## PHYS345 Electricity and Electronics

Student Requests for Clarification

My responses are italicized. I sometimes edit slightly, truncate, or abridge the submissions. Listed in reverse chronological order. Need a clarification?

1. The complex conjugate of the voltage is just the voltage with the sign of the angle switched, right? [12/10 1:15am]

Correct! In rectangular representation, the complex conjugate is found by switching j with -j throughout. In polar representation, change the sign of the phase angle. [12/10]

2. How do you find the power dissipated by the motor in problem 3 from the first exam? [12/9 9:46pm]

Find the impedance for the motor (inductance and resistance in series). Then use the applied voltage and Ohm's law to find the current through the motor. The complex power associated with the motor is the the product of the voltage and the complex conjugate of the current. [12/10]

3. I just wanted to know how much we need to know about dc motors for the final exam. [12/9 8:45pm]

I announced at the review session that there would be no detailed coverage of motors on the final exam this year. [12/10]

1. What is the difference between Zeq and Zeff?

2. What do you do if you are given an ac RLC circuit and the value of w is not given? Is there a way to calculate it?

3. In a lot of your problems we have to convert from rectangular to polar, to get a phase angle. We keep getting the right angle but with the wrong sign, and we can't figure out this sign problem. [10/26 9:31pm]

1. No difference; they are synonymous.

2. If w is not specifically stated, then you may have been given specific values of reactance; that is, j5 ohm rather than some value of inductance and also a frequency from which the j5 ohm could be calculated.

3. There might be two possibilities.
• You are forgetting to enter the sign of the imaginary part.
• You are dividing a voltage by the complex impedance in polar form and forgetting that in division, you subtract the phase angle from the quantity in the denominator from the one in the numerator (which for the driving voltage is often set to zero).[10/27 6:15pm PST]

4. Could you please compute the Norton Equivalent circuit for quiz 3? [10/26 5:57pm]

Excellent question! Since I am in a problematic remote session, I will keep the answer short. You already have the Norton resistance since it is identical in value to the Thévenin resistance, shown to be 2.2 ohm in the quiz solution. So all that remains is to evaluate the short-circuit current. You may analyze the resulting short-circuit condition using the circuit analysis tools available to you. OR, since you already have the Thévenin voltage, 5.6 V, you may use the fact that Voc = Isc RTH to find that IN = 2.55 A. [10/27 5:57pm PST]

5. URGENT!!! How do we calculate the magnitude of the impedance, Z, for use in calculating Pav? [10/25 8:03pm]

There are two ways to go:

1. If you have an expression for the impedance in rectangular form (the standard form), convert to the polar form. The polar form of a complex number is reported as the magnitude followed by the phase angle.
2. For any expression of a complex number, you may find its magnitude by calculating the square root of the product of the number and its complex conjugate.
[10/27 5:13pm PST]

6. In the expression v(t) = A cos(wt-q), q represents the phase shift when referenced to a typical cosine function graph. Is that correct? O.K., Why would a sinusoidal wave be shifted in the first place? What causes that? Is is simply due to a difference in the frequency w? [10/25 7:41pm]

No, the phase shift is not due to a change in frequency. In the linear ac circuits that we have been considering in PHYS345, there is only one frequency for all of the voltages across circuit elements and and the currents through them in a circuit when driven by sources all working at one particular frequency. However, there is the likely possibility that there are many phase shifts among the various circuit parameters at various locations depending on the reactances and their locations in the circuit.

The physical origin of the phase shifts comes from the relationship between current and voltage for the three circuit elements that we have considered:

1. The voltage across a resistor is perfectly in phase with the current through it.
2. Since the charge separation on a capacitor is directly proportional to the voltage across it, the time rate of change of charge, the current, is the time rate of change of voltage. The derivative of a cosine is the sine, or a cosine shifted in phase by 90 degrees. Thus we find after consideration of some details that the ac current through a capacitor leads the voltage across it by 90 degrees.
3. Similarly for an inductor there is a relationship between the voltage across an inductor and the time rate of change of the current through it. Consideration of a few details leads to the realization that the current through a pure inductance lags the voltage by 90 degrees.
[10/27 5:08pm PST]

7. Thanks, that response hit the mark! [10/23 12:58pm]

8. Thank you for your prompt responses. However, the w/wc and wc/w thing is still troubling. You explained it, I am sure. I just can't follow your explanation. It is probably because there is a vast difference between our respective knowledges. Could you post two contrasting examples? Analytically, I can follow it--please, don't forget all the algebra/complex number steps. Also, for the gain situation this would be preferred. [10/22 8:33pm]

Let's say the independent variable is x. If I have an expression such as 1 + abx/d, then I could rewrite that as 1 + x/xc, where xc = d/ab.

For the expression 1 - fg/x, I would rewrite that as 1 - xc/x, where xc = fg.

Note the algebraic position of x in each expression, and that I assigned the remainder of the term to a characteristic "x" such that that the dimensionless character of the term became evident. I do not believe that it is the complex number or the particular function of the gain that is troubling you, so the simpler examples that I picked eliminated these features of a typical gain expression.

[10/23]

9. I have a question about CH-7. Are we responsible on the exam for material found in 7.3 Transformers, 7.4 Three-Phase Power, 7.5 Residential Wiring; Grounding and Safety; and 7.6 Generation and Distribution of AC Power. I don't recall spending much class time on this material, with the exception of a few safety points. Thanks. [10/22 11:27am]

We have not had the time yet to consider three-phase power. I hope to return to it near the end of the semester. So you may delete 7.4 from your responsibility for the time being.

We did not consider transformers in the detail presented in the textbook in 7.3 beyond the first page and a half, so skipping over the later portion of that section is fine.

I would recommend that you have read 7.5 and 7.6 for familiarity with the topic, supported by the presentation of the material on our website. [10/22]

10. Thanks for your explanation, however still a few holes.

a) Once I write the transfer function say jwL/[R+jwL], do I just re-write it in the form 1/[1 +/- j(something)], then look at that something and put it in the form w/wc. Because in one case we had wc/w. How do I know which way to write it. For example, in the above example 1/[1-jwc/w]; where wc = R/L, but in H(jw) = 1/[1+jwRC] we have 1/1+jw/wc; where wc = 1/RC. Do you see the reason for confusion?

b) Also, you said the gain is equal to the magnitude of H(jw), for the first equation above, we then expressed H(jw) = 1/1-j. What happened to wc/w? Don't we use it in SQRT[H(jw)x H*(jw)]? Then again, why can't we just use H(jw) = 1/SQRT[1+(wRC)2]?

c) Without sounding picky, it seems like there are always a couple of ways to arrive at the same result--For us beginners this is what is leading to the confusion! I need to learn one way, well, before learning others. Also, for next years class, why don't you just write a packet and use that instead of using our book, or just teach the method in the book. Since the book leaves something to be desired, I suggest writing a packet. Be glad to help this winter if it could count as credit toward one of my ME tech. electives. Consider it? [10/21 9:45pm]

a) After writing the transfer function, look for w. Its algebraic position does not change when the characteristic frequency is set to make the frequency factor dimensionless. So R / [jwL] becomes -jwc / w and jwRC becomes jw / wc.

b) When I wrote that the gain was 1/1-j, I was considering the case when w = wc, a special case. That is, what is the gain when the frequency is equal to the cutoff frequency? The formula that you wrote for the magnitude of the gain is correct; you choose to work with a formula expressed directly for the magnitude. I prefer to find the gain and then find the magnitude nuemrically. It's not a huge difference, just a personal choice.

c) The webpages for the course are a step in the direction of independence from a textbook. I recognize that the personal differences that I bring to the course are potentially confusing when contrasted to the preferences of the author in coverage and style. It is not easy to pick and choose your way through a survey textbook designed for two semesters in just one semester. However, we are making progress! [10/22]

11. O.K., two more problems? How do we determine the cut-off frequency, especially in the high pass filter cases? Also, how do we determine the gain? [10/21 6:00pm]

After writing the transfer function for the circuit in question, look for the frequency dependence. When the transfer function is written in a way that the frequency dependent term is added to a dimensionless quantity (usually unity if written in the simplest form), then the remaining factors in the frequency term must serve as a characteristic frequency, the one we have been calling the cutoff frequency for the simple first-order filters that we have been considering.

I generally refer to the transfer function as the gain, perhaps more precisely we should use the term "gain" to refer to the magnitude of the transfer function at a particular frequency. So speaking more precisely, eavluate the gain of a circuit at a particular frequency by calculating the magnitude of the (generally complex) transfer function. Do not forget to specify the gain in dB! [10/21]

12. I think I have it! Is this right for Filter Analysis?
1. If as w approaches 0, the transfer function approaches 1, it is a Low Pass Filter.
2. If as w approaches infinity, the transfer function approaches 1, it is a High Pass filter. [10/21 5:28pm]

That is correct! [10/21]

13. Hey, the review was good today. However, still having trouble with a few details. I am outlining the CH-6 sec. 6.1 dealing with resonance and I don't see where the web notes "Frequency Dependence of ac Circuits" fit into the analysis of sinusodial frequency response. Please clarify. Is the web discussion just a separate (albeit related) topic or do I need to know how to calculate the Magnitude of Current, Resonance Current, Natural Frequency, Quality Factor to analyze the frequency response of a circuit. Also, could you explain how the universal resonance formula differs from Hv(jw) in the book... [10/21 4:37pm]

The concept of resonance in an ac circuit need not be treated separately. With the techniques of ac circuit analysis, we can write down an analytical expression for a voltage at one point in a circuit relative to the voltage at another point in the circuit; we've been calling that gain. Often in a circuit (the specific example that we considerd was an RLC circuit) the response of the resonance will have the universal resonance form (as stated on the web page). I have used the same symbol H for the universe resonance formula as we have been using for the transfer function. [10/21]

14. Can you show me how to do problem 5.6, I can't get the right answer. If you could, show me with out using diff eq's. [10/13 11:08pm]

I can do all the Ch.5 homework problems assigned without explicitly considering differential equations; rather I rely on considering initial and final conditions and also on the forms of the solutions to the diff eqs that I presented in class.

For 5.6, when the switch is in the upper position, after a long time, the circuit has reached its final steady-state current. The voltage across an inductor with a steady current is zero. The current is determined by the amount of resistance over which the emf is being dropped. In this case, the steady-state current with the switch in the upper position is 12 V / 0.7 ohm = 17.1 A.

When the switch is moved to the lower position, the voltage source is removed and the inductor begins to deenergize through the 22 k resistor. The current present in the inductor (17 A) before the switch is moved is the initial current for the deenergization. The voltage across a 22 k resistor when 17 A is flowing through it is 380 kV. [10/14]

15. I was outlining Ch 6 and became troubled by the differences between the frequency response equations in that chapters and those on your web sight. Could you please clarify (see pg 233 , eq. 6.1)? Also, could you please explain the difference between the equations on pg 237; eq 6.11? In addition, I could not follow example 6.2 on pg 238, could you explain, specifically, how do you go from the I(L)using the current divider to VL/IS x (jw) = ..... [10/12 9:40pm]

All good calls for clarification...

Refer to Fig. 6.1 when considering Eq. 6.1. VL , the voltage across the load, is what I would call the output voltage, vo. VS , the source voltage, is what I would call the input voltage, vi. I generally suppress the j when I write H(w) = vo / vi.

There are different ways to consider the gain. We have been considering gain to be the ratio of two voltages, listed as HV in Eq. 6.11, and will continue to do so in PHYS345. Current gain HI , the ratio of two gains, is also commonly encountered outside of the course. HZ is defined as the ratio of the output voltage to the input current and looks like an impedance, thus the Z subscript; similarly HY defined as the ratio of output current to input voltage looks like an admittance.

Finally, VL / IS(jw) is meant to indicate that the ratio is a function of jw , not a product, and is arguably set in a poor notation. [10/12]

16. Is the answer to question 5.6 correct? It is a two part question. [10/12 8:49pm]

Thanks for alerting me to the missing portion. I added it. [10/12]

17. I was having difficulty with what might be a really important concept in our physics class. As I was going through the class notes, I kept coming across an equation referred to as the "Transfer Function." I remember you going over it in class. I am confused as to exactly what the transfer function is. Can you please e-mail me a basic explanation of exactly what the function is and what it tells us and a general description of exactly how one might use it. I apologize for any inconvenience. Thank you for your time. [10/10 6:00pm]

As I use it most frequentlyin PHYS345, the transfer function H(w) is the frequency-dependent ratio of two voltages. Its functional dependence depends on the network elements (resistances, reactances) and their interconnections. Using the transfer function and knowing the input voltage to a circuit, you may determine the output voltage. That is, the transfer function describes how signals propagate through a system, from input to output. [10/11]

18. For the Electronics Workbench lab, are we supposed to complete the exercises BEFORE or DURING the lab session? [10/6 1:15pm]

My intention was that you would work out the analysis ahead of time, leaving the lab time for learning how to use Electronics Workbench and confirming your calculations. [10/6]

1) How to put a circuit in phase?
2) The difference between Pav and Q?
3) Power factor and power factor correction?
[10/5 6:48am]

You can put the current through a portion of a circuit in phase with the voltage across that portion of the circuit by eliminating the effective reactance in that part of the circuit; that is, make the net reactance zero by changing the frequency or by adding an additional reactive element.

Pav is the actual average power dissipated, the real part of the complex power; Q is the reactive power, the imaginary part of the complex power. Q is a measure of the effect of the reactance in the circuit. Generally, you wish to bring eliminate Q via power factor correction.

The power factor is the cosine of the phase angle; once again, it is a measure of the effect of reactance in a circuit. For a purely resistive circuit, the power factor is unity. For a purely reactive circuit (no resistance present) , the power factor would be zero. For a general impedance, the power factor is somewhere between. To correct the power factor, make it equal unity! [10/7]

20. Would you please recheck your answer to problem 7.17. I got an answer of 2074V at an angle of 49.9 degrees. My Pav for the three items are: 716, 325, 293. My Q for the three items are: 902, 385, 310. Totals (1334,1588) in rectangular and (2074 VA at 49.9 degrees) polar.

My equations for Pav, i.e., (9.6 * 120)cos(.9) = 716 and for Q, i.e., (9.6 * 120)sin(.9) = 902

Also, if these equations are wrong, please indicate so. Thank you. [10/4 12:45 am]

No, my answer is correct as it stands. Where you went wrong was the way the power factor was incorporated. pf = 0.9 means that the cos of the phase angle is 0.9, cos -1(0.9) = 25.8 deg; for Q use sin(25.8) = 0.436. Recalculation using the corrected trig factors should bring your numbers into agreement with mine. [10/4]

21. I am having trouble writing the procedure for the Oscilloscope Lab. The steps are in order the way you have written them with the discussion in between, but the steps taken from the discussion don't make any sense. Any suggestions of how we could write the procedure so it doesn't sound silly? Thanks. [10/3 2:38pm]

I'm not sure what to say; I guess I still do not understand what the problem is. Perhaps with the oscilloscope lab there is not much of a discussion of the procedure required. I would check with the lab instructor to be sure. [10/3]

22. ...I am having trouble solving two equations with complex numbers. I was wondering if you could give me a few hints to simplify the equations. [10/2 1:24pm]

Solving a system of equation involving complex numbers uses the same approach is with real numbers. For example, if the two equations are

ax + by = c
ex - fy = g
then I multiply the first equation through by f and the second eqn through by b and I add them to eliminate y, then solve for x. You just need to be careful with the complex algebra, that's all. You may use any other technique you may have also learned, such as Kramer's method. Alternatively I would recommend a computer algebra package such as MathCAD or Maple to minimize the mathematical effort required to solve the equations. [10/3]

23. What is resonance? What is the relationship between resonance, the source voltage, the effective impedance, and the current in the circuit? None of this is clear to me. I spent several hours on Saturday and Sunday puzzling over today's homework and got nowhere. I went to your office hour yesterday and you were not there. I waited about twelve minutes. After that, I asked an upperclassman who had this course last fall for help with 4.30. We worked on it for about an hour, and finally she told me that they did these kinds of problems with their calculators and that she didn't understand what you were trying to teach us. Last night and this morning I was able to figure out some of the homework problems, but I have no clue about 6.7. [10/28 10:30am]

Sorry that I missed you at the beginning of office hour. I had agreed to make a presentation to new faculty from 1:20 to 1:55 on Monday and I forgot to notify the class that my office hour would be starting late.

Regarding resonance, there are several ways to proceed. Generally it is synonomous with the current through a section of circuit being in phase with the voltage across the section being considered. This will happen when the net reactance is zero; this can generally be achieved by changing the frequency so that the inductive and capacitive elements shift their reactance values, or by adding additional reactance if the frequency is fixed. I plan on presenting a detailed example in class on Sept. 30. [10/29]

24. About the lab, we measured the resistance of our batteries and the current and performed the in-lab experiment thoroughly, but the TA said that the flashlight made out of everday things only had to light up.  We asked if we had to measure resistance of ordinary things in order to construct the flashlight and he said that it only had to include resistors of some sort and to be functional.  Now I keep hearing that we have to give a presentation on the resistors we chose with mathematical basis.  I guess I am just not sure what is expected of us tomorrow.  Could you please clear this up, and if we do need to have measured several resistances, what should my lab group do? [10/27 6:15pm]

It's probably OK to ignore the resistance of any conductors that you use. However, if you needed a resistance of a particular value then that might be a problem. A number of students have told me that they included extra resistance in the circuit beyond the lightbulb -- I would not recommend that approach since the energy dissipated by the extra resistor is wasted (no light) even though it may help limit the current. [10/27]

25. Is there any special discussion or analysis that you would like us to include in out lab write ups besides the recorded data? [10/27 5:41pm]

Yes, everything that you think is relevant should be included. Especially, you need to convince me that your flashlight will last four hours. Sorry to be vague about the format, but I would like to see what the groups come up with on their own... [10/27]

26. On 4.30, I can get to the part in the hint, but how do I go about getting the equivalent impedance? [10/26 8:33pm]

The two branches are in parallel, so Zeq-1 = Z1-1 + Z2-1. Carry out the complex algebra properly for the answer. [10/26]

27. In Problem 4.30, are you sure the equation for the Resistor/Inductor Branch is 2.3k - j0.57k? I see no reason for the negative sign in front of the imaginary part of the number. If you could help me out with that I'd be very appreciative. [10/23 11:24pm]

You are quite right about the sign error; I have corrected the hint. Thanks for reporting the sign error. [10/24]

28. Are you sure the resistance for the Thévenin generator in Problem 3.33 is 4 ohm? not five? [10/20 1:18pm]

Yes, the two 2 ohm resistors that are in parallel are not seen when "looking back" from the Thevenin terminals. This is because the one end of the two in parallel is left dangling when the current source is killed.

Another way to check that it is 4 ohm rather than 5 is to evaluate the short circuit current (which turns out to be -5/4 A) along with the open circuit voltage (which is -5 V); the ratio is 4 ohm. [10/20]

29. I am still having trouble finding the voltage for the Thévenin equivalent in Special problem 3.1c. I understand how to get the resistance, but the Voltage and Current source combo is killing me. Are there any other hints you can give? [9/19 2:31pm]

Looking at the drawing that I added to the hint, if you can find the voltage across the 4 resistor, then that plus the voltage across the voltage source will be the open circuit voltage. Where must the 2 A from the current source go??? [9/19]

30. For class next tuesday, do we have to reconstruct a flashlight 'that will last for four hours' or just construct a flashlight with the batteries. [9/15]

It should be the flashlight that you designed that would last for four hours. [9/15]

31. After attempting the assignment due next week I am having quite a bit of trouble finding the Thévenin voltage source for 3.1c that corresponds with your hint. Thanks. [9/15 4:05pm]

I double-checked the problem and the hint and they are both correct. I added a drawing to the hint that may help clarify the situation. [9/15]

32. I think on the hint page for question 3.5, V2 should be multiplied by 4 not 2. I used 4 when I solved it in maple and my answers matched yours for the voltages and current. The hints and answers section has given me a huge advantage over last week's hmwk, me not knowing it was there! [9/13 6:52pm]

Many thanks for pointing out the error! [9/13]

33. I have absolutely no idea (alright, maybe a little) about how to go about nodal analysis. I am looking at special problem 3.2a and I have seen your hints page on how to go about solving the problem and I can come up with the same equation as what you have, but for some reason I cannot go beyond that. It is the same problem with all the other problems (except the problems out of the book - those I can't even get the equations started). Could you please give me a clue as to how to go about solving these (farther than what you have already helped with)... [9/10 1:11am]

For 3.2a, once you have the equation for the one node (that's the only one of interest), then you solve the algebraic equation for V1. That's all there is. For the other problems, you may have to consider additional nodes as they are present, and solve the resulting system of equations.

And then once the nodal voltage are determined, you may have to find relevant currents to answer specific questions.

It's a little hard to assess exactly how to help via electronic communication. I encourage you to visit during office hours or make an appointment so that we can work through setting up a circuit analysis or two. [9/10]

34. I was looking at the Detailed Nodal Analysis Example in the class notes and at the bottom it shows how to find V1 and then states something about being able to calculate the current through each of the three resistors. Do you use V1 for all of the resistors when finding the current? I guess I just don't completely grasp this concept yet... [9/9 10:43pm]

Yes, V1 is the voltage on one side of all three resistors. However the voltage on the other side of each resistor is different. So the voltage difference across each resistor is different. That, and the different resistance for each leads to different currents in each branch. [9/10]

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Last updated Dec. 10, 1999.
Copyright George Watson, Univ. of Delaware, 1999.