How will the power of the hotplate change if the spiral length is reduced by 15% of the original length?

l2 = l1 – l1 * 0.15 = l1 * 0.85 – the length of the electric stove spiral was reduced by 15%;

It is required to determine W2 / W1 – how the power of the electric stove will change with a decrease in the length of its spiral.

Since the problem statement is not specified, we assume that the mains voltage in both cases will be the same and equal to U.

The power of the electric stove in the first case will be equal to:

W1 = U * I1 = U * U / R1 = U ^ 2 / (r * l1 / s) = U ^ 2 * s / (r * l1), where s is the coil specific section area, r is coil resistivity.

The power of the electric stove in the second case will be equal to:

W2 = U * I2 = U * U / R ^ 2 = U ^ 2 / (r * l2 / s) = U ^ 2 * s / (r * l2) = U ^ 2 * s / (r * l1 * 0 , 85) = U ^ 2 * s / (0.85 * r * l1).

W2 / W1 = (U ^ 2 * s / (0.85 * r * l1)) / (U ^ 2 * s / (r * l1)) = 1 / 0.85 times, that is, will increase by 15% …

Answer: if the length of the spiral of the electric stove is reduced by 15%, its power will increase by 15%.