Fair Game! Topic open: everything

At this point in the year, everything is fair game for you, so any question you have is fair game for me. Feel free to ask me anything you have covered in the course (or even things from last semester that might come up).

On that note, a piece of wisdom from my 3rd year mechanics professor:

You don't have problems, you have exercises! ~Dr. Melvin Calkin

Question: difference between RC, RL, LC

How do you tell the difference between RC, RL, LC circuits?

Each of these circuits, and what they qualitatively do in DC and AC circuits...

RC - a resistor and capacitor in series. Exhibits charging behaviour with characterisitic time constant with DC voltage source. Acts as a high pass filter (allows high frequency currents, but not low frequency currents) in AC circuits.
RL - a resistor and inductor in series. Acts as a short with a DC voltage source, but smooths out rapid variations in current. Acts as a low pass filter (allows low frequency currents, but not high frequency currents) in AC circuits.
LC (and RLC) - an inductor and capacitor (and resistor) in series. If initially charged, has oscillitory behaviour (damped if also has a resistor). Has resonant behaviour with AC driving voltage (damped if also has a resistor).

Question: LC in parallel

How to calculate the resistance of a capacitor and an inductor connected in parallel?
(-XcXL/(XL-Xc))^2 ?

This is a little bit trickier that considering elements in series because the inverse of the impedances have to be added. I'm pretty sure it is beyond the scope of this course to need to do this calculation, but I will put it here anyway. I'm also not sure how you would arrive at the answer without using the appropriate complex represenations of the impedance, so I will use them but try to be careful to explain.

So, for elements in parallel:

1/Z_tot=1/Z_C+ 1/Z_L+ 1/Z_R ...[1]

We will consider just an LC circuit (with ideal components), so R=0. Now, in complex notation Z_Cand Z_L are:

Z_C=1/iωC ...[2]
Z_L=iωL ...[3]

So substituting [2,3] into [1]:

1/Z_tot=iωC+1/iωL

or, using the fact that 1/i=-i:

1/Z_tot=iωC-i/ωL
1/Z_tot=i(ωC-1/ωL) ...[4]

so, we can now replace 1/ωC with X_C and ωL with X_L:

1/Z_tot=i(1/X_C-1/X_L) ...[5]

Now we can simplify the part in the brackets, giving us:

1/Z_tot=i(X_L-X_C)/X_CX_L ...[6]

or taking the inverse of the fraction on both sides (and once again using 1/i=-i):

Z_tot=-iX_CX_L/(X_L-X_C) ...[7]

Now, if we had also had a resistor in parallel in the circuit (R), we would end up at equation [7] with both real and imaginary parts. To find X_tot we would have to find the magnitude of the resultant phasor. Since Z_tot is purely imaginary, we need only to find the magnitude of this number, which we can express as:

X_tot=-X_CX_L/(X_L-X_C) ...[8]

Which is pretty much what I think you suggested. Note that if we hadn't used the imaginary numbers we would not have been able to get the right signs and would have come up with a different answer.

Looking back at eqn. [7], let's think about what this represents... In the imaginary plane Z_tot would be a phasor pointing along the y-axis. For values of X_L and X_C which make Z_tot positive, the impedance could be considered "inductive", and inversely if Z_tot were negative, the impedance could be considered "capacitive".

Let's consider two extreme cases...

If the frequency is high, X_L is very large, and X_C is very small (let's say negligible). In the numerator of Z_tot the frequency cancels out, so this is a constant. If we look at the denomenator, X_L >> X_C, so the denomenator is approximately X_L. Since X_L is large, Z_tot will be small and negative (capacitive). In fact if you substitute in equations [2,3] ignoring X_C in the denominator, you will recover Z_C. This is consistant with the inductor providing a break in the circuit, and the capacitor providing a short.

Similarly, if the frequency is very low, X_C becomes very large, and X_L very small. In this case, the result is small and positive (inductive), and indeed again, if you work it out you will recover Z_L as Z_tot. This is consistant with the capacitor providing a break in the circuit, and the inductor providing a short.

To summarize:

ω→∞ Z_tot→Z_C and current flows through the capacitor
ω→0 Z_tot→Z_L and current flows through the inductor

These limiting cases can be worked out simply as a thought experiment (consider which branch of the circuit has a small resistance and which has a large resistance) and are often more insightful than getting through the algebra.

Hope this helps.

Topic Open: RC, RL, LC, LRC circuits

The following topic is now open for questions: RC, RL, LC, LRC circuits (and any AC circuits)

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Question: Impedance of LR circuit with "real" inductor

Just had a question about LR circuits. Just say there was a circuit consisting of a resistor and inductor. And just say the inductor had a resistance of its own. How do you find the total impedance of the circuit. This was similar to one of the CAPA questions. My friend told me to find the impedance of just the resistor and inductor and then add the resistance of the inductor to Z. I am confused on why you do this? Why couldn't you just add the resistance of the inductor and resistor first, then use the formula Z = sqrt(R^2 + XL ^2), R being the resistance of the inductor and resistor combined?

Actually, you are right! You can (and should) just add the resistance of the inductor to the resistance of the resistor then find the impedance including the inductive term.

Here's how it works. A real inductor will have some inherent resistance due to the fact that it is a (usually fairly large) coil of wire. To analyze a circuit, we would replace a "real" inductor with and "ideal" inductor and a resistor (to represent the pure resistance of the inductor). So, in an LR circuit you would go from having 1 resistor and 1 "real" inductor in series to having 2 resistors in series and one "ideal" inductor also in series. You would then find the net resistance of the circuit, and this would be "R" in the formula you quote above.

From this you would find the magnitude of the inductance from Z=sqrt(R²+X_L²)
(this formula comes from the fact that R and X_L are 90^o out of phase, so the magnitude is the hypotenuse of the two "phasors" see: Basics of AC circuits/RLC circuit example). Since R and X_L do not add linearly, I do not think you should get the right answer if you go about it in the opposite order (except maybe in some special circumstance).

EDIT: A similar question was also asked regarding an LRC circuit. In this case, the methodology is the same. The inductor with a resistance is replaced by a resistor and an inductor in series. The resistance of the whole cicuit is calculated, and then the magnitude of the impedance is determined. In the case of an LRC circuit, Z is given by:

Z=sqrt(R²+X_L²+X_C²)

Otherwise, the whole problem is the same.

Hope this helps!

Question: Solenoid in a coil

If a solenoid is placed inside a coil, how does the current flowing through it affect said coil? How would you find the inductance value for the coil?

That's a very good question. Qualitatively, the solenoid (with some current running through it) establishes a magnetic field parallel to the axis of the solenoid. If a coil is then placed on the outside of it, the coil will "sense" the magnetic field established by the solenoid. If the current in the solenoid is time-varying (changing in time) then the magnetic field established by it will also change with time. The coil will then "sense" this changing magnetic field and respond to oppose the change (Lenz's law).



Quantitatively, we will have to look at what the magnitude of the magnetic field produced by the solenoid is, and how that may depend on time. The magnetic field of a solenoid is given by:

B_sol=μ₀I[N/l]

Thus, if the current, I is a function of time, the magnetic field will have the same time dependence. If, say, the time dependence is linear (eg. we ramp up the current to the solenoid at a constant rate), then we can use the slope as ΔB/Δt, which is related to the EMF:

EMF=-ΔΦ_B/Δt=-A_{perpendicular}ΔB/Δt

Chapter 21, Question #80 is an example of this, and might be good to look at.

I hope this is helpful and answers your question.

Topic open: Charges and currents in magnetic fields

The following topic is now open for questions: Charges and currents in magnetic fields.

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Topic open: Magnetic fields

The following topic is now open for questions: Magnetic fields.

To pose a question, please post a comment to this post by clicking on the comments link below.

Status check

Just wanted to check in with everyone using the webTA site. I have a couple of questions for you!

1. Am I keeping pace with the topics in class? (I have access to the notes and the CAPA's but without feedback from you I can't tell exactly where you are with the material).

2. Are people finding the material both here on the blog and on the main website useful? Any suggestions regarding what is good/bad, could be improved/added/discarded?

Any feedback you have for me would be very much appreciated! Hope your courses are going well, I know this can be a busy time of year!

Question: Current and a switch

I'm just wondering. If there is an open switch AFTER a resistor, the current still goes through the resistor but does not continue in the circuit, right?

If there is an open circuit, there is no current flow. For current to flow there has to be a continuous path from a high potential to a low one, otherwise charge would have to build up somewhere... which cannot happen. If the charge built up in this way, the charges flowing in behind them would feel more and more repulsive force until the potential required to pile in more equaled the potential of the battery and there was no driving force left. This is basically what happens in a capacitor, and a real circuit will have some of this occuring, but the end result in both cases is that the current stops flowing.

Ever been on a really crowded bus? There is somewhat of a driving force from the front of the bus to the back of the bus as people get on and the bus driver yells to everyone to go to the back. As the bus fills up, and fills up, people at the front push people in the middle who push people at the back closer and closer together, but eventually there is nowhere for them to go... at some point, the flow of people towards the back has to stop until people get off and make room again.

A wire is like an already crowded bus. The charges close to the battery (the front of the bus) feel a driving force from the battery, those charges move slightly, and because they are displaced in the wire, there is a net charge which repels like charges near them and pushes them ahead a bit. This continues all around the wire. But, if there is a break in the loop, there is nowhere for the charges to go (they can't just jump out of the wire), so they just sit there and wait until there is a way for them to move.

The short answer to your question is: if there is an open switch anywhere in a loop, there will be no current flowing through any part of that loop. If there are alternate closed paths in the circuit, then current can flow through those parts of the circuit, but they must constitute a closed loop, with some source of potential a.k.a. electromotive force (EMF) (a battery, DC or AC power supply, a charged capacitor...).

Hope this helps. Please post a comment to this if you need clarification

Topic open: Anything DC circuits

The following topic is now open for questions: Anything DC circuits. Including: Resistor networks, Kirchoff's laws, Capacitor networks, RC circuits...

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Using loop rules (Chapter 19 #29)

I am stuck at # 29 (chapter 19) and I really don't know where I went wrong. I made 2 loops, 1 with R1 R2 and another with R2 and R3 and then I had one junction equation.

Well, from what you said it seems like you are on the right track. Using two loop rules, and one junction rule, I get the following equations:

loop 1: E1-I1R1-I2R2=0 loop 2: E2-I2R2-I3R3=0 junction: I1+I3=I2

This does give you 3 equations with 3 unknowns. Solving these systems by substitution can be a bit tricky, but with perserverance can be done. I show an example on the help home solving both by substitution and a matrix method (which can be a quick way to solve if you have access to a graphing calculator, or are good with matrix reduction). To see the example and a review of Kirchoff's laws go to: Kirchoff's laws and their application.

Let me know if these are not the equations you got, and we will backtrack some more and find out where the problem is. (Remember, you can post follow-up questions as comments to this post.)

Question: Adding another resistor

I was wondering if you could tell me if, say you have five resistors in a circuit, and you add another one, does the power of the battery have to increase? Or does the system just slow down or something...

Ok, I'm first going to have to clairfy and simplify the system we are talking about. I'm going to reduce the problem to a battery and a resistor. The only difference in having 5 as opposed to 1 is reducing that network of 5 resistors. We will also assume that we have a battery providing a constant voltage source (this is the normal case), and that it can provide whatever current we demand of it (it has no maximum current output).

So, let's take our battery and resistor. Using Ohm's law, we can determine that the current is I=V/R (if we had 5 resistors of equal resistance in series to start the current would be I=V/5R since the equivalent resistance of those 5 resistors in series would be 5R). And then the power drawn from the battery would be P=IV, or P=V²/R.

Now suppose we add another resistor, of equal resistance, in series. The equivalent resistance of the two resistors would be 2R, and the new current would be I=V/2R, and the new power would be P=V²/2R (if we had added one more in series to our 5 resistors we would have I=V/6R, and P=V²/6R).

The voltage provided by the battery does not change due to the addition of another resistor, however the power output decreases (because the current drawn decreases).

We can go through the same analysis assuming the resistors are in parallel. The current with just one resistor is I=V/R, and power is P=V²/R. If we add a resistor in parallel, the current will become I=2V/R, and the power will be P=2V²/R. Here, again the voltage of the battery does not change (it is a constant source of potential), but the addition of another resistor in parallel increases the current drawn from the battery. Since the current increases, the power supplied by the battery also increases.

So, what happens when we "add" a resistor to a circuit? That depends on how we add the new resistor. The voltage provided by the battery stays constant in all cases. The current may change depending on the arrangement of the resistors. To determine how the current will change you must reduce the resistor network before and after adding in the new resistor, and use Ohm's law to determine the current before and after. If the current changes, the power will change, since Power is proportional to current. Again, compare before and after to determine the change.

Finally, let's be clear about the difference between "voltage" and "power". Voltage, or potential difference, is increase/decrease in potential energy per unit of charge. Power is the energy per unit time, in this case electrical energy per unit time flowing through the circuit. It is the battery which supplies both. That energy has to come from somewhere, right? You can think of power as being an energy transfer, the battery is providing energy at some rate to move charges around the circuit, at some rate.

Hope this helps clarify some things. You might also be intrested in a previous student's question on Voltage, EMF and resistors (about half way down the page). Let me know if this answers your question or not (by posting a comment to this post).

Topic open: Resistor networks

The following topic is now open for questions: Resistor networks.

To pose a question, please post a comment to this post by clicking on the comments link below.

Solution to first homework: a ring of charge

The solution to my first "homework" exercise is posted on the homework page.

Please let me know if you have questions about the solution, or if you think / don't think similar exercises would be helpful in future.