Find the Capacitance when the switch is closed

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Homework Help Overview

The problem involves determining the capacitance of capacitor C2 when a switch is closed, resulting in a potential drop across capacitors C1 and C2, which are connected in parallel. The initial conditions include known values for C1 and its charge before the switch is closed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between charge and capacitance, questioning how to find the charge on C2. There is also a focus on the configuration of the circuit and the implications of the switch's position.

Discussion Status

The discussion includes various interpretations of the circuit setup and the behavior of the capacitors when the switch is closed. Some participants have offered insights into the charge conservation principle, while others have raised questions about the clarity of the problem statement.

Contextual Notes

There is ambiguity regarding the switch's configuration and its effect on the circuit, which may influence the understanding of the problem. Additionally, the initial charge values and voltage conditions are provided, but the exact arrangement of the components is not fully detailed.

SalsaOnMyTaco
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Homework Statement



When the switch is closed, Potential drops to 40v across C1 & C2(Parallel series). Find the Capacitance of C2

Homework Equations


Before switch is closed
C1=300uF and holds a charge of Q1=3e4 uC V=100vAfter switch is closed
V'=40v

The Attempt at a Solution


V1=V2

So Q1/C1=Q2/C2
Q=Q1+Q2
How could I find the charge on C2?
 
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SalsaOnMyTaco said:

Homework Statement



When the switch is closed, Potential drops to 40v across C1 & C2(Parallel series). Find the Capacitance of C2

Homework Equations


Before switch is closed
C1=300uF and holds a charge of Q1=3e4 uC V=100v


After switch is closed
V'=40v


The Attempt at a Solution


V1=V2

So Q1/C1=Q2/C2
Q=Q1+Q2
How could I find the charge on C2?

Might be useful if there was an indication how the capacitors and the switch are connected to each other.
 
jaIxytR.jpg
 
SalsaOnMyTaco said:
jaIxytR.jpg

SO where is the switch?
 
2ciTXkP.jpg
 
That's a two way switch so the concept of it being "open" or "closed" is meaningless. However in this case it's clear they mean it starts off to the right and moves to the left.

V1=V2 is incorrect (they even tell you it falls to 40V). It's the charge that stays constant.

Edit: Ah I see you what you mean about V1=V2. You are correct but that's irrelevant really. The charge is constant but the total capacitance changes when the switch is made.
 
Last edited:
I've solve it

Q=q1+q2

if q1=12,000

30,000-12,000=q2

q2=18,000

so C=q2/v

C=450uF
 

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