Capacitor problem - Calculate the heat in connecting wires

[LHC_23]

Capacitor problem -- Calculate the heat in connecting wires

Homework Statement

A $5.0$$\mu$$F$ capacitor is charged to $12 V$ . The positive plate of this capacitor is now connected to the negative terminal of a $12 V$ battery and vice versa. Calculate the heat developed in the connecting wires.

Homework Equations

$W = QV$

$E = 0.5CV^2$

The Attempt at a Solution

I know from the above 2 equations that half of the energy supplied by the battery should be lost as heat . But I don't know how to proceed with this problem ?

 

[rude man]

Homework Statement

A $5.0$$\mu$$F$ capacitor is charged to $12 V$ . The positive plate of this capacitor is now connected to the negative terminal of a $12 V$ battery and vice versa. Calculate the heat developed in the connecting wires.

Homework Equations

$W = QV$

$E = 0.5CV^2$

The Attempt at a Solution

I know from the above 2 equations that half of the energy supplied by the battery should be lost as heat.

That statement is incorrect since there is an initial charge on the capacitor.
In this particular case, what is the capacitor stored energy before (t<0) and after (t=∞)? Which says what about the fraction of battery energy lost to heat?

So, back to fundamentals:

You need to write the KVL around the loop including consideration of the initial voltage on C.
This starts as an integral equation in current i but you can make it a 1st order differential equation in i with the appropriate initial condition, getting you i(t), then integrate R∫(i^2)dt from 0 to infinity for the total power dissipation where R is the wire resistance.

There may be a shortcut here using only initial and final capacitor stored energy but I don't see one.

 

[Tanya Sharma]

Hello rude man...

Do you get 2CE² as the heat dissipated as well as the work done by battery ? Here C is the capacitance and E is the EMF of the battery .

 

[ehild]

Hello rude man...

Do you get 2CE² as the heat dissipated as well as the work done by battery ? Here C is the capacitance and E is the EMF of the battery .

It is right.

ehild

 

[rude man]

Hello rude man...

Do you get 2CE² as the heat dissipated as well as the work done by battery ? Here C is the capacitance and E is the EMF of the battery .

Yes I do Tanya.