circuit questions wacky in TBR

[echoyjeff222]

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For this one, there is no diagram and the answer key explains it as though they assume you know it's a resistor/capacitor in parallel. HUH? I don't get that because if there were another resistor in series, the voltage drop would definitely not be emf.

Also, along those same lines:

Wouldn't the equivalent resistance affect the max charge, since Q = CV and voltage drop wouldn't be the same if there were another resistor before the capacitor? Kind of weird ... Am I supposed to assume that it's getting all the voltage from the battery?

Wacky questions ...

 

[pepocho]

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[echoyjeff222]

What I'm saying is that if we put a resistor in series behind the capacitor, then it would affect 1) the max charge on the capacitor and 2) voltage across the capacitor would no longer be emf. The question is ambiguous because it didn't specify.

Voltage is lost before reaching the capacitor, so resistors in series behind the capacitor would affect the overall charge ...

 

[pepocho]

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[echoyjeff222]

Well since voltage drops across the first resistor, that means that there is less voltage available to drop across the capacitor. Since Q = CV, if V goes down then Q should go down. ?

 

[deleted388502]

Well since voltage drops across the first resistor, that means that there is less voltage available to drop across the capacitor. Since Q = CV, if V goes down then Q should go down. ?

Okay, I think you're confusing how resistance works in relation to a capacitor.

So through ohm's law, we know V = IR, right? We know how this works in a circuit where we don't have capacitors? You seem to have those concepts down decently, so in applying this to a capacitor, think about what a resistor actually does. It's preventing as much current as we want from flowing through. So current is flowing, but not as much current as the battery is capable of providing is flowing as quickly as we want because we have this resistance.

Thus, because of this resistance, charge does not build up as quickly. That's why for capacitors, we look at Q = CV and not resistance. Through Q = CV we don't account for resistance because all of that charge will build up, eventually. It'll take a while, but all the charge will build up. Resistance just makes it take a while.

So now going back to your question, question 23, the potential difference is literally the definition of emf - that's what the V supplying the voltage of your battery is. I don't know if you were overthinking this, but if it is fully charged and you look at Q = CV, that means that you have Qmax and Cmax, and therefore Vmax. If you have Vmax and it is in a circuit, then you have the max potential difference, and therefore it's equal to the emf of the battery.

I'm assuming the answer here is C, but literally just look at Q = CV. I think you're really just overthinking this. Capacitance does not involve resistance in the larger sense. Resistance is just determining the time it takes for that charge to build up, it's not affecting how much charge or affecting the capacitance. Q = CV tells us that the maximum charge a capacitor can accumulate is dependent on the capacitance of the capacitor itself (determined by C=A/d x Eo), and the potential difference of the battery. Resistance would simply affect how quickly that charge would be being transferred due to the flow of current being smaller, explained by V = IR. That's all it is.

Let me know if this does not make sense and I can try to help out more!!! 🙂

 

[BerkReviewTeach]

Well since voltage drops across the first resistor, that means that there is less voltage available to drop across the capacitor. Since Q = CV, if V goes down then Q should go down. ?

You have a completely valid point that the voltage drop across a capacitor may or may not be equal to the emf of the voltage source. With that acknowledged, it still comes down to determining the best answer.

23. It still won't be 0 V, because it's charged. It may or may not equal the voltage of the battery, because as you point out there may or may not be another circuit element (another capacitor or resistor) in series, taking away some of the circuit's voltage drop. From the equation Qmax = VC, we know that both C and D cannot be correct. So our choices are a) no, b) maybe, c) no, and d) no. Not all questions on the MCAT will be perfect, so it is important to develop the skills of (1) choosing the better of two right answers and (2) choosing the best of four wrong answers. It's not going to happen a lot, but this question is within the realm of viability. For what it's worth, I think this question moved from a passage to become a free-standing question, so that's where the issue in the explanation likely stems from. But that aside, B is still the best answer.

24. The maximum charge on a capacitor is found using the equation Qmax = VC, so changing either voltage across the capacitor or the capacitance will affect the maximum charge. Choice I is a given from the equation. Choice II will affect the voltage across the capacitor whether or not there is anything else in the circuit, series or parallel. Another circuit element in parallel will not affect the voltage across the capacitor and another circuit element in series will absorb some of the voltage of the circuit and thereby reduce the voltage drop. However, if you were to increase the emf of the battery, then it would increase all of the voltage drops experienced throughout the circuit, including the capacitor. So choice II will affect the maximum charge across the capacitor. Choices C and D can be tossed out, because those deal with resistors and not capacitors.

The wording of question 24 is such that it simply wants to know what the total charge depends on. Perhaps you are overthinking here, because whether or not there is anything else in series does not change the fact that the voltage of the battery will influence the amount of stored charge on a capacitor.

 

[echoyjeff222]

Thanks for the replies! Yeah, I was overthinking it after the fact -- I got the questions right based on "best of the four" but I was still a bit confused (those counterexamples I had).

Wait, so for 24 ... I wasn't wondering about the emf, I was wondering about C, the number of resistors in circuit. Like you said, if it's in series, wouldn't it reduce the voltage drop and therefore the max charge built up on the capacitor?