Topic open: Circuits

The following topic is now open for questions: Circuits. Anything about resistors, capacitors, RC, RL, RLC circuits.

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Question: Two resistors, Two capacitors (Ch 19 #50)

Ch 19 #50 part b

I am not sure how to distinguish whether a capacitor is charging or discharging from this problem. Sometimes, I can't tell through which resistor will a capacitor will discharge...is it through the one in parallel or in series?

Let's just look at simplifying what we have when the switch is open, then closed.

When the switch is open there are two resistors in series which are in parallel with two capacitors in series. The voltage across the set of resistors will be the same as the voltage across the set of capacitors (24 V in this case). At point "a" there will be some voltage/potential between 0 and 24 V due to the voltage dropped (IR) across one of the two resistors. Likewise at point "b" there will be some voltage/potential between 0 and 24 V as there is some voltage dropped (Q/C) across one of the two capacitors. (Note that to determine this you will have to use the equivalent resistances and capacitances of each branch of the circuit to determine the net current and charge, respectively.)

Now, when the switch is closed, charge will flow from a to b or vice versa because the potential at a and b were not equal initially. Once the switch is closed this potential difference cannot be maintained. Without working through to see which is greater and thus which direction the charge will flow I can't say whether the capacitors are charging or discharging. However, I can tell you if the potential at "a" is greater than at "b" the current (charge) will flow toward the capacitors and thus the capacitors will charge (partially). Conversely if the potential at "b" is greater than at "a" the current (charge) will flow away from the capacitors and the capacitors will discharge (partially).

Good luck. This is a tricky problem.

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).

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: 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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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.

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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).