To solve this problem we will apply Ohm's law. From there we will calculate the current with the given power. The current will allow us to calculate the resistance of the lost power. Later in state two of 345kV we will repeat a similar procedure, in order to find the lost power.
According to Ohm's law the power can be defined as
[tex]P = VI[/tex]
Here
V = Voltage
I = Current
Replacing,
[tex]150kW = 10kW I[/tex]
[tex]I = \frac{150}{10}[/tex]
[tex]I = 15Amp[/tex]
Now we can find the resistance
[tex]R = \frac{P_{loss}}{I^2}[/tex]
Replacing,
[tex]R = \frac{1000}{15^2}[/tex]
[tex]R = 4.444\Omega[/tex]
At the second state when the voltage is 345kV we have that,
[tex]I = \frac{150*10^3}{345*10^3}[/tex]
[tex]I = 0.437A[/tex]
Now power loss,
[tex]P_l = I^2 R[/tex]
[tex]P_l = (0.437)^2(4.444)[/tex]
[tex]P_l = 0.84W[/tex]
Therefore the power loss to transmit the same amount of power is 0.84W
