Finding charge of a capacitor in RC circuit

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Homework Help Overview

The discussion revolves around finding the charge on a 4 µF capacitor in an RC circuit with given resistor values. Participants are exploring the relationship between voltage, current, and capacitance in the context of capacitors connected in parallel with resistors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the total resistance in the circuit and question whether it should be treated as series or parallel. There are inquiries about the appropriate voltage to use in the capacitance formula and the implications of the capacitor being charged.

Discussion Status

Some participants have provided calculations and confirmations regarding current and voltage across components. There is an ongoing exploration of the relationship between the voltage across the resistors and the capacitor, with some guidance offered on understanding the connections in the circuit.

Contextual Notes

Participants are navigating the complexities of circuit analysis involving capacitors and resistors, with some uncertainty about the definitions and relationships between voltage and current in this configuration.

Teptip
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Thanks, solved!
 
Last edited:
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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
 

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