A capacitance $C$, a resistor $R$ and an emf $\epsilon$ are connected in series at $t=0$. What is the maximum value of the power delivered by the battery?

I got the following image from an answer:

But I don't understand how to determine the power delivered by the battery. I understand that power is the rate of energy delivered. We know the energy delivered. But we don't know the time to determine the rate. Could you guide me in this regard, sir?

@JohnRennie: Hi sir. I notice that you're busy in the CodeClub chat room; Can you clarify the above doubt once you're free sir?

@GuruVishnu I'll ping you when I'm free ...

Thank you sir.

@GuruVishnu power is just IV. Yes?

@JohnRennie Yes sir.

The voltage across the battery is always just $V$, so if the current through the battery is $I$ then the power being supplied by the battery is $P = IV$.

Where it gets complicated is that the voltage across the resistor and capacitor change with time so the power dissipated by the capacitor and resistor change with time.

@JohnRennie But isn't battery a non-ohmic device, sir? If so how could this result be applicable here?

The battery supplies power while resistors etc dissipate power, so yes they are different. But conservation of energy requires that the power supplied is equal to the power dissipated. Yes?

@JohnRennie Yes sir. So power supplied by the battery equals the power delivered to the capacitor plus the power dissipated by the resistor. Are we just calculating the power based on the energy losses? And it seems this coincides with the usual result for Ohmic devices.

Yes

@JohnRennie: Ok sir. Thank you for your help. I understood this :-)

:-)

Sir, was this the silliest doubt anyone has asked you? :-) I suspect this was very easy and I didn't realise it and was thinking about it for a long time.

It's often the apparently simple ideas that are hard to grasp. They look simple but are actually really fundamental. A lot of the principles in relativity are actually very simple, but students find them hard to grasp.

:-)