How Is Energy Loss Calculated in Connected Capacitors?

AI Thread Summary
The discussion focuses on calculating energy loss in connected capacitors, specifically C1 (4 µF) and C2 (3 µF), with initial charges of 200 µC and 300 µC, respectively. Participants clarify that the energy loss can be determined by first calculating the energy stored in each capacitor using the formula E=1/2 QV and then finding the common potential difference after connecting them in parallel. The effective capacitance is calculated as 7 µF, leading to a final energy calculation that initially yields incorrect results. Ultimately, the correct energy loss is confirmed to be approximately 0.02 J, emphasizing the importance of using the right formulas and understanding the system configuration.
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Homework Statement


A capacitor C1 (the one on top in the diagram) of capacitance 4 mu F is charged until its charge is 200 mu C and another capacitor C2 (the one below) of capacitance 3 mu F is charged until its charge is 300 mu C . Then both capacitors are connected in the circuit as shown in the figure . What is the total enerygy loss in the capacitors ?


Homework Equations



E=1/2CV^2 or E=1/2 QV

The Attempt at a Solution



I an not sure its asking for energy loss as a whole or energy loss for each capacitors .

so i know how to calculate the energy stored in each capacitors using the formula , what should i do after that >
 

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Find the common potential difference across the capacitors.
They are connected in parallel.
find the equivalent capacitor.
Find the final energy stored in the combination of the capacitors.
 
rl.bhat said:
Find the common potential difference across the capacitors.
They are connected in parallel.
find the equivalent capacitor.
Find the final energy stored in the combination of the capacitors.

ok let me try

the effective pd would be 100+50=150 V

Effective capacitance = 7 mu

energy loss = 1/2 x 7 mu x 150^2 = 0.07875 J ?

but the answer given is 0.002 J

btw , is energy stored in the capacitors the same as energy loss in the capacitors
 
Common voltage V = (Q1 + Q2)/(C1 + C2)
 
rl.bhat said:
Common voltage V = (Q1 + Q2)/(C1 + C2)

thanks , so let me retry ,

E= Q^2/2C

=[(500 x 10^(-6))^2]/2(7 x 10^(-6))

=0.018 which is approximately 0.02 J

Am i correct nw ?
 
That is correct.
 
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