My responses are italicized. I sometimes edit slightly, truncate, or abridge the submissions. Listed in reverse chronological order. Need a clarification?
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]
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]
I announced at the review session that there would be no detailed coverage of motors on the final exam this year. [12/10]
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 V_{oc} = I_{sc} R_{TH} to find that I_{N} = 2.55 A. [10/27 5:57pm PST]
There are two ways to go:
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:
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/x_{c}, where x_{c} = d/ab.
For the expression 1 - fg/x, I would rewrite that as 1 - x_{c}/x, where x_{c} = 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]
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]
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/w_{c}. Because in one case we had w_{c}/w. How do I know which way to write it. For example, in the above example 1/[1-jw_{c}/w]; where w_{c} = R/L, but in H(jw) = 1/[1+jwRC] we have 1/1+jw/w_{c}; where w_{c} = 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 w_{c}/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 -jw_{c} / w and jwRC becomes jw / w_{c}.
b) When I wrote that the gain was 1/1-j, I was considering the case when w = w_{c}, 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]
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]
That is correct! [10/21]
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]
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]
All good calls for clarification...
Refer to Fig. 6.1 when considering Eq. 6.1. V_{L} , the voltage across the load, is what I would call the output voltage, v_{o}. V_{S} , the source voltage, is what I would call the input voltage, v_{i}. I generally suppress the j when I write H(w) = v_{o} / v_{i}.
There are different ways to consider the gain. We have been considering gain to be the ratio of two voltages, listed as H_{V} in Eq. 6.11, and will continue to do so in PHYS345. Current gain H_{I} , the ratio of two gains, is also commonly encountered outside of the course. H_{Z} is defined as the ratio of the output voltage to the input current and looks like an impedance, thus the Z subscript; similarly H_{Y} defined as the ratio of output current to input voltage looks like an admittance.
Finally, V_{L} / I_{S}(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]
Thanks for alerting me to the missing portion. I added it. [10/12]
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]
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]
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.
P_{av} 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]
My equations for P_{av}, 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]
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]
Solving a system of equation involving complex numbers uses the same approach is with real numbers. For example, if the two equations are
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]
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]
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]
The two branches are in parallel, so Z_{eq}^{-1} = Z_{1}^{-1} + Z_{2}^{-1}. Carry out the complex algebra properly for the answer. [10/26]
You are quite right about the sign error; I have corrected the hint. Thanks for reporting the sign error. [10/24]
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]
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]
It should be the flashlight that you designed that would last for four hours. [9/15]
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]
Many thanks for pointing out the error! [9/13]
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]
Yes, V_{1} 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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