Circuit olrig.f'ceoltage Vs(1)!.45 M-p1ase withi lfe su(1)sab4 mH108.33 iFi(t)50 ΩFigure P7.47: Circuit for Problem 7.47Ds(t)0.2 H0.2 mF7.48 Determine the Thévenin equivalent of the circuit inFig. P7.48 at terminals (a, b), given thatb0.2 mFUs(t)12 cos 2500t VFigure P7.45: Circuit for Problem 7.450.5 cos(2500r - 30°) A.is(t)450CHAPTER 7AC ANALYSIS2 kΩ3 k25 QW| IL4 mH4 mH15/0 Vj6 kZL)Us(t)is):80 μF10ΩbFigure P7.51: Circuit for Problem 7.51Figure P7.48: Circuit for Problem 7.48.Sections 7-7 and 7-8: Phasor Diagrams and Phase Shifters000

Question
Asked Nov 13, 2019
111 views

Determine the Th´evenin equivalent of the circuit in
Fig. P7.48 at terminals (a, b), given that
υs(t) = 12 cos 2500t V,
is(t) = 0.5 cos(2500t − 30◦
) A.

 

 

Circuit olrig.
f'ce
oltage Vs(1)!
.45 M-p1ase withi lfe su
(1)sa
b
4 mH
108.33 iF
i(t)
50 Ω
Figure P7.47: Circuit for Problem 7.47
Ds(t)
0.2 H
0.2 mF
7.48 Determine the Thévenin equivalent of the circuit in
Fig. P7.48 at terminals (a, b), given that
b
0.2 mF
Us(t)12 cos 2500t V
Figure P7.45: Circuit for Problem 7.45
0.5 cos(2500r - 30°) A.
is(t)
450
CHAPTER 7
AC ANALYSIS
2 kΩ
3 k2
5 Q
W
| IL
4 mH
4 mH
15/0 V
j6 k
ZL
)Us(t)
is)
:80 μF
10Ω
b
Figure P7.51: Circuit for Problem 7.51
Figure P7.48: Circuit for Problem 7.48.
Sections 7-7 and 7-8: Phasor Diagrams and Phase Shifters
000
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Image Transcriptionclose

Circuit olrig. f'ce oltage Vs(1)! .45 M-p1ase withi lfe su (1)sa b 4 mH 108.33 iF i(t) 50 Ω Figure P7.47: Circuit for Problem 7.47 Ds(t) 0.2 H 0.2 mF 7.48 Determine the Thévenin equivalent of the circuit in Fig. P7.48 at terminals (a, b), given that b 0.2 mF Us(t)12 cos 2500t V Figure P7.45: Circuit for Problem 7.45 0.5 cos(2500r - 30°) A. is(t) 450 CHAPTER 7 AC ANALYSIS 2 kΩ 3 k2 5 Q W | IL 4 mH 4 mH 15/0 V j6 k ZL )Us(t) is) :80 μF 10Ω b Figure P7.51: Circuit for Problem 7.51 Figure P7.48: Circuit for Problem 7.48. Sections 7-7 and 7-8: Phasor Diagrams and Phase Shifters 000

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Expert Answer

Step 1

The given circuit is shown below:

4 mH
4 mH
'z,"
vs(C)
: 80 F
10Ω
(.2500)
V 12 cos 2500V
I 0.5cos (2500-30° )A
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4 mH 4 mH 'z," vs(C) : 80 F 10Ω (.2500) V 12 cos 2500V I 0.5cos (2500-30° )A

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Step 2

From the circuit:

Z jL
z, j2500(4x103
Z j10n
Similarly
Z j100
and
1
joC
1
Z3
j2500(80x10
Z j5
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Image Transcriptionclose

Z jL z, j2500(4x103 Z j10n Similarly Z j100 and 1 joC 1 Z3 j2500(80x10 Z j5

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Step 3

Now, converting the T-network into the Pie-...

3
Z2
2
1
Z,Z Z.Z Z,Z
Z2
Z3
Similarly
Z.Z2 Z,Z Z
Z:3
Z2
Z,Z ZZ ZZ
ZZ
N
help_outline

Image Transcriptionclose

3 Z2 2 1 Z,Z Z.Z Z,Z Z2 Z3 Similarly Z.Z2 Z,Z Z Z:3 Z2 Z,Z ZZ ZZ ZZ N

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