Download Solution Set 7

Solution Set 7
7.1 Determine the phase angles by which v1(t) leads i1(t) and v1(t) leads i2(t), where:
v1(t) = 6 sin (377t + 25o) V
i1(t) = 0.04 cos (377t – 10o) A
i2(t) = -0.2 sin (377t – 75o) A
Solution:
i1(t) = 0.04 cos (377t – 10o) A = 0.04 sin (377t – 10o + 90o) A = 0.04 sin (377t + 80o) A
It means that v1(t) leads i1(t) by 25o – 80o = -55o
i2(t) = -0.2 sin (377t – 75o) A = 0.2 sin (377t – 75o + 180o) A = 0.2 sin (377t + 105o) A
It means that 25o – 105o = -80o
7.2 Calculate i(t), the time-domain current in the resistor in the following circuit if the
input voltage is:
a) v1(t) = 12 cos (377t + 180o) V
b) v2(t) = 16 cos (377t + 45o) V
Solution:
This is a purely resistive network: i(t) and v(t) are in phase
1) i1(t) = v1(t) / R =
2) i1(t) = v2(t) / R =
!" !"#!!""!!!"#! !
!
!" !"#!!""!!!"! !
!
A = 3 cos(377t + 180o) A
A = 4 cos(377t + 45o) A
7.3 Calculate i(t), the time-domain current in the capacitor shown in the following circuit
if the voltage is:
a) v1(t) = 8 cos (377t – 30o) V
b) v2(t) = 4 cos (377t + 60o) V
Solution:
This is a purely capacitive network: i(t) leads v(t) by 90o.
! ! !!
!" !
, however, this equation can be rewritten as I= !"#$ (eq. 7.29 Irwin)
!"
1) I1 = !"# !" ! !"! !, where v1(t) = 8 cos (377t – 30o) V
I1 = 377*1440*10-6 ! 90o * !" ! !"! A = 4.34 ! 60o A
i1(t) = 4.34 cos (377t + !"! )
2) I2 = !"# !"!!"! !, where v2(t) = 4 cos (377t + 60o) V
I2 = 377*1440*10-6 ! 90o *!"!!"! A =2.17!!!"#!
i2(t) = 2.17 cos (377t + !"#! )
7.4 Calculate i(t), the time-domain current in the inductor in the following circuit for the
following voltage inputs:
a) v1(t) = 24 cos (377t + 12o) V
b) v2(t) = 18 cos (377t + 48o) V
Solution:
This a purely inductive network: i(t) lags v(t) by 90o
!" !
! ! ! !! !" This can be simplified to V= !"#$
1) v1(t) = 24 cos (377t + 12o) V
!!
!!
!"#!"!
!
! !" !"! ! !"! ! !" ! !"!
!"# !"" ! !"!!" ! !"!! !!"!
i1(t) = 6 cos (377t - !"! ) A
2) v2(t) = 18 cos (377t + 48o) V
!!
!!
!"#!"!
!
! !!!! !"! ! !"! ! !!!! ! !"!
!"# !"" ! !"!!" ! !"!! !!"!
i2(t) = 4.5 cos (377t - !"!) A
7.5 Find the frequency-domain impedance, Z, of the following network:
Solution:
This is a series RLC network:
Impedance Z Z R Z C Z L
Z
ZR ZC ZL
Z
3.04‘80.54$
0.5: j 2: j 5:
0.5 j 3(:)
0.52 32 ‘ tan 1 (
3
)
0.5
3.04‘80.54$
7.6 Find the frequency-domain impedance Z of the following network:
Solution:
This is a Parallel RLC network
1
1
1
Impedance Z (
) 1
ZR ZC ZL
1
1
1
1
1 j 0.25 j 0.5 1 j 0.25
Z 1: j 4: j 2:
1
1
1
Z
:
:|
: 0.97‘14.04$ :
$
2
1
1 j 0.25
1.03‘ 14.04
1 (0.25) ‘ tan (0.25 / 1)
Z
0.97‘14.04$ : or Z 1 Y | 1.03‘ 14.04$
7.7 The voltages vR (t ) and v L (t ) in the following circuit can be redrawn as phasors in a phasor
diagram. Use a phasor diagram to show that vR (t ) v L (t ) vs (t ) where
vs (t ) 120 cos(377t )V
Solution:
Vs
120‘0q
I
11.79‘ 10.67 $.
R jZL 10 j (377 u 5m)
VL jZLI 377 * 5 *103 ‘90$ *11.79‘ 10.67$ 22.22‘79.33$
v L (t ) 22.22 cos(377t 79.33$ )V
Solution Set 7 Rev05_2011
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'‘ tan 1 (
VL
22.22
) tan 1
10.67
117.9
VR
VS 2 VR 2 V L 2 117.9 2 22.22 2 14394.14
VS
‘VS
120
‘VR '‘ 10.67 10.67 0
7.8 The currents i R (t ) and i C (t ) in the following circuit can be drawn as phasors in a phasor
diagram. Use such a phasor diagram to show that i C (t ) i R (t ) i s (t ) where:
is (t )
20 cos(377t 30$ ) A
Solution:
This is a parallel network:
1
1
Zc
j13.26 13.26‘ 90
jZ C j *377*200*106
Using a current divider,
ZC
13.26‘ 90
IR IS *
20‘30
ZR ZC
2 j13.26
13.26‘ 90
13.26
20
‘30 90 81.42
13.41‘ 81.42
13.41
19.78‘21.42
ZR
2
20‘30
IC IS *
13.41‘ 81.42
ZR ZC
20‘30
20
Is
2
‘30 81.42
13.41
Ic 2 I R 2
19.782 2.982
Solution Set 7 Rev05_2011
20;
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'L
‘I S
tan 1 (
IC
2.98
) tan 1 (
) 8.57
19.78
IR
‘I R 'L
21.42$ 8.57$ | 29.9 | 30$
7.9 The currents i L (t ) and i C (t ) of the inductor and capacitor, respectively, in the following
circuit can be drawn as phasors in a phasor diagram. Show in the phasor diagram that
i L (t ) i C (t ) i S (t ) where:
vs (t ) 10 cos(103 t 30$ )V
Solution:
Zx is a parallel network Z x Z L || Z C
1
Z C 1/ jZ C
j103
6
3
j10 *10
ZL
jZ L
j103 (10*103 )
j10
Z C * Z L j103 * j10
j10.1
Z L Z C j103 j10
Voltage divider:
ZX
j10.1
10‘30
Va Vs
1 j10.1
Z X ZR
Zx
10‘30
10.1‘90
10.15‘84.35
9.95‘35.65
Solution Set 7 Rev05_2011
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V
ZL
IL
9.95‘35.65
j10
Ic V / Z C
9.95‘35.65
0.995‘ 54.35 ( A)
10‘90
1
1
9.95‘35.65 (
) 9.95‘35.65 ( 3
)
3
j10
10 ‘ 90
9.95*103 ‘125.65
Is 0.995‘ 54.35 9.95*103 ‘125.65
0.995‘ 54.35 9.95*103 ‘ 54.35
0.985‘ 54.35
Verification:
Vs 10‘30 10 cos 30 j10sin 30 8.66 j 5.00
Va 9.95‘35.65 8.09 j 5.80
Vs Va 8.66 j 5.0 _(8.09 j 5.8)
Is
R
1
0.982‘ 54.53
0.57 0.8 j
Isphasor = Iscalculated
7.10. Repeat 7.9 for vs (t ) 10 cos(10 4 t 30$ )V VXUSULVLQJUHVXOWFDOOHGµUHVRQDQFH¶
Solution:
ZC
1/ jZ C
ZL
jZL
1
j10 *10 6
4
j10 4 (10 *10 3 )
Solution Set 7 Rev05_2011
j100
j100
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j100 * j100 10 4
j f
j100 j100
0
Zx
1
Voltage divider: Va Vs
Vs
Zx ZR
1 Z R / Z x
Va Vs
LC is in resonance
ZX
Zc * ZL
ZC ZL
Vs (1/(1 0))
This is very large impedance (Zx) ±it is like an open circuit
Since Va Vs there is no current through the resistor
10‘30$ 10‘30$
I L Vs / Z L
0.1‘ 60$
j100
100‘90$
IC
Vs / Z C
10‘30$
j100
10‘30$
100‘ 90$
0.1‘ 120$
Vs Va
( A) 0 / R( A) 0( A)
R
Note: the currents are identical in magnitude, but 180º out of phase. This means the Is is zero.
This is the same value as the current which passes through the resistor.
The currents flow between L and C, but never escape the resonant network. The energy is
exchanged between the electric field of the capacitor and the magnetic field of the inductor.
Is
Solution Set 7 Rev05_2011
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