conflicting concepts in berkeley review physics: resistors' effect on capicators

[ihatebluescrubs]

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In BR physics Chapter 9, # 31:

-we are given a circuit and asked what happens with a new resistor is added in front of the capacitor. The paraphrased question: "If a new resistor was inserted into the circuit before the capacitor, then the maximum charge that could be stored on the capacitor would:

a) inc cuz voltage across capacitor would inc
b) dec cuz voltage across the capacitor would inc
c) inc cuz the voltage across the capacitor would dec
d) decrease cuz the voltage across the capacitor would dec

The answer is D. The explaination given is that: Voltage and charge for a capacitor are related through Q=VC. If voltage inc, then charge will inc. With the new resistor in the cirucit some of the Voltage has to go on the resistor and some on the capacitor. Since the whole voltage is spread across two circuit elements instead of just the capacitor, we can conclude the voltage must dec across the capacitor.

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The second similar question: BR physics Chapter 9, # 56

Passage gives a circuit battery --> Resistor --> Capacitor

The question asks: The max charge on the capacitor depends on:
I: the capacitance of the capacitor
II: the emf of the battery
III: the Resistor

The answer is both I and II. Because "Capacitance is define as C= Q/V, where Q is the charge on the plates of the capacitor and V is the voltage diff betw the plates. There is no dependence on resistance of any kind"

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Because of these two answers, I now even more confused as to how a resistor can affect a capacitor. Why is it that in the first question, they say that putting in a new resistor will change the max charge of the capacitor while in the second question they say that the resistor doesn't affect the max charge? Shouldn't the resistor affect max charge because Q = CV and the V reaching the capacitor depends on how much a resistor eats up the charge before it gets to the capacitor?

 

[milski]

You have a different version of the book - #31 is not about capacitors in mine and the last question is #52, so #56 is not even there.

Without the question, just some thoughts:

You are correct, if the circuit is just a capacitor and a resistor in series, increasing the resistor will not change the maximum charge, only the time it will take to reach it.

If I have to guess about #31 there are other elements in the circuit which interact with the new resistor and result in change of the voltage across the capacitor. For example, if the capacitor was a part of a voltage divider, adding the resistor in front of it will change the ration of the divider and give you a new voltage across the capacitor. Without seeing the circuit, there is not much more that I can say.

 

[HotHamH2O]

just looking at the equation Q/V=C you can determine the fact that as voltage is increased capacitance is decreased and visa versa right? It's a bit tricky because Capicitance just like resistance is a set value not based on q or v but on the area and separation of the plates. Much like how resistance is a set value determined by the material of the resistor. Q/V=C is just a tool to relate the values to each other much like Ohms law is a tool to relate V,I, and R, even though R is not actually determined by V or I but can be found by knowing V and I.

 

[BerkReviewTeach]

The difference between the two questions is that in one case (question #56) the resistor of interest is in series with the capacitor and there is nothing in parallel with the capacitor versus the other case (question #31) where the resistor of interest is in series with the capacitor and there are two additional resistors in parallel with the capacitor. In one case the filled capacitor will stop the flow of current through the entire circuit while in the other case the filled capacitor does not stop the flow of current through the entire circuit.

In question #31, the new resistor is in series, so it drains voltage, thereby reducing the voltage drop across the capacitor. Because there are two resistors in parallel with the capacitor, there is current constantly running through the circuit, even after the capacitor is full. This is because there is a path from cathode to anode that doesn't pass through the capacitor, so there will always be current through the new resistor and therefore it will always be draining some of the battery's emf. The more voltage that resistor drains, the less voltage that will pass across the capacitor, which in turn reduces the charge that builds up (Q_stored = V_dropC).

In question #56, there are two resistors to consider (one for the discharging loop and one in the simple filling loop). R_d is in the discharging loop, so it does not have any impact on the charge that builds up on the capacitor. So we need only consider the impact of R_c on the build up of charge on the capacitor. What makes this tricky is that as the capacitor fills, the current through the circuit drops, because there is only one path from cathode to anode, and it must pass through the capacitor. Over time, as the capacitor fills, it opposes the emf of the battery, resulting in less current through the filling portion of the circuit. This means that there is less current across R_c, and therefore less of a voltage drop across R_c. As the current across R_c drops to zero, the voltage drop across R_c drops to zero, meaning that the capacitor eventually gets all of the battery's emf. So R_c affects the fill rate, but once the capacitor is full, R_c has no impact on the filling loop of the circuit and takes away none of the battery's emf.

I'm not sure what version of the book you have, but the one I'm looking at (from 2006) explains this fairly well. What is the copyright year for your book? As milski pointed out, you have a different book from what most people are using (it changed drastically a little over three years ago). I fully get trying to save money, but the BR physics book is one area where skimping is not a great idea. Brand new the pair costs $65 and used you might be able to find them for as low as $50, which is a drop in the bucket when you think about the cost of the MCAT itself, the amount you paid for the online course you took, and the cost of applying to medical school.