BR Physics, Electric Circuits, Question 31 vs 56

[Catburr]

---

Members do not see this ad.

In TBR Physics, Section IX on Electric Circuits there are two questions, from two different passages, that seem to contradict each other. Or more likely, I'm missing something!

Q#31, page 197: If both switches were closed, and a new resistor were inserted into the circuit before S1, then the maximum charge that could be stored on the capacitor would:
A. Increase, because the voltage across the cap. would increase
B. Decrease, because the voltage across the cap. would increase
C. Increase, because the voltage across the cap. would decrease
D. Decrease, because the voltage across the cap. would decrease

This question goes with a circuit drawing which I'll scan tomorrow if anyone demands, but it's simple "with both switches closed" as this problem states. It's a battery, with one capacitor, and two resistors. All are in parallel to one another.

Answer, summarized explanation: D., this new resistor, inserted into the circuit in series with the capacitor, has a voltage drop across it. This means less voltage drop for the capacitor, so according to Q = VC, decreased voltage across same capacitor must mean decreased maximum possible charge. "Since the whole 10mV is spread across two circuit elements, instead of just the cap., we can conclude that the voltage must decrease across the capacitor."

OK, that's all well and good. Fast forward to question 56.

Q#56, page 203: The maximum charge on a capacitor depends on:
I. the capacitance of a capacitor
II. the emf of the battery
III. Resistor c
IV. Resistor d

This is also in reference to a circuit, I'll try to describe the pertinent parts. There are basically two sides to it, and a switch that can be flipped to connect all of one side OR all of the other, based on whether you want to charge or discharge the capacitor.

Side 1 - Battery, resistor c, capacitor, all in series
Side 2 - Capacitor, resistor d (which has some other funny stuff given that this is a passage about a pacemaker, but that's not relevant to this problem, as far as I can tell), in series.

So my reaction to this problem went something like this: "Well duh, of course the capacitance affects Qmax. And because Q = VC, the emf of the battery of course matters too. Oh, and I remember learning back when I did problem 31 that since a resistor in series with a capacitor has a voltage drop across it, it's like it eats some of the voltage that the capacitor would otherwise have. So resistor c must affect Qmax for this capacitor!"

Wrong.

Answer, summarized explanation: B.; "I and II only" because C = Q/V. "There is no dependence on resistance of any kind"

Umm, what? Didn't it just say that resistors eat voltage, and voltage is part of C = Q/V, therefore resistors have some (small? I don't know...) effect on C?

Sad face...hope someone can help me out. This is the first time I've stumbled across something like this in TBR...sure some parts are worded awkwardly, but nothing like this. What am I missing here?

For what it's worth, I'm using the 2009 blue TBR physics. I hear tell there's some fancy new orange one, but I guess I didn't dot my i's right or something on my order form, and they stuck me with this old one. : )

 

[Rabolisk]

I have a vague idea of what's going on, but to confirm, I will need the images scanned.

 

[movax]

Answer, summarized explanation: D., this new resistor, inserted into the circuit in series with the capacitor, has a voltage drop across it. This means less voltage drop for the capacitor, so according to Q = VC, decreased voltage across same capacitor must mean decreased maximum possible charge. "Since the whole 10mV is spread across two circuit elements, instead of just the cap., we can conclude that the voltage must decrease across the capacitor."

Yep, you've got that down correctly, there's a voltage drop across the resistor, hence the potential across the cap will now be lower as well.

Q#56, page 203: The maximum charge on a capacitor depends on:
I. the capacitance of a capacitor
II. the emf of the battery
III. Resistor c
IV. Resistor d

I've bolded what I believe to be the key in this question. Say you have a 0.1uF capacitor. It will hold 0.1 microfarads of charge across its plates.

Now put a resistor in series with this. As the voltage has now dropped, looking at the previous question and Q = C*V, with V being lower, Q has also dropped.

But, the maximum charge of the capacitor has not changed. It's still a 0.1uF capacitor. If the resistors go away, it will see the full emf of the battery (and hence the highest voltage) and hit its maximum capacitance. I'd need the circuit to be sure, but I'm pretty sure that's what going on.

 

[Catburr]

Sorry for the delay, but I've finally attached the circuits here.

I think your post makes a lot of sense movax. I need to reread it, but just wanted to get the pics up while I had a chance.

Thanks a lot for taking the trouble to write all that out! These forums are really an awesome resource.

 

[Rabolisk]

In an RC circuit in which the resistor and the capacitor are in series, the maximum charge that can be held by a capacitor is simply CV. The resistor has an effect on how quickly it charges or discharges, which is related to the time constant, but it doesn't have an effect on the maximum charge.

In the first question, though, they say that both switches are closed. When this happens, the voltage drop across the parallel elements (the two resistors and the capacitor) have to be equal by definition. This voltage drop is also equal to the emf of the battery. When you add an extra resistor in series with all of these, there will be some voltage drop across this resistor (V = IR), since there will be some current. So the voltage drop across the other two resistors and the capacitor will be less than what it was before (they will still be equal to each other), and that's why the maximum charge held by the capacitor is not equal.

The crucial difference is that in the first problem, you have resistors in parallel with the capacitor. This means that current will always flow. In the second problem, current stops flowing when the capacitor is fully charged, and thus there is no voltage drop across the resistor (V = IR), and this its resistance has no effect on the voltage drop across the capacitor and the maximum charge stored by the capacitor.

 

[Catburr]

Hmm, I've read and reread your post Rabo, and I'm not sure I have a good grasp on what the rule is, if there is one, I guess.

So if you have a capacitor and resistor and battery all in series, nothing else, then the resistor (and it's associated voltage drop?) DO NOT decrease the maximum charge the capacitor is capable of attaining? And this is because once the capacitor is fully charged, the whole circuit stops seeing current? Or looking at it from the other direction, until the capacitor reaches it's max charge as defined by Q = CV, current doesn't have anywhere else to go so it just keeps pushing through the filling capacitor?

But, if there's another part to the circuit, some resistors in parallel like in problem 31, then adding more resistance in series ahead of everything DOES cause a decrease in the maximum possible charge? And this is because once the capacitor is fully charged, current just goes solely to those parallel resistors? In other (probably iffy) words, once the charge on the capacitor gets high enough, the current prefers the parallel resistors?

Am I repeating you, or totally misinterpreting you? : )

And I'm not sure how to correlate all that with movax's input, but I hesitate to try since that post was offered before I bothered to post the circuit pics.

 

[Rabolisk]

Basically, you're repeating me.

In a simple series RC circuit, the current charges the capacitor, until the capacitor can hold no more charge. Then current stops. As a side note, this current is not constant, but decreases exponentially to 0, while charge on the capacitor increases exponentially to CV (Qmax).

If you have a resistor in parallel with a capacitor, then there is no reason for current to stop, because it has a pathway back to the battery. Your reasoning is correct. As a side note, a capacitor acts as a short circuit (0 resistance) when it is fully discharged, and an open circuit (infinite resistance) when it is fully charged.

To be fair, this would be extremely easy (at least the second question) to someone who knows about RC circuits. The whole spiel about the time constant should be background knowledge to someone who has studied EM. I don't know if RC circuits are are officially part of the MCAT content, although it appears that "discharge of a capacitor through a resistor" is, however vague that sounds.

 

[Catburr]

RC circuits are fair game for the MCAT for sure. I guess it just wasn't immediately obvious to me that extra serial resistance only affects maximum charge on a capacitor when the current has an alternate path available. Sort of makes sense now in relation to the whole time constant thing; the idea that it's the time constant that varies and not the max charge with a series circuit. If that's spelled out in the BR book somewhere, I for sure missed it, so your input has been really helpful, thanks!

Here's a hypothetical question then, if you don't mind...if you increase the resistance of the resistor that is parallel to a capacitor, does that increase the maximum charge on the capacitor (as in increase it closer to what it would be if there were no parallel path)? My reasoning there is, now the capacitor gets a little MORE charged before the current starts preferentially picking the parallel path and no longer adding charge to the capacitor. So as you approached infinite resistance (no current) in that parallel path, you'd return to the same Qmax that the capacitor has when it's in a basic series circuit?

 

[movax]

Here's a hypothetical question then, if you don't mind...if you increase the resistance of the resistor that is parallel to a capacitor, does that increase the maximum charge on the capacitor (as in increase it closer to what it would be if there were no parallel path)? My reasoning there is, now the capacitor gets a little MORE charged before the current starts preferentially picking the parallel path and no longer adding charge to the capacitor. So as you approached infinite resistance (no current) in that parallel path, you'd return to the same Qmax that the capacitor has when it's in a basic series circuit?

You can (mostly) treat infinite resistance as an open circuit (i.e, air gap) and 0 resistance as a short circuit. In reality, with high enough voltages (above dielectric breakdown of air) you can get sparks through the air, and of course wires have resistance.

A capacitor blocks DC; it acts as an open-circuit. For AC (higher frequencies), it acts as an short-circuit. By putting a capacitor between a power rail and ground (bypass capacitor), you allow DC to pass through, but AC (noise) gets shunted to ground. When freq is 0 (i.e. DC), open-circuit, no current flows.

Likewise, putting a capacitor in-line with a signal will AC couple it, blocking DC voltage, but allowing through AC signals.

A single cap and a single resistor in parallel will see the same voltage. Current will always flow through the resistor (congratulations, you made a heater). The cap, once charger, can basically be ignored unless you remove the battery, in which case it will discharge into the resistor.

The second figure doesn't indicate a system ground. When the switch is flipped, the battery essentially drops out because it's negative terminal (where electrons would leave, the anode) isn't connected to the circuit anymore. The cap will discharge voltage into Rd as per the diffEQ C dv/dt.

If you have trouble visualizing circuits involving switches/transistors, just redraw the circuit as it would look when the switch is in a given position (and drop out nets with no connections).

As an aside, electrons do not travel through the capacitor. For each electron that arrives at one plate, another is depleted from the opposite plate.