4 ms·
Resistance is proportional to the square of current. This would mean 1/16th the loss instead of 1/4th.
by loufe 3y ago
Resistance is proportional to the square of current. This would mean 1/16th the loss instead of 1/4th.
- 05 3y agoNope. Consumed power is voltage*amperage, and it’s constant, so 4x more voltage means 1/4 the amps. Wire heating is I squared R, so with both I reduced and R increased the losses are just 1/4, not 1/16 of the 12V system. Because you wouldn’t keep the same wire gauges you’ve had in the 12V system, you’d reduce them to actually have some benefits from conversion.
- ahepp 3y agoIsn't the resistance of a wire usually modeled as a constant value? Power dissipated in the wire as heat is proportional to the square of the current. It's equal to the voltage drop times current (P = I * V). Voltage drop is current times resistance (V = I * R). So P = I^2 * R.
- deleted 3y ago[deleted]