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).