1 j 2 ) 3
( 1 j) 3
o +
I =
13 j 4
20 j 40 I o
I (^) o = 3.288 ∠ 99.46 ° A
P.P.10.8 When the voltage source is set equal to zero,
Z th = 10 +(-j 4 )||( 6 +j 2 )
6 - j
(-j4)(6 j2) th^10
Z = +
Z th = 10 + 2. 4 −j 3. 2
Z (^) th = 12.4 – j3.2 Ω
By voltage division,
6 j 2
(-j 4 )( 30 20 ) ( 30 20 ) 6 j 2 j 4
V =
V th
V th = 18.97 ∠ -51.57 ° V
P.P.10.9 To find V th, consider the circuit in Fig. (a).
At node 1, 8 j 4
4 j 2
− V V V
- ( 2 +j) V 1 (^) = 50 +( 1 −j 0. 5 )( V 1 − V 2 )
50 = ( 1 −j 0. 5 ) V 2 (^) −( 3 +j 0. 5 ) V 1 (1)
At node 2, (^0) 8 j 4
1 2 o =
V V
V ,^ where^ V o^ = V 1 − V 2.
Hence, the equation for node 2 becomes
V 1
V 2
4 – j
8 + j
a
b
0.2Vo
(a)
+ V −
o
4 – j
8 + j
a
b
0.2Vo
(b)
−
+ V −
o
Is VS
P.P.10.10 To find Z N, consider the circuit in Fig. (a).
13 j
( 4 j 2 )( 9 j 3 ) N (^4 j^2 )||(^9 j^3 ) −
Z = + − =
Z (^) N = 3.176 + j0.706 Ω
To find , short-circuit terminals a-b as shown in Fig. (b). Notice that meshes 1 and 2
form a supermesh.
I N
For the supermesh, - 20 + 8 I (^) 1 +( 1 −j 3 ) I 2 −( 9 −j 3 ) I 3 = 0 (1)
Also, I (^) 1 = I 2 +j 4 (2)
For mesh 3, ( 13 − j) I (^) 3 − 8 I 1 −( 1 −j 3 ) I 2 = 0 (3)
Solving for I 2 , we obtain
9 j 3
50 j 62 I N I 2
I N = 8.396 ∠ -32.68 ° A
Using the Norton equivalent, we can find I oas in Fig. (c).
4 Ω j 2 Ω
8 Ω 1 Ω - j 3 Ω
ZN
a
b (a)
4 Ω j 2 Ω
8 Ω 1 Ω - j 3 Ω a
(b) b
I I^ N
2
−
20 ∠ 0 ° I 1
I 3
- j 4 Ω
IN
10 – j 5 Ω
ZN
(c)
Io
By current division,
176 j 4. 294
176 j 0. 706
10 j 5
N N
N o ∠ ° −
= I
Z
Z
I
I o
I (^) o = 1.971 ∠ -2.10 ° A
P.P.10.
× ×
ω
⎯⎯→ -j20k j( 5 10 )( 10 10 )
j C
10 nF 3 - 9 1
× ×
ω
⎯⎯→ -j10k j( 5 10 )( 20 10 )
j C
20 nF 3 - 9 2
Consider the circuit in the frequency domain as shown below.
As a voltage follower, V 2 (^) = V o
At node 1, 10 - j 20 20
2 V 1 (^) V 1 V o V 1 − V o
4 = ( 3 +j) V 1 (^) −( 1 +j) V o (1)
At node 2,
1 o o−
V − V V
V 1 (^) =( 1 +j 2 ) V o (2)
Substituting (2) into (1) gives
4 = j 6 V o or = ∠- 90 ° 3
V o
−
−
10 k Ω 20 k Ω
- j 20 k Ω - j 10 k Ω
Io
Vo
V 2
V 1
2 ∠ 0 ° V
P.P.10.13 The schematic is shown below.
Since. Setup/Analysis/AC Sweep as
Linear for 1 point starting and ending at a frequency of 447.465 Hz. When the schematic
is saved and run, the output file includes
ω= 2 πf= 3000 rad/s ⎯⎯→ f= 477. 465 Hz
Frequency IM(V_PRINT1) IP(V_PRINT1)
4.775E+02 5.440E-04 -5.512E+
Frequency VM($N_0005) VP($N_0005)
4.775E+02 2.683E-01 -1.546E+
From the output file, we obtain
V o = 0.2682∠-154.6° V and I (^) o =0.544∠-55.12° mA
Therefore,
v (^) o (t) = 0.2682 cos(3000t – 154.6 ° ) V
i (^) o (t) = 0.544 cos(3000t – 55.12 ° ) mA
P.P.10.14 The schematic is shown below.
We select ω = 1 rad/s and f = 0.15915 Hz. We use this to obtain the values of
capacitances, where C = 1 ωXc, and inductances, where L = XL ω. Note that IAC does
not allow for an AC PHASE component; thus, we have used VAC in conjunction with G
to create an AC current source with a magnitude and a phase. To obtain the desired
output use Setup/Analysis/AC Sweep as Linear for 1 point starting and ending at a
frequency of 0.15915 Hz. When the schematic is saved and run, the output file includes
Frequency IM(V_PRINT1) IP(V_PRINT1)
1.592E-01 2.584E+00 1.580E+
Frequency VM($N_0004) VP($N_0004)
1.592E-01 9.842E+00 4.478E+
From the output file, we obtain
V x = 9.842 ∠ 44.78 ° V and I (^) x= 2.584 ∠ 158 ° A
P.P.10.15 ⎟(^ × =
×
×
6
1
2 eq 10 10 10 10
C 1
R
R
C 1 )^ 10 μ F
P.P.10.16 If R = R 1 =R 2 = 2. 5 kΩ and C =C 1 =C 2 = 1 nF
π × ×
π
2 RC
f (^) o 3 - 9 63.66 kHz
