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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.

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  • $\begingroup$ 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. $\endgroup$ – DKNguyen May 24 at 3:51
  • $\begingroup$ @DKNguyen. If we still think about it, then even power delivered to the load is converted completely into heat. $\endgroup$ – Ammaarah Fatima May 24 at 3:53
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    $\begingroup$ 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. $\endgroup$ – DKNguyen May 24 at 3:54
  • $\begingroup$ Related DC motor. $\endgroup$ – Farcher May 24 at 7:11
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You are right, the flow of electricity (specifically: voltage x current) performs work in electric motors, as used in appliances and whatnot. Let me try to clear up your confusions:

Let us imagine an electric motor, with a certain internal resistance R. We connect it to a battery of voltage V but instead of letting the motor shaft turn, we lock it so although it produces a torque on the shaft, it cannot turn it. Since shaft power = (torque x RPM), and since it is not turning (RPM = zero), the motor is doing no work and delivering no power.

But now we notice the wire inside the motor is getting hot, because there's still a current flowing through it in the amount I = V/R. So in this case, all the power being delivered to the motor by the battery is being wasted as heat.

If we unlock the motor shaft, it will start to turn and as it does, it is now capable of performing work. For now we just let it spin as fast as it wants. Since its coils are now moving through the magnetic field of the motor, a reverse voltage is generated in the coils which "fights" the current flowing from the battery. At the motor's maximum speed, the back voltage equals the battery voltage and the current through the motor falls almost all the way to zero. The only work the motor is doing is overcoming a little bit of friction in its bearings, which is provided by a very small amount of (current x voltage). In this case, since the motor is not connected to a load of any sort (it's just spinning freely) its power output is zero and is power consumption is almost zero because the current flowing through it is almost zero.

Inbetween these two extremes of either no RPM or no torque, the motor is capable of performing useful work against a load (like a saw blade cutting wood, for example) and it will be drawing current from the battery. Part of the (current x voltage) is converted into shaft horsepower and part of it is still wasted as heat.

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  • $\begingroup$ Thanks a lot for clarifying my confusion. $\endgroup$ – Ammaarah Fatima May 24 at 5:05

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