voltages

[Bumbl3b33]

So, I was always under the impression that voltages in a circuit had to add up to the voltage of the battery source, meaning that the voltage of the battery source = voltage across resistors + voltage across capacitors. I didn't know this, but it has come up a lot of times in Kaplan stuff, and even an AAMC test. This is true, yes?

However, a Kaplan question is: If instead of one, 2 resistors are added in series before a capacitor, the total time taken for the capacitor to charge is found to take longer. It can be deduced that: I had it between

b) the presence of resistors affects the final voltage across the capacitor plates
and
d) the presence of resistors hinders the flow of charge, thus reducing currrent in the circuit


I knew d was right, but why isn't b true in general? The kaplan explanation says " actually a capactitor charges to the same voltage whatever the resistance of the other parts of the circuit. That doesn't mesh at all with what I've seen before, because as I said ealier, I was under the impression that the voltages of all the components in the circuit had to add up; so if the ohms of resistance is an inherent property of a resistor then that won't change so if you decrease current by adding more resistors, then you decrease voltage of the resistors meaning, meaning that the voltage of the capacitors has to increase.....i know my thinking is wrong since Q=CV doesn't support this, but i don't know how to disprove myself/where my thinking is wrong....

Help?

nevermind, i got it; i answered my own question....

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[gabdolce]

So everything is hooked up in series:

Let's break it down (you should be stepping through this basic kind of methodology in these types of questions - a lack of clear reasoning can end up confusing very quickly):

Let's say we have a 10V battery supplying the potential for the circuit.

You will agree with me that there will be a voltage drop across that resistor right? V=IR. Let's say the V(resistor)=2.

Now because the capacitor is in SERIES with the resistor, the hi-potential terminal of the capacitor is now 8V in respect to "ground" which in this case, would be the (-)terminal of the battery.

If you now for the sake of simplicity, add another identical resistor in front of the capacitor, the voltage drop across those two resistors add up to 4 V, leaving the poor capacitor with only 6V now. The voltage across the capacitor has indeed changed.

Had this resistor(s) been in PARALLEL with the capacitor, you could add or removed all the resistors you wanted to, and the voltage across the capacitor would not have been altered.

 

[Bumbl3b33]

So everything is hooked up in series:

Let's break it down (you should be stepping through this basic kind of methodology in these types of questions - a lack of clear reasoning can end up confusing very quickly):

Let's say we have a 10V battery supplying the potential for the circuit.

You will agree with me that there will be a voltage drop across that resistor right? V=IR. Let's say the V(resistor)=2.

Now because the capacitor is in SERIES with the resistor, the hi-potential terminal of the capacitor is now 8V in respect to "ground" which in this case, would be the (-)terminal of the battery.

If you now for the sake of simplicity, add another identical resistor in front of the capacitor, the voltage drop across those two resistors add up to 4 V, leaving the poor capacitor with only 6V now. The voltage across the capacitor has indeed changed.

Had this resistor(s) been in PARALLEL with the capacitor, you could add or removed all the resistors you wanted to, and the voltage across the capacitor would not have been altered.

Hi, I think I answered my own question, but hte part that i've bolded, that's only true until the the capacitor is not fully charged right? Because my understanding is that as the capacitor charges, it's voltage would eventually become the same as the battery's and the resistor's voltages will get to 0 because current will eventually be 0?

 

[gabdolce]

Yes, if the only complete path is through the resistor and capacitor in series, after enough time, V(cap)=V(battery).