How Is Energy Loss Calculated in Connected Capacitors?

In summary, the total energy loss in the capacitors is 0.02 J. This is calculated by finding the common potential difference, equivalent capacitor, and final energy stored in the combination of the capacitors. The formula used was E= Q^2/2C, with a resulting value of 0.02 J.
  • #1
thereddevils
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0

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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  • #2
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.
 
  • #3
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
 
  • #4
Common voltage V = (Q1 + Q2)/(C1 + C2)
 
  • #5
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 ?
 
  • #6
That is correct.
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device that stores electrical energy by creating an electric field between two parallel plates. It consists of two conductive plates, separated by a dielectric material, and is used to store and release electrical charge.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by accumulating opposite charges on the two plates, creating an electric field between them. When a voltage is applied, one plate becomes positively charged while the other becomes negatively charged. The electric field between the plates stores energy in the form of electrostatic potential energy.

3. What is the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is determined by the area of the plates, the distance between them, and the dielectric constant of the material between the plates. It is directly proportional to the area of the plates and inversely proportional to the distance between them.

4. What is the equation for calculating the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the equation C = ε0A/d, where C is the capacitance in farads (F), ε0 is the permittivity of free space, A is the area of the plates in square meters (m^2), and d is the distance between the plates in meters (m).

5. What are some practical applications of parallel plate capacitors?

Parallel plate capacitors have a wide range of practical applications, including filtering, energy storage, power factor correction, and electronic tuning. They are also commonly used in electronic circuits, such as power supplies, amplifiers, and filters, to store and regulate electrical energy.

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