If an electric motor is rewound to cut its resistance by half, what effect does this have on motor power?

When the resistance of an electric motor is cut in half by rewinding, the impact on power can be understood through the relationship between voltage, current, power, and resistance as described by Ohm's Law and the power formula.

Ohm's Law states that ( V = I \times R ), where ( V ) is voltage, ( I ) is current, and ( R ) is resistance. The power ( (P) ) consumed by the motor is given by the formula ( P = V \times I ).

To analyze the scenario where the resistance is halved, we can assume that the voltage supplied to the motor remains constant. If the resistance is halved, the current drawn by the motor would increase. Specifically, if original resistance is ( R ), then halving it yields ( \frac{R}{2} ). According to Ohm's Law, the current can be expressed as ( I = \frac{V}{R} ). With the new resistance, the current becomes ( I' = \frac{V}{\frac{R}{2}} = \frac{2V}{R} ), which indicates that the current has doubled.

Now, knowing that power is also