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Suppose I have a simple circuit with capacitors, a battery and a key. There Is a change in energy stored in capacitors when the key is closed. How will I find the heat developed in the system?

Is the use of work energy theorem below correct?

Work done by all forces = Change in kinetic energy = 0

(Work done by heat energy) + (work done by chemical energy in battery) + (world done by stored energy in capacitors) =0

Hence, we get

Heat developed = (Initial stored energy - Final stored energy) - (work done by battery)

But the solution to one problem gives the formula as

Heat developed = (work done by battery) + (Initial stored energy - Final stored energy)

I hope I am not simply messing with the signs.

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In the case of an ideal capacitor, it initially looks like a perfection short circuit when the switch is closed (I assume by "key" you mean switch). The battery charges the capacitor until the capacitor voltage equals the battery voltage. No heat is dissipated in an ideal capacitor, in ideal (no resistance) conductors, and in an ideal battery having no internal resistance.

Ideal capacitors do no exist. For that matter ideal conductors (except for perhaps super cooled conductors) and ideal batteries don't exist. There will always be some resistance in the circuit that dissipates heat. The amount of heat can be determined by considering the energy delivered by an ideal battery versus the energy stored in the capacitor.

The work done (energy delivered) by a battery in moving charge $Q$ between the capacitor plates is

$$E_{batt}=QV$$

Where $V$ is the battery emf.

The energy stored in the capacitor is

$$E_{C}=\frac{CV^2}{2}$$

The relationship between voltage, capacitance and charge is

$$C=\frac{Q}{V}$$

Substituting for $C$ in the previous equation we have

$$E_{C}=\frac{QV}{2}$$

This tells us the energy stored in the capacitor is one half the energy delivered by the battery. For conservation of energy, the the difference in energy is that dissipated as heat in the circuit is

$$E_{heat}=\frac{QV}{2}$$

Hope this helps.

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  • $\begingroup$ Ok got it. The energy provided by the battery is being converted to energy of the capacitor and heat. Am I right? But is there a way to derive the mentioned formula from work energy theorem? $\endgroup$
    – Mathboi
    Commented Jul 2, 2019 at 18:01
  • $\begingroup$ @Mathboi That's the gist of it, yes. $\endgroup$
    – Bob D
    Commented Jul 2, 2019 at 18:03
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The work energy theorem is a special case of the first law of thermodynamics, which says, "Heat given to the system + work done on the system = change in internal energy of the system"

When heat given to the system is zero, this becomes "change in internal energy (kinetic energy) = work done by external forces"

In this case, if we take the battery and the capacitor as a single system, no external work is being done on it. The heat given to the system is negative of the net heat released = -$CV^2/2$, which is also the change in internal energy of the battery + capacitor system. (Battery is losing energy $CV^2$, capacitor is gaining energy $CV^2/2$)

You can also consider the battery and the capacitors as different systems and use the above arguments to find the heat released, work done, and change in internal energy in each of them separately.

Also, there is no convention of "work done by heat energy".

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