Charge and Voltage on Capacitors.

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Homework Help Overview

The discussion revolves around determining the charge and voltage across capacitors in a circuit, specifically focusing on a scenario involving a battery voltage of 4.0 V and capacitors with a capacitance of 3 μF. Participants are examining the relationships between charge, voltage, and equivalent capacitance in series and parallel configurations.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the calculation of equivalent capacitance and the charge distribution among capacitors in series. There is confusion regarding the derivation of values for combined capacitors and the assumptions made about charge equality.

Discussion Status

Some participants are seeking clarification on the initial conditions and values of the capacitors, while others are discussing the validity of the equations presented. There is acknowledgment of differing interpretations of the circuit configuration and the relationships between the components.

Contextual Notes

There is a request for clearer definitions of the initial capacitor values and circuit conditions, as well as a need to address the assumptions regarding charge distribution in series capacitors.

Yut
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Homework Statement


Determine the charge on each capacitor and the voltage across ech, assuming C=3yF(micro) and the battery voltage is V=4.0 Vhttps://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11006374_10152682348496806_273021698509210726_n.jpg?oh=128f2d03e7e0ad275cb982e11d827c2c&oe=54F5E264&__gda__=1425403951_252b026b8105e21db6642c5d28a9f753

Homework Equations


so,
C23 = 2C/3 = 2microF
Qeq = Q1 + Q23 = Ceq V <--- why is Qeq is the sum of the series capacitors, i thought they had the same charge
Q eq = 8. micorC
Q1 = Q23 = 4.00 microC
V1= Q1/C = 1.33 V
V23 = V - V1 = 2.67 V
Q2 = Q3 = CV23 = 2.0 micro C <--- not sure how did they get this value, when I plug in the found valued I don't get 2.

The Attempt at a Solution

 
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The problem statement is unclear to me. You state that C = 3 μF, does that mean the circuit is fundamentally made up of a several 3 μF capacitors? If so, how do you arrive at C23 = 2 μF?

Capacitors in series, unless there are circuit changes when capacitors are already charged, should bear the same charge as you indicate.

Is the solution that you present as Relevant equations your solution or someone else's?
 
This is an example problem from one of my lectures, and I was not sure how they arrive to the solution.
Originally the C23 capacitor, was C2 and C3 capacitors in parrallel. They combined them into Ceq by additing them two.
 
Can you list all the initial capacitor values and the initial circuit conditions?
 
https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11042003_10152682449531806_1029796180_n.jpg?oh=fcaa90dacb0f79caf862a01ccfe11094&oe=54F5C0C6&__gda__=1425387735_717d66c414ad18a91e2dbebad58aa77e
 
Okay, it's much more clear now that I can see the whole circuit. Their derivation of the equivalent capacitance looks fine.

In your first post you showed and questioned the step:

Qeq = Q1 + Q23 = Ceq V

which I agree is incorrect. Series connected capacitors should have the same charge. So what would be your version of a solution?
 

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