How Is Energy Loss Calculated in Connected Capacitors?

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SUMMARY

The discussion focuses on calculating energy loss in connected capacitors, specifically a 4 µF capacitor (C1) charged to 200 µC and a 3 µF capacitor (C2) charged to 300 µC. The total energy loss is determined by finding the common potential difference and the equivalent capacitance when the capacitors are connected in parallel. The final calculations yield an energy loss of approximately 0.02 J, confirming the correct approach to using the formula E = Q²/(2C) for energy stored in capacitors.

PREREQUISITES
  • Understanding of capacitor charge and capacitance (e.g., 4 µF and 3 µF capacitors)
  • Familiarity with energy formulas for capacitors (E = 1/2 CV² and E = Q²/(2C))
  • Knowledge of parallel circuit connections and equivalent capacitance calculations
  • Ability to calculate potential difference across capacitors
NEXT STEPS
  • Research how to calculate equivalent capacitance in parallel circuits
  • Learn about energy conservation in capacitor systems
  • Explore advanced capacitor configurations and their energy calculations
  • Study the impact of initial charge on energy loss during capacitor discharge
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone interested in understanding capacitor behavior in circuits.

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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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