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...

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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.

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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.

Question: E=V/d or V=-Ed? (to minus or not to minus)

I was wondering why the notes from class say that for uniform Electric fields E=V/d, while the textbook and WebTA say that V=-Ed. Why is there no negative sign in the one case, but the negative is included in the other?

That is a very good question. I can first tell you that the reason my formula and the formula in the book are the same is that I copied mine out of the book. ;) I can also say that Prof. Altounian's notes are correct.

To get to the heart of this matter, really we have to think about what V is and what E is. The electric potential, V, is a scalar quantity. That is, it has a magnitude at a particular point in space, but no direction associated with it. But the electric field, E, is a vector quantity. This means it has both a magnitude and a direction. So if we look at a simple formula for the electric field, like E=V/d, we have to think this isn't the full story... how do we relate a vector and a scalar?? where is the direction part of E??? Well, it isn't there. Really all we get is the magnitude of E. So, whether you stick a negative sign there or not, you have to determine the direction of the electric field in another way (by determing where a high potential is and where a low potential is).

So, why did the text book (and myself) bother putting a negative sign there? Well, it comes from the calculus relation between E and V. If you have a map of V over an area, the electric field points downhill, so when you look at it from within a differential formalism you need the negative sign (so I'm used to seeing it that way).

In any case, I mean to change my webTA pages to agree with the class notes on this point. And the moral of the story is that since E is a vector you have to determine the direction as well as the magnitude.

Server unavailable Jan 27 05:00-07:00

I received this message this morning regarding the physics department servers:

please be advised that tomorrow (friday) the webserver (and its filesystem) will be unavailable between 05:00 and 07:00.

This will effect the webTA home (site with all examples etc), but not the blog site.

Posting questions

Hi all, I would like to suggest that when you ask a question related to a text book problem that you include the problem with your question. It makes it easier for everyone reading your question, and much easier for me if I happen to not be near my text book (or too lazy to look it up right away).

Also, I think everyone is getting the idea about asking questions by posting comments. Don't worry if they don't appear right away, I have to clear them first. I would prefer, unless you are having a problem with the site, that you don't email me with questions. It's just easier to have them all in one place on the blog, and possibly helpful to others that might have been wondering the same thing.

Thanks everyone for your continued participation, so far I think this is working out quite well. :)

Question: 2 Charges Dropped in an Electric Field (Chapter 17, #68)

I am confused on what the equation for conservation of energy should be for this problem....Should we consider potential energy due to the height?

For example I assumed the equation to be: mgh(before)+PE(before)=KE(after)

The question states: "Near the surface of the Earth there is an electric field of about 150V/m which points downward. Two identical balls with mass m=0.540kg are dropped from at height of 2.00m, but one of the balls is positively charged with q₁=650μC, and the second is negatively charged with q₂=-650μC. Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground."

Whoo-ee! That's a nice problem, it incorporates several different ideas, just the kind I hated as a student, and see the value of now when I teach. :)

Your energy equation is mostly correct, but we need to figure out the signs of the energies.

Let's first draw a diagram, because I find that problem a mouth full. (I will put in the electrostatic forces, but not gravitational... we all know that points downward anyway).

So we have the positive charge (drawn on the left) experiencing a downward force due to the electric field, thus as it falls (down) it's electric potential energy will be decreasing, just as it's gravitational potential energy will be decreasing. However, for the negative charge (drawn on the right), the electrostatic force is pointing upward, opposing the falling motion, so it's electric potential will increase as it falls.

We will have to look at the conservation of energy each charge separately. Let's start with the positive charge:

PE(electric)+PE(gravitational)=KE
qEh+mgh=(1/2)mv₁²

(note that I used h for the distance the charge travels through the electric field).

Now let's look at the negative charge:

PE(gravitational)=KE+PE(electric)
mgh=(1/2)mv₂²+qEh

When plugging in the numbers, disregard the sign of the charge. I have included it already by determing whether the electric potential energy increases or decreases. Alternatively you could have used the energy equation for the first charge and plugged in the sign of the charge to get the same thing. Personally, to avoid confusion with signs, I think it's better to work through the logic of what's going on than to plug in signs, but that doesn't make the other way wrong.

So, using the energy equations for each ball, you can solve for the final velocities. I have confidence that you can all plug in numbers, so I will leave it at that. Hope this helps, and if you have any further questions feel free to post a comment.

Question: Increase plate separation, increase energy stored in capacitor

If two plates of a capacitor with constant charge have their separation doubled, the energy stored also doubles. I'm confused as to why this happens, since I thought E-stored was related to C, and C decreased as distance of separation increased. I'm looking at the equation Energy Stored = 0.5CV^2.

You are right that the energy stored in a capacitor if related to C, the capacitance. However, you have to consider what is being held constant as the separation doubles. If you look at the equation: U=(1/2)CV², you will notice that U is proportional to C, however you have a factor of V² that changes in a way you can't determine directly. If instead, you re-write the equation substituting in V=Q/C, so that you have: U=(1/2)Q²/C, then you know that Q is a constant, so we can talk about U being proportional to 1/C. Thus, as C decreases, U increases.

It might help you to look at the example I posted on this if you haven't already: Chapter 17, #50. Note also that since the amount of energy stored in the capacitor is increased, work must be done to increase the separation between the plates.

First homework: a ring of charge

Well, I've posted my first "homework" exercise. I hadn't gotten around to it because I've been trying to polish up some other things, and keep up with your questions. It should be a short exercise, so before you dive into doing a bunch of math, think about ways to simplify the problem.

If you're interested you can get to it from the homework page.

Good luck, I will post an answer in a few days or so.

Topic open: Resistance and Resistivity

Resistance is futile... if less than 1Ω. ~ unknown author

The following topic is now open for questions: Resistance and Resistivity.

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

I realize you may still have questions regarding capacitors, feel free to continue asking questions on old topics (by posting comments to the appropriate topic post).

Popups?!?

I would like to know if any of you have been experiencing pop-ups on this page. I have pretty good pop-up blocking, so I had no idea that the "free" counters I use on all my webpages have recently started creating popups on the webpages they are placed on.

If anyone has seen this happen on the blog, or the static tutorial page, please let me know and I will remove the counter immediately.

Thanks, and my appologies to anyone experiencing these popups.

Edit: I have removed the webstats4u counter and replaced it with another that claims it will not put ads on the sites that use it. Still, let me know if anything strange happens. (2:14 PM Jan 23 2006)

Question: Capacitor miscellany

here are my questions about capacitor:

1 #51 b)Why doubling?

The energy stored in a capacitor is given by:

U=(1/2)QV=(1/2)CV²=(1/2)Q²/C.

In Ch. 17, #51 (b) we are told that the amount of charge on each plate is doubled (ie. Q is doubled), and the capacitor remains connected to a battery (ie. V is held constant). So, we can look at an initial energy, U₀, and a new energy after the charge is doubled, U':

U₀=Q₀V
U'=Q'V=(2Q₀)V

The ratio of the new energy to the old is then:

U'/U₀=Q'V/Q₀V=2

So, indeed if the charge on the capacitor is doubled, the energy stored in the capacitor is doubled.

2 Say, we have 2+Q on the left plate and 2 -Q on the rite, can I just add 1 +Q to the left and add nothing to the rite? Or it is a must for me to add the same amount of charges to both sides?

You must add the same amount of charge to both sides. (Check out the Ch. 17 summary on where it describes a capacitor.) If you added more of one charge to one side than the other, the net charge of the capacitor would be non-zero. I'm not sure what exactly would happen, but it wouldn't be a simple capacitor anymore and wouldn't follow the behaviour we've been discussing. Sorry I can't give a better explanation than that.

3 If I added 1 +Q to left and 1 -Q to rite, the total charge should be +2Q, rite?

Well, the net charge would be zero, the total amount of charge would be |2Q| (those || are to show we are talking magnitude/absolute value and disregarding the signs), but the charge on the capacitor (the "Q" you use in the capacitor equation) would be 1Q. The measure of charge that matters in a capacitor is how much you have separated off to each plate, so it is this charge you consider.

4 #77b) what i did was PE/2=1/2*m*v^2, what is wrong with my solution( i know it must be wrong -_-||)?

I had a student ask this question last year, I think the answer given there will help you also. Check out electrostatics: previous student questions, it is the second item on the page.

I believe your problem lies in the factor of 1/2 you are using for the potential energy... I'm not sure where this comes from, and I'm pretty sure it isn't necessary. ;)

Good Luck, hope this helps. As always you can ask follow-up questions.

Topic open: Anything Capacitor

The following topic is now open for questions: Capacitors.

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