Physics question about power and voltage

You have Power, Voltage, Current & Resistance. You need to know two of these. Then you apply the appropriate formula to compute the other two unknowns.

Since you know voltage (400kv) and power (250MW), then you can computer current.
P = IV. I = P/V. 250x10^6/400x10^3 = 250/400 x 10^3 = 625 amps.

Knowing that you have 625 amps flowing through a 5 ohm cable, you have 3125 volts (E=I x R) across the cable length. So Power = I x E = 625 x 3125 = 1.95 MW loss.

A power station generates electricity at 250 MW and 400kV. The electricity I'd transmitted to a distant town along cables with a resistance of 5.0 ohms.


250 MW is the power
400 kV is the voltage

Power = Volts * Amps


1 MW = 1 megawatt = 1 * 10^6 watts
250 megawatt = 250 * 10^6 = 2.50 * 10^8 watts

1 kV = 1 kilovolt = 1 * 10^3 volt
400 kV = 400 * 10^3 volt = 4.00 * 10^5 volt

2.50 * 10^8 = 4.00 * 10^5 * Amps
I = 2.50 * 10^8 ÷ 4.00 * 10^5

I = 625 amps

This is the maximum amount of current that will flow through the long distance cables.

As current flows through the wire filament in a light bulb, the wire gets hot. This means some the electric energy is being converted into heat energy. The amount of heat energy produced each second is the amount of electric energy which is not converted into light energy.

As the current flows through the aluminum wires in the cable, the electrons collide with the aluminum atoms. This causes the aluminum atoms to vibrate at a higher velocity, so the average kinetic energy of the atoms increases. Temperature measures the average kinetic energy of the atoms, so the temperature of the wire increases. This means some of the electric energy is being converted into the kinetic energy of the atoms. Since the temperature of the aluminum wire is higher than the temperature of air around the wire, heat energy is transferred from the wire to the air.