Voltage in an AC circuit

Homework Statement

Now suppose the power supply is 1200 V, and the motor is rated at 100 W at this higher voltage. If the wires have a resistance of 7.0 ohms, what is the voltage drop across the wires? The voltage at the motor?

Power= Irms*Vrms
V=IR

The Attempt at a Solution

In terms of the voltage drop across the wires I did: V= IR, I= 100W/1200V= .083 A therefore V= .083A*7.0 ohms = .58 V. However, why is Irms instead of maximum current used to calculate the voltage drop? Also, should the voltage at the motor simply be 1200V, or should it be 1200V-IR (from the resistor)? Thank you!

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gneill
Mentor
Unless otherwise stated, usually the circuit voltages, currents, and ratings for components like motors assume RMS values.

The rating for a component like a motor assumes ideal conditions (no resistance in the wires from the power source). Your circuit conditions are not ideal, since the wires have resistance. I would suggest first determining an equivalent impedance for the motor from the rating information, and then drawing the circuit. The impedance of the wire and the impedance of the motor will form a voltage divider.

Perfect, thanks!