Physics question

[blankguy]

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What is the charge on the capacitor in the circuit below after the circuit has been on a long time?

The circuit has a battery of 12V with a 2 Ohm resistor parallel to a 1uF capacitor with the capacitor being on the outer of the two branches which merge and before going back to the battery there is another 2 Ohm resistor

A. 1.2X10^-10 C
B. 2.5X10^-6 C
C. 6.0X10^-6 C
D. 1.7X10^-7 C

The answer is C because the charge is equal to the voltage times the capacitance. I don't get how they arrived at that answer. 😕

 

[theblastopore]

What is the charge on the capacitor in the circuit below after the circuit has been on a long time?

The circuit has a battery of 12V with a 2 Ohm resistor parallel to a 1uF capacitor with the capacitor being on the outer of the two branches which merge and before going back to the battery there is another 2 Ohm resistor

A. 1.2X10^-10 C
B. 2.5X10^-6 C
C. 6.0X10^-6 C
D. 1.7X10^-7 C

The answer is C because the charge is equal to the voltage times the capacitance. I don't get how they arrived at that answer. 😕

First, find the current going through the second resistor you mentioned. This is done by the Kirchhoff's Rule chosing the loop going from the "0-terminal" of the battery trhough the second resister u mentioned through the first one throught the "12-terminal" and back to the "0-terminal":
I*R2 + I*R1 - 12 = 0
I*2 + I*2 = 12
I = 3A

Now consider the bigger loop (going through R2 and the capacitor), as the current going through R2, the votage drops by
dV = I*R = 3*2 = 6V

This means the voltage at the capacitor is only 6V (12V - 6V). And you are right, Q = C*V = (1*10^-6)*(6) = 6*10^-16

Good luck

 

[grapeflavorsoda]

above poster's solution is pretty good, but it might be too time consuming since this kind of question is more likely given as a discrete.

just simply treat any capacitors after a long time as a resistor with an infinite resistance, like an open cirucit so no current goes through.

so your circuit will act like a simple series circuit where your 2 resistors are connected in series, thus the current through the circuit will be

I= V/R = 12V/(2 ohm + 2 ohm) = 3 amps.

so the voltage across the first resistor would be V = IR = 6 V.

since the wire connecting the resistor in the parallel circuit is also connected to the capacitor the voltage drop across capacitor should be same 6 V.

so the charge on the capacitor is Q = CV = 6 micro-V.



whenever you see capacitor in a circuit and explanation saying that the swtiched was on for a long time, treat the capacitor point as an open circuit(just think as if they have an infinite resistance so no current goes through it). So if a capacitor is in a circuit where it is in series with a resistor and connected to some battery, there will be no current flow will be going through the circuit after a long time.

 

[Shrike]

Grape's solution is better, obviously. Quick 'n dirty -- it's more than just a recipe for an evening of fun, it's also the right way to solve MCAT problems.

Another quick way here: Having decided the capacitor doesn't permit current and you therefore have two resistors in series, you can skip V=IR. You notice that their resistance is the same so their voltage drops are the same, hence 6V (half of 12V). The cap is parallel to one of those, hence sees 6V. Q=CV. Done.

 

[maxim3L]

Hi, what is the capacitor was in series with the resistors instead of parallel to one of them? How would that change the resistance on this capacitor?

 

[Vihsadas]

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