Another one of them circuit questions

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The discussion focuses on calculating the charge on a capacitor in a circuit containing two resistors (2.0 ohms and 4.0 ohms), a capacitor (9.0 µF), and a 12V battery. The user successfully derived the current (I) as 2A using the equation 12V = 2I + 4I. Subsequently, they calculated the voltage across the 4-ohm resistor as 8V, leading to the charge (q) on the capacitor being 7.2 x 10^-5 C using the formula q = CV.

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The circuit in the drawing shows two resistors, a capacitor and a battery. When the capacitor is fully charged, what is the magnitude of q of the charge on one of its plates?

Here's what I got so far:

So 12= 2I + 4I

Since q=CV and I have C=9.0uF, all I need is the voltage. And since the voltage drop between parallel circuits is the same then all I have to find is the voltage at the node going into 4 ohm resistor? Sounds like a good method but I can't seem to get the voltage drop.

Any ideas? :rolleyes:
 

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okay for those who can't see the attachment because it's waiting approval:

Code:
|-R=2.0ohm-|--------|
|                |           |
V=12V        R=4ohm  C=9.0uF 
|                |           |
|__________|_______|

Ok where I placed the letters is where you'll get a resistor or capacitor or voltage etc.
ugh, the code thing doesn't work so well, but the R=4ohm is in the middle branch and the C=9.0uF is at the end branch.
 
Ack, I got it nvm.

Easy really, almost had it too

12=2I+4I, I=2

V=IR at the 4
V=2*4
V=8 Volts

q=CV
=9.0x10^-6Fx8Volts
=7.2x10^-5 C.

Thanks notaphysicsman :)
 

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