BR example question #9.7b

[pineappletree]

---

Members do not see this ad.

What is going on here?

It is obvious from the graph that A has a larger capacitor
But why does B have a larger resistor??


This doesn't seem to make any sense because of the formula of q=VC
A larger resistor should increase the voltage (V=IR) and INCREASE the value of q



Why am I wrong here??
It is also clearly not a factor of charging time as B has plateaued out

 

[lorenzomicron]

post the problem?

 

[pineappletree]

There is a graph of Q vs. t
There are two lines on the graph

The first one (A) immediately reaches a very high charge and plateaus quickly
The Second line (B) has a much smaller slope, and plateaus later at a charge half that of A

The question is:

Two different capacitors, each connected to an identical voltage supply but a different resistor, are charged. Their charge, as a function of time, is plotted to the left. What is true of the two RC circuits?
A.) A has a larger capacitor; B has a larger resistor
B.) B has a larger capacitor; A has a larger resistor
C.) A has a larger capacitor; A has a larger resistor
D.) B has a larger capacitor; B has a larger resistor

The answer is A, but why?

I know that the problem says the voltage supply is the same, what relevance does this bit of information have to the problem? Considering that if V would remain constant, and R rises, then current should decrease. But considering that Current doesn't affect the capacitor whatsoever (in either charging time nor full charge), I have no real idea where the problem is going....


As a side note:
Also, there is the fact that a larger capacitor takes LONGER to charge than a smaller capacitor. I only know that A has a larger capacitor from the fact that it ends up at a higher charge.
But considering that B took a lot longer to charge, if they eventually ended up around the same charge, can I assume that B has a larger capacitor??





Please help!!
I am completely lost~~

 

[fizzgig]

since VC=Q, the one with higher Q has higher capacitance
-so height difference accounted for.

now time difference.
tau = RC for RC circuits
since B takes longer, its time constant is larger. we already established its capacitance is lower, so its resistance has got to be higher to raise tau.

 

[ilovemedi]

I have the new TBR physics, and I still don't get this.. is this a typo? Answer: Resitor A must have a lower reistant than Resitor B" and the answer says it's C: A has a larger resistor... typo?

 

[ilovemedi]

Anyone?????? 🙁

 

[Chuka115]

yes, typo

 

[gtbROX]

It's not a typo. Your error is that you assumed voltage can change, but it CANNOT CHANGE UNLESS YOU CHANGE THE BATTERY (VOLTAGE SOURCE). I caps-ed that phrase because it is something that we as premeds tend to skip over. Voltage depends ONLY ON YOUR BATTERY/SOURCE. In any equation with Voltage, especially proportionality, you have to change other variables in order to keep voltage constant.
If R increases, then I should decrease. What is I? It is current, but what is current? It is Q/t (charge over time), now look at the title of your graph (Q vs. t). What is the slope of your graph? Change in charge over the change in time deltaQ/deltat, which is CURRENT! ---> I
This is why B has a greater resistor and a smaller slope. The point at which it plateaus is its maximum charge that it can store (Q), and we know that Q=CV, but since we established that V SHOULDNT CHANGE, we know that instead, capacitance(C) of B must be smaller than that of A since C is directly proportional to Q and the Q is larger in A. As such, A has a larger capacitance, and B has the larger resistance.

Also, the Q-stem even said that the voltage is identical for either. Make sure you work through this problem on your own too and aim to understand this, because capacitor/circuits are common on the mcat.

 

[BerkReviewTeach]

It's not a typo. Your error is that you assumed voltage can change, but it CANNOT CHANGE UNLESS YOU CHANGE THE BATTERY (VOLTAGE SOURCE). I caps-ed that phrase because it is something that we as premeds tend to skip over. Voltage depends ONLY ON YOUR BATTERY/SOURCE. In any equation with Voltage, especially proportionality, you have to change other variables in order to keep voltage constant.
If R increases, then I should decrease. What is I? It is current, but what is current? It is Q/t (charge over time), now look at the title of your graph (Q vs. t). What is the slope of your graph? Change in charge over the change in time deltaQ/deltat, which is CURRENT! ---> I
This is why B has a greater resistor and a smaller slope. The point at which it plateaus is its maximum charge that it can store (Q), and we know that Q=CV, but since we established that V SHOULDNT CHANGE, we know that instead, capacitance(C) of B must be smaller than that of A since C is directly proportional to Q and the Q is larger in A. As such, A has a larger capacitance, and B has the larger resistance.

Also, the Q-stem even said that the voltage is identical for either. Make sure you work through this problem on your own too and aim to understand this, because capacitor/circuits are common on the mcat.

You are 100% correct there, but that's choice A. The explanation in the Berkeley book is correct (like yours), but they then list choice C as the correct answer. The letter in the key is the typo.