# Is power delivered( or work done) by battery completely converted into heat?

I am currently a $$9^{th}$$ grade student.
I recently learned about power delivered by batteries and also the heating effect of current.Consider a circuit consisting of a battery of e.m.f $$E$$ and a resistor of resistance $$R$$.(Assume the battery to be ideal i.e. no internal resistance)
Now this is how the author derived power delivered by the battery:
The current in the circuit is $$I=E/R$$. Due to the flow of current $$I$$ for time $$t$$, the amount of charge passed is $$Q=I*t$$.
By the definition of potential difference, work needed to move a charge $$Q$$ through a potential difference $$E$$ is $$W=QE$$. But $$Q=It$$
$$\implies W=EIt=I^2Rt$$.
So this is the work done by battery. Therefore power delivered by it is $$P=EI$$.
After this the author further explains the heating effect of current(Joule's law of heating) where he says that heat produced in a wire is proportional to square of current passing through it, it's resistance and also the time for which the current is passed.Then he writes this:
$$H\propto I^2Rt$$$$and hence H=I^2Rt$$.
This is where I am confused.
As it can be concluded from above that work done by battery is completely converted into heat of the resistor(since $$H=W$$), then how are the electrical appliances working. For example, If I replace that resistor in the circuit with a d.c. fan, then the fan has some wires in it which has some resistance. Then if the battery's work is completely converted into heat then from where is the fan receiving that extra energy to rotate.I would be glad if someone explains it to me.

• The voltage dropped across the wire is different than the voltage dropped across the load. Think leaky water pipes. Some of the water is lost along the way and not all of it makes it to your faucet. The same current flows through the wire to get to the load. That means different amounts of power are lost in the wire versus the load. – DKNguyen May 24 at 3:51
• @DKNguyen. If we still think about it, then even power delivered to the load is converted completely into heat. – Ammaarah Fatima May 24 at 3:53
• If your load is a resistor (like a heater) then yes. But again, like the water pipes, do you want most of your water to leak away in the pipes before it gets to your faucet? Or do you want it o leak away in your sink drain? Where matters. Not to mention that, not all loads are heaters or resistors. – DKNguyen May 24 at 3:54
• Related DC motor. – Farcher May 24 at 7:11