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- Feb 12, 2011
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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....
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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