Consider having a circuit which consists of a battery and one resistor.
$V = 10$ volts, $R = 5$ ohms, so $I = 2$ Amperes, and
$P = 20$ watts.
If we double the voltage and resistance, the current will be the same and the power will be equal to 40 watts, hence the resistor will be hotter than the first case.
Now here is the silly question.
The current in each of the two cases is constant and equals the current of the other. The speed which charges move across the wire is constant. So why is more heat generated while the speed of charges is constant? I know that the potential is doubled, but the potential is potential, and we can't make use of potential energy unless it is converted to kinetic energy. How can the resistor make use of this (potential) energy, the mechanism which the resistor turns potential energy into heat.
Best Answer
Your initial circuit is like this:
So you get 2A flowing and a power of 20W dissipated in the resistor.
Then you double the voltage and the resistance:
The two batteries add up to single source of 20V. The two resistors add up to a total resistance of 10$\Omega$. So, as you correctly state, the current is the same as before (2A). Therefore the power is now 40W
But this power is shared between the two resistors: 20W each, exactly as before.
You might also note that the potential at the point between the resistors is 10V, so each resistor has 10V across it, exactly as before.
Really all you've done is doubled up the circuit so that you have twice of what you had before.