Capacitors in parallel question

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Homework Help Overview

The problem involves two capacitors, a 4μF capacitor charged to 200μC and a 3μF capacitor charged to 300μC, connected with their positive plates together. The discussion centers on calculating the total loss of energy during this process.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial and final energy calculations, questioning whether the energy equations used are for initial or final states. There is also confusion regarding the configuration of the capacitors, whether they are in series or parallel, and how this affects capacitance and energy calculations.

Discussion Status

Some participants have provided guidance on calculating the total capacitance and energy stored in the capacitors. There is an ongoing exploration of the implications of the problem statement and whether it presents a trick question. Multiple interpretations of the arrangement of the capacitors are being considered.

Contextual Notes

Participants note the importance of unit consistency, specifically regarding the use of μF and μC as valid units for capacitance and charge. There is also a mention of the problem statement's wording and its potential implications for understanding the setup.

Shiuanchiam
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Homework Statement


A 4μF capacitor is charged to a charge of 200 μC,and a 3μF capacitor is charged to a charge of 300μC. Both the capacitors are connected so that the positive plates are connected together. What is the total loss of energy in the process?

Homework Equations

The Attempt at a Solution


Loss of energy= E initial -E final
By using Energy =1/2 (Q^2)/c
The energy I get is the final energy or initial energy?
I'm kind of confused.

Thanks.
 
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Think of the final situation as one ‘big’ capacitor C=c1*c2/(c1+c2). Q=q1+q2. Now you can calculate energies…
 
Start by finding the initial energy stored in the system. That is, the sum of the energies stored in the two separate capacitors.

I note that the problem statement specifies that the positive plates of the two capacitors are connected together, but doesn't mention the negative plates. Is the problem statement quoted exactly or have you presented it in your own words? The reason I ask is that it is possible that it is meant to be a bit of a trick question...
 
Replusz said:
Think of the final situation as one ‘big’ capacitor C=c1*c2/(c1+c2). Q=q1+q2. Now you can calculate energies…
By using that equation what we get is the capacity in series, but when positive plate is connected together, isn't it suppose to be parallel? Capacity =C1+C2
 
gneill said:
Start by finding the initial energy stored in the system. That is, the sum of the energies stored in the two separate capacitors.

I note that the problem statement specifies that the positive plates of the two capacitors are connected together, but doesn't mention the negative plates. Is the problem statement quoted exactly or have you presented it in your own words? The reason I ask is that it is possible that it is meant to be a bit of a trick question...
Thanks.Yes it is quoted exactly.I assume that the statement is trying to show that the capacitors are arranged in parallel?
 
Shiuanchiam said:
Thanks.Yes it is quoted exactly.I assume that the statement is trying to show that the capacitors are arranged in parallel?
Okay, if that's the case then assuming the capacitors are connected in parallel, what's the total capacitance of the combination?
 
gneill said:
Okay, if that's the case then assuming the capacitors are connected in parallel, what's the total capacitance of the combination?
Total Capacitance = 3μ+4μ =7μ
So the final energy =1/2 (200μ+300μ)^2/7μ
Then do the initial energy separately = (1/2 (200μ)^2/4μ )+(1/2(300μ)^2/3μ)
 
Okay, what are your results for initial and final energy? BY the way, "μ" alone is not a unit. You should get into the habit of attaching the actual units so that you can trace issues with unit consistency should they arise. Thus μF and μC are valid units for capacitance and charge.
 
gneill said:
Okay, what are your results for initial and final energy? BY the way, "μ" alone is not a unit. You should get into the habit of attaching the actual units so that you can trace issues with unit consistency should they arise. Thus μF and μC are valid units for capacitance and charge.

Ok take note of that, Initial energy =0.02Joule
Final energy =0.01786Joule

Total loss of energy=0.02-0.01786 = 2.14×10^-3 Joule

Is that correct?
 
  • #10
Looks good.
 

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