# Ohm's law and power equation

#### rakker579

Joined Jul 5, 2021
2
Hello I have a question about electricity. I know that transmitting power over long distance is better in AC. I also heard that the voltage needs to be high to reduce losses. I wanted to calculate this for myself.

Imagine there is a windmill that produces 100 KW of power. This 100 KW needs to be transported over a 1 km wire that has 1 ohm of resistance. We can transfer this 100kW of power at 1kV or 10kV. But we are going to look which one is most beneficial.

When we send the 100kW at 1kV:
- This means we have 100A (Power/Voltage) with a loss of 10kW (Current^2*Resistance).

When we send the 100kW at 10 kV:
- This means we have 10A (Power/voltage) with a loss of 0.1kW (Current^2*Resistance).

This proves that the losses are less when the voltage is high than when the voltage is low. However I noticed something odd. When we send the 100kW at 1 kV I calculated that the current is 100A via the power equation (100 000 / 1 000). But when I calculate the current via ohm's law I get 1kA (1 000 * 1). What am I doing wrong?

#### rakker579

Joined Jul 5, 2021
2
Hello I have a question about electricity. I know that transmitting power over long distance is better in AC. I also heard that the voltage needs to be high to reduce losses. I wanted to calculate this for myself.

Imagine there is a windmill that produces 100 KW of power. This 100 KW needs to be transported over a 1 km wire that has 1 ohm of resistance. We can transfer this 100kW of power at 1kV or 10kV. But we are going to look which one is most beneficial.

When we send the 100kW at 1kV:
- This means we have 100A (Power/Voltage) with a loss of 10kW (Current^2*Resistance).

When we send the 100kW at 10 kV:
- This means we have 10A (Power/voltage) with a loss of 0.1kW (Current^2*Resistance).

This proves that the losses are less when the voltage is high than when the voltage is low. However I noticed something odd. When we send the 100kW at 1 kV I calculated that the current is 100A via the power equation (100 000 / 1 000). But when I calculate the current via ohm's law I get 1kA (1 000 * 1). What am I doing wrong?
EDIT: found it, can't use ohms law because not the whole 1kV is used on the transmission wire

#### MrChips

Joined Oct 2, 2009
24,609
Ohm’s Law still works.
You are using the wrong value for the load resistance.
The resistance of the load is not 1Ω.

#### RBR1317

Joined Nov 13, 2010
674