Finding charge of a capacitor in RC circuit

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Teptip
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Thanks, solved!
 
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Teptip said:
Here is the circuit in question: http://imgur.com/glr2C4f

Homework Statement


Find the charge on the 4 µF capacitor if
R1 = 11 Ω and R2 = 4 Ω .
Answer in units of µC


Homework Equations


V = IR
C = Q/V


The Attempt at a Solution


We've never had to deal with capacitors in parallel with resistors before so I'm not sure how this works.

Do I assume the total resistance of the circuit is in series and thus = 16Ω ?

Yes, after a short time that the circuit was connected the capacitor is charged and does not take current any more.

Teptip said:
What voltage do I use for the C=Q/V equation?

The capacitor is connected parallel to the 4 ohm resistor. What do you know about the voltage across parallel connected circuit elements?

ehild
 
Okay so I can calculate current I=V/R = 19/16 = 1.1875 A

V total = (V of Capacitor) + (V of 11Ω Resistor)

V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625

19 = (V of Capacitor) + 13.0625
V of Capacitor = 5.9375

Then plugging into the capacitance formula:
Q = CV = (4µ)(5.9375) = 23.75µ

Does this work look good?
 
Teptip said:
Okay so I can calculate current I=V/R = 19/16 = 1.1875 A

The current is correct.

Teptip said:
V total = (V of Capacitor) + (V of 11Ω Resistor)
No, what do you mean of Vtotal? And you do not need it.

Teptip said:
V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625
You know the voltage across the 11 Ω resistor. Look at the figure. Where are the plates of the capacitor connected to?

ehild
 
Is the voltage I want then just the voltage across the 4Ω resistor?
 
Teptip said:
Is the voltage I want then just the voltage across the 4Ω resistor?

Yes. And what current flows though it?ehild
 
Ah is it the same current as throughout, since the capacitor is quickly charged?

(I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)
 
Teptip said:
Ah is it the same current as throughout, since the capacitor is quickly charged?

(I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)

Yes, they are connected to the same points, so the voltage must be the same across both. (The voltage does not "go through", it is across two points)

ehild