Download EET310 - Circuit Analysis Unit 4 Circuit Theorems in AC Analysis By

EET310 - Circuit Analysis
Unit 4 Circuit Theorems in AC Analysis
By: Steve Youchnik
Instructor: Dr. Setoodehnia
Online EET Department
ECPI University
Date: 7/20/13
ECPI’s Honor Pledge: I pledge to support the Honor System of ECPI. I will refrain from
any form of academic dishonesty or deception, such as cheating or plagiarism. I am
aware that as a member of the academic community it is my responsibility to turn in all
suspected violators of the honor code. I understand that any failure on my part to
support the Honor System will be turned over to a Judicial Review Board for
determination. I will report to a Judicial Review Board hearing if summoned.
Abstract
Although an AC circuit differs from a DC circuit, the same theorems that we used for DC
circuits can be applied to AC circuits. AC circuits used impedance rather than resistance to find
unknown values of a circuit. Thevenin’s theorem can be used when you open two terminals of a
given circuit and replace the components with its equivalent impedance. This lab is going to
show how these theorems apply to AC circuits.
Introduction
This lab consists of five total circuits, the first two show how frequency affects the maximum
overall power output as well as Thevenizing a circuit. The final three experiments are circuits
that may have a fault in them and my job is to find them using calculations and then using
Multisim to verify where the fault is.
Circuit E19-04
With this experiment I have to find Vth of the circuit with respect to RL.
Method
The first step is to remove RL from the circuit.
The voltage from A to B is the same as voltage across XL. This is determined using the voltagedivider method.
VL = (XL<90o/R1 + jKL)Vs
= (59<90o ohms/100 ohms + j50 ohms)
= (50<90o ohms/ 112<26.6o ohms)25<0o V = 11.2<63.4o V
Vth = VAB = VL = 11.2<63.4o V
Here are the measurements from Multisim.
Related Problem
Determine Vth if R1 is changed to 47 ohms.
Method
The first step is to remove RL from the circuit.
The voltage from A to B is the same as voltage across XL. This is determined using the voltagedivider method.
VL = (50 < 90o/47 + j50)25 < 0o
= √472 + 502<tan-1 (50/47) = 68.6 < 1.06o
= 50<90o/11.8 < 26.6o = .729 < 88.9o
= (.729 < 88.9o)(25 < 0o) = 18.2 < 88.9o V
Vth = VAB = VL = 18.2 < 88.9o V
Here are the measurements from Multisim
Conclusion
Finding Vth in an AC circuit is similar to finding the equivalent in a DC circuit but I believe that
it is more difficult. AC circuits involve imaginary numbers and to find the final answer, you have
to constantly convert from rectangular to polar form to make it easier on yourself. Since the
voltage from A to B is the same as voltage across XL I used the voltage-divider method to find
the voltage at L1. This circuit gave me a good understanding of Thevenin’s theorem.
Circuit E19-16
The purpose of this experiment is to find the frequency at which maximum power is provided
from a power amplifier. I will construct a graph of the given frequencies of 10 kHz, 30 kHz, 50
kHz, 80 kHz, and 100 kHz. Here is the circuit in Multisim.
Method
For f = 10 kHz
Xc = 1/2pifC = 1/2pi(10 kHz)(0.01 uF) = 1.59 kohms
XL = 2pifL = 2pi(10 kHz)(1 mH) = 62.8 ohms
Ztot = √(Rs + RL)2 + (XL – Xc)2 = √(20 ohms)2 + (1.53 kohms)2 = 1.53 kohms
I = Vs/Ztot = 10 V/1.53 kohms = 6.54 mA
PL = I2RL = (6.54 mA)2(10 ohms) = 428 uW
For f = 30 kHz
Xc = 1/2pifC = 1/2pi(30 kHz)(0.01 uF) = 531 ohms
XL = 2pifL = 2pi(30 kHz)(1 mH) = 189 ohms
Ztot = √(Rs + RL)2 + (XL – Xc)2 = √(20 ohms)2 + (342 ohms)2 = 343 ohms
I = Vs/Ztot = 10 V/343 ohms = 29.2 mA
PL = I2RL = (29.2 mA)2(10 ohms) = 8.53 mW
For f = 50 kHz
Xc = 1/2pifC = 1/2pi(50 kHz)(0.01 uF) = 318 ohms
XL = 2pifL = 2pi(50 kHz)(1 mH) = 314 ohms
Ztot = √(Rs + RL)2 + (XL – Xc)2 = √(20 ohms)2 + (4 ohms)2 = 20.4 ohms
I = Vs/Ztot = 10 V/20.4 ohms = 490 mA
PL = I2RL = (490 mA)2(10 ohms) = 2.40 W
For f = 80 kHz
Xc = 1/2pifC = 1/2pi(80 kHz)(0.01 uF) = 199 ohms
XL = 2pifL = 2pi(80 kHz)(1 mH) = 503 ohms
Ztot = √(Rs + RL)2 + (XL – Xc)2 = √(20 ohms)2 + (304 ohms)2 = 305 ohms
I = Vs/Ztot = 10 V/305 ohms = 32.8 mA
PL = I2RL = (32.8 mA)2(10 ohms) = 10.8 mW
For f = 100 kHz
Xc = 1/2pifC = 1/2pi(100 kHz)(0.01 uF) = 159 ohms
XL = 2pifL = 2pi(80 kHz)(1 mH) = 628 ohms
Ztot = √(Rs + RL)2 + (XL – Xc)2 = √(20 ohms)2 + (469 ohms)2 = 469 ohms
I = Vs/Ztot = 10 V/469 ohms = 21.3 mA
PL = I2RL = (21.3 mA)2(10 ohms) = 4.54 mW
Here are my readings from Multisim
For f = 10 kHz
For f = 30 kHz
For f = 50 kHz
For f = 80 kHz
For f = 100 kHz
Maximum Power Transfer
Maximum Power Transfer
3
2.5
2.4
PL (W)
2
1.5
1
0.5
0
0.000428
10k
0.00853
30k
50k
0.0108
80 k
0.00454
100k
Frequency
Related Problem
If R = 47 ohms and C = 0.022 uF in a series circuit, what is that complex conjugate of the
impedance at 100 kHz?
Method
Xc = 1/2pi(100 kohms)(0.022 uF) = 72.3
Ztot = 47 + 72.3j
Complex conjugate = 47 + 72.3j
Conclusion
This experiment displayed how frequency in an AC circuit affects the maximum power output.
This proves that the load impedance has to be the complex conjugate of the circuit’s output
impedance to get maximum power transfer out of it. As the impedance of the capacitor and the
inductor became closer, the more power was seen at PL. The graph shows how the frequency
affected the overall power output in watts and I also verified it with pictures from Multisim.
Circuit P09-18
Determine the fault if there is one. Here is the circuit from Multisim
First I have to apply power to the circuit.
I will take readings to see if there is a fault inside of the circuit.
There is no current going through R2. As you can see all of the other components seem to be
working fine because there is current flow through each of them. I am going to replace R2 and
take more readings to see if the problem is fixed.
As you can see I replaced R2 and there is now current flowing through it.
Conclusion
The first problem for this circuit is that there was no power supply hooked up to it. Once I
hooked up a power supply I realized there was no current flowing through R2 so therefore it had
an open. As soon as I replaced R2 the circuit started to work fine.
Circuit P19-19
I need to find the fault in the circuit if there is one. Here is a picture of the circuit.
The first step is to connect a power source to the circuit and then I will find the total current.
C1 = 1/2pi(60)(1e – 9) = 2.65 Mohms
C2 = 1/2pi(60)(1e – 7) = 26.5 kohms
Z = √1002 + 265225822 = 2.65 Mohms
I1 = 120/26526 = 4.52 mA
I2 = 120/2655609 = 45.24 uA
Itot = I1 + I2 = 4.53 mA
As you can see the values do not add up to our calculations. I am going to isolate each
component and see if I can find the problem.
As you can see it looks as if there is a problem with C2. I am going to replace it and see if that
gets our readings to where we want them.
Those numbers look much better. Our problem was C2.
Conclusion
Again in this circuit there was no power connected to the circuit so that was where I started.
After the calculations were made I made the measurements in the circuit and they did not match
up. I dug further into the circuit by isolating each component and testing them. After further
tests, I found that C2 was faulty and I replaced it and the circuit went back to normal operation.
Circuit P19-20
Determine if there is a fault in the circuit, if so, find it and fix it. Here is the circuit I will be
working on.
Method
The first step is to connect a power supply
The impedance of the capacitors should be 1/2pi(60)(3.3e – 8) = 80.4 kohms since they are both
the same in value.
I will test the circuit and see what values I get.
As you can see, the amperage is not getting through the circuit. There is a problem somewhere. I
am going to isolate the components and see if I can find the problem.
These readings show that C1 is open. I am going to replace it and measure the circuit.
Now my numbers match up because I replaced the faulty C1 capacitor with a new one.
Conclusion
The first problem that I found was connecting a power source to the circuit because there was not
one there to begin with. After that I took readings of the circuit and found that C1 was faulty so I
replaced the component and the circuit went back to normal operation. These labs helped me
gain understanding of using Thevenin’s theorem and the maximum power theorem. It also
furthered my understanding of how these circuits should work so when I am troubleshooting
them I know what to look for.
References
Floyd, T. (2008). Circuit Theorems and Conversions Electric Circuits. (2nd ed. custom). Upper
Saddle River, NJ: Pearson Education.
NI Multisim 11.0 (2010) National Instruments. [Computer software].