A3) Simulate the circuit with the above values and obtain the plots for VF, VC, VR and I of the circuit. Obtain the plots using the same reference system so that the phase difference between the plots can be observed. vf f=4 kHz R=1kΩ ww C=10 nF Figure 1. Measurement circuit #1. ORC
A3) Simulate the circuit with the above values and obtain the plots for VF, VC, VR and I of the circuit. Obtain the plots using the same reference system so that the phase difference between the plots can be observed. vf f=4 kHz R=1kΩ ww C=10 nF Figure 1. Measurement circuit #1. ORC
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.7P: Let a 100V sinusoidal source be connected to a series combination of a 3 resistor, an 8 inductor,...
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I need help with part A3 please :)
.
Transcribed Image Text:A3) Simulate the circuit with the above values and obtain the plots
for VF, VC, VR and I of the circuit. Obtain the plots using the same
reference system so that the phase difference between the plots
can be observed.
Vf
f=4 kHz
R=1kΩ
C=10 nF
Figure 1. Measurement circuit #1.
X
ORC
Transcribed Image Text:A1) Analyze the circuit in Figure 1, consider R = 10 KQ and C= 10 nF. Obtain the
mathematical equation for the equivalent impedance seen by the generator as a function
of frequency. By voltage division obtain the mathematical relationship for the voltage
across the capacitor and resistor as a function of VF and frequency. Perform all the circuit
analysis (voltages, impedances, currents, waveforms).
Equivalent impedance,
Zeq = R + jaC = 10³ + jx8000xx10x10-⁹⁹
Zeq
1000 -j3978. 87,
=
Voltage across resistor,
1000
1000-j3978.87 X V₁,
R
VR = Zea
⇒ VR =
1000
1000-j3978.87
1+j3.978
16.831
⇒ VR =
1+j3.978
⇒ VR =
16.831
V₁ = VR + VC,
⇒ Vc = V₁ - VR.
⇒ Vc = Vj -
⇒ Vc =
⇒ Vc =
Current, I =
108
> /2xX5\
x Vf,
X
1+j3.978
16.831
x V₁,
16.831-1-j3.978
16.831
15.831-13.978
16,831
VC
Zea
|XC| = R₁
⇒ | jx2xxxc| = R₁
|jx2xfx10x10-⁹|
= 1000,
=
1000+j3978.87
1000+j3978.87
X V₁.
X Vf,
A2) For the conditions of the previous point, deduce the value of the frequency that meets
|XC| = R.
x V₁,
V₁
1000-J3978.87
= 1000,
10⁰
⇒ f =
2000
⇒ f = 1.57 x 105 Hz
x Vf,
= 157079.63 Hz
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