Calculating Heat Generated from Capacitors

[cupid.callin]

I was studying capacitors when I came across these three questions:
Please refer the jpg with the topic.

Note: I wrote the quoted text before I realized I could post the pic of questions … but I had written all this anyway and didn’t wanted it to go to waste … so I posed it anyway!

1st: A parallel plate capacitor is connected with a battery and after it is fully charged the separation b/w the plates is increased, --- of course capacitance decrease and charge flows backwards --- I am supposed to find heat generated in the process …

2nd: A capacitor is charged to some voltage V and then disconnected from it and connected to another battery of voltage V’ (<V) … so this time the current flows backwards … question is again to calculate the heat.

3rd: this time a capacitor is connected to a battery, removed after long time, connected to a capacitor of smaller capacitance! And again I need to calculate the heat generated!

Answer to 1st=
1st= 0
2st= 7.2 mJ
3st= 1.77 mJ

Now…
How am I supposed to know how to find the heat generated? What basis do I use… well my questions are insane I know... but just because I have no idea what to do!
Please help me find a way to find heat generated … not only for these cases but also for any other random case!

 

[collinsmark]

To start, note that the energy stored in a capacitor (of a given, constant capacitance C and voltage V) is

E = ½CV².​

You'll have to use this equation a little differently for each part of the problem. But essentially, it means finding the equivalent capacitance before and after; and also finding the voltage across the equivalent terminals, before and after. Then compare the energies before and after.

 

[cupid.callin]

But for Ques 3 work done on battery will be QV'
Shouldn't that be heat produced?

 

[cupid.callin]

Well I'll try what u said and get back if it works!

Thanks for help

 

[rwisz]

Hello,

Calculating the heat generated from capacitors involves using the formula Q = CV^2/2, where Q is the heat generated, C is the capacitance, and V is the voltage. This formula can be used for all three of the scenarios you mentioned.

For the first scenario, where the separation between the plates is increased after the capacitor is fully charged, the capacitance decreases and the charge flows backwards. In this case, you can use the formula Q = (C1 - C2)V^2/2, where C1 is the initial capacitance and C2 is the final capacitance. The difference in capacitance represents the amount of charge that has flowed backwards, and this can be used in the formula to calculate the heat generated.

For the second scenario, where the capacitor is initially charged to voltage V and then disconnected and connected to a lower voltage V', the formula Q = (V - V')^2/2C can be used. This represents the difference in voltage and the initial capacitance, which will determine the amount of charge that flows backwards and thus the heat generated.

For the third scenario, where the capacitor is connected to a battery, removed after a long time, and then connected to a capacitor of smaller capacitance, you can use the formula Q = (C1 - C2)V^2/2, where C1 is the initial capacitance and C2 is the final capacitance. Again, the difference in capacitance represents the amount of charge that has flowed backwards and can be used to calculate the heat generated.

In general, when calculating the heat generated from capacitors, you will need to know the initial and final capacitance, the difference in voltage, and the formula Q = CV^2/2. I hope this helps you in solving your problems and understanding how to calculate heat generated from capacitors.