A 20-m-long extension cord used to power a 1000-W circular saw has a resistance

Question

A 20-m-long extension cord used to power a 1000-W circular saw has a resistance of 1.0 ohm (higher than normal, so it is worn). (a) Assuming that 120-V electricity is used, calculate the voltage that actually gets to the saw. (First calculate the resistance of the saw assuming it gets the full voltage; then add the two resistances together and find the current in the extension cord.) (b) What is the actual power consumption of the saw if its resistance is constant? (c) Repeat both (a) and (b) assuming that the saw can be modified to use 220-V electricity. This is often done to save on power lost in extension cords and is a major advantage of using 220-V electricity.

Explanation / Answer

power = V^2 / resistance

1000 = 120^2 / resistance

resistance of the saw = 14.4 ohm

both the resistances are in series so,

total resistance = 1 + 14.4

total resistance = 15.4 ohm

by ohm's law

V = IR

120 = I * 15.4

I = 7.79 A

voltage = 1 * 7.79

voltage that reaches to the saw = 120 - 7.79

voltage that reaches to the saw = 112.21 V

actual power consumption = 112.21^2 / 14.4

actual power comsumption = 874.38 W

when voltage is 220 V

resistance of the saw = 220^2 / 1000

resistance of the saw = 48.4 ohm

current = 220 / (48.4 + 1)

current = 4.453 A

voltage that reaches the saw = 220 - 4.453 * 1

voltage that reaches the saw = 215.547 V

actual power = 215.547^2 / 48.4

actual power = 959.92 W

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