Given, f=84 Hz .: W=2πf =2πT x 84 = 527.7876 for Capacitor (xc) = 1 1 jwc j527.7876 x 2×15-6 =-3947.35082 for Inductor (x) = jwL = j527.7876 x 50 x×10 Now Redraw the circuit in =j26-38942 given frequency 3640 -1947.35082 th V + {2k-2 √x3j26.3894 + 44233 Vo 1VX Apply Nodal √x -36400+ Vx Vx Vx-Vo + + 2000 j26.3894 j947.3508 360° = x[ Vx [2000 4 + j26.3894 + 0947.3505 0947.3500 == Apply Nodal on Vo, 3610° = 10.036842 L-89.2224) Vx+ Vo 0947-3508 V₂ Vo-Vx + 0.1Vx+ 4000 947.3508 4000 j947. -]+ Vx [0.1+ 1947. 308 13508 / +1 (0.001085 £76.6757) №6 + ·(0. from quation ①and ② 1947.3508 =O • 100006 L_0.605°) Vx=0 =O Vx=372.4317 L9.174567° Volts 16=34334.5034 2111.894° VOHS LTspice XVII - [Draft2] File Edit Hierarchy View Simulate Tools Window Help Aa .op I1 AC 36 ← SINE(0 36 84 0.0119) .ac lin 1 383,2743 383,2743 This is ground. ○ Ara C1 2μ 0.1 Vx R1 Vx 2k 50m G1 + > 0 0 Vo 4k х 5 14:38 9.11.2024
Given, f=84 Hz .: W=2πf =2πT x 84 = 527.7876 for Capacitor (xc) = 1 1 jwc j527.7876 x 2×15-6 =-3947.35082 for Inductor (x) = jwL = j527.7876 x 50 x×10 Now Redraw the circuit in =j26-38942 given frequency 3640 -1947.35082 th V + {2k-2 √x3j26.3894 + 44233 Vo 1VX Apply Nodal √x -36400+ Vx Vx Vx-Vo + + 2000 j26.3894 j947.3508 360° = x[ Vx [2000 4 + j26.3894 + 0947.3505 0947.3500 == Apply Nodal on Vo, 3610° = 10.036842 L-89.2224) Vx+ Vo 0947-3508 V₂ Vo-Vx + 0.1Vx+ 4000 947.3508 4000 j947. -]+ Vx [0.1+ 1947. 308 13508 / +1 (0.001085 £76.6757) №6 + ·(0. from quation ①and ② 1947.3508 =O • 100006 L_0.605°) Vx=0 =O Vx=372.4317 L9.174567° Volts 16=34334.5034 2111.894° VOHS LTspice XVII - [Draft2] File Edit Hierarchy View Simulate Tools Window Help Aa .op I1 AC 36 ← SINE(0 36 84 0.0119) .ac lin 1 383,2743 383,2743 This is ground. ○ Ara C1 2μ 0.1 Vx R1 Vx 2k 50m G1 + > 0 0 Vo 4k х 5 14:38 9.11.2024
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter23: Resistive-inductive-capacitive Series Circuits
Section: Chapter Questions
Problem 1PA: You are an electrician working in a plant. A series resonant circuit is to be used to produce a high...
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Could you please edit the circuit with the solution so that it is displayed on LTSpice as in the attachment? So, simulation on LTSpice is required.
![Given, f=84 Hz
.: W=2πf =2πT x 84 = 527.7876
for Capacitor (xc) = 1
1
jwc
j527.7876 x 2×15-6
=-3947.35082
for Inductor (x) = jwL = j527.7876 x 50 x×10
Now Redraw the circuit in
=j26-38942
given frequency
3640
-1947.35082
th
V
+
{2k-2 √x3j26.3894
+
44233 Vo
1VX
Apply Nodal
√x
-36400+
Vx
Vx
Vx-Vo
+
+
2000 j26.3894 j947.3508
360° = x[
Vx [2000
4
+
j26.3894
+
0947.3505
0947.3500
==
Apply Nodal on Vo,
3610° = 10.036842 L-89.2224) Vx+
Vo
0947-3508
V₂
Vo-Vx
+ 0.1Vx+
4000
947.3508
4000
j947.
-]+ Vx [0.1+ 1947. 308
13508 / +1
(0.001085 £76.6757) №6 +
·(0.
from quation ①and ②
1947.3508
=O
• 100006 L_0.605°) Vx=0
=O
Vx=372.4317 L9.174567° Volts
16=34334.5034 2111.894° VOHS](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb42373c1-1a13-480b-8fa8-c5f396e48c85%2F207e50b1-d099-4369-8153-6e7309ecedf7%2Fim78zsw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given, f=84 Hz
.: W=2πf =2πT x 84 = 527.7876
for Capacitor (xc) = 1
1
jwc
j527.7876 x 2×15-6
=-3947.35082
for Inductor (x) = jwL = j527.7876 x 50 x×10
Now Redraw the circuit in
=j26-38942
given frequency
3640
-1947.35082
th
V
+
{2k-2 √x3j26.3894
+
44233 Vo
1VX
Apply Nodal
√x
-36400+
Vx
Vx
Vx-Vo
+
+
2000 j26.3894 j947.3508
360° = x[
Vx [2000
4
+
j26.3894
+
0947.3505
0947.3500
==
Apply Nodal on Vo,
3610° = 10.036842 L-89.2224) Vx+
Vo
0947-3508
V₂
Vo-Vx
+ 0.1Vx+
4000
947.3508
4000
j947.
-]+ Vx [0.1+ 1947. 308
13508 / +1
(0.001085 £76.6757) №6 +
·(0.
from quation ①and ②
1947.3508
=O
• 100006 L_0.605°) Vx=0
=O
Vx=372.4317 L9.174567° Volts
16=34334.5034 2111.894° VOHS
![LTspice XVII - [Draft2]
File Edit Hierarchy View Simulate Tools Window Help
Aa .op
I1
AC 36
←
SINE(0 36 84 0.0119)
.ac lin 1 383,2743 383,2743
This is ground.
○ Ara
C1
2μ
0.1 Vx
R1
Vx
2k
50m
G1
+
>
0
0
Vo
4k
х
5
14:38
9.11.2024](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb42373c1-1a13-480b-8fa8-c5f396e48c85%2F207e50b1-d099-4369-8153-6e7309ecedf7%2Fgokr6u_processed.png&w=3840&q=75)
Transcribed Image Text:LTspice XVII - [Draft2]
File Edit Hierarchy View Simulate Tools Window Help
Aa .op
I1
AC 36
←
SINE(0 36 84 0.0119)
.ac lin 1 383,2743 383,2743
This is ground.
○ Ara
C1
2μ
0.1 Vx
R1
Vx
2k
50m
G1
+
>
0
0
Vo
4k
х
5
14:38
9.11.2024
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