Power Loss in a 10kV, 5ohm Transmission Line

PERE = POWER / VOLTAGEIn summary, to calculate the power loss in a transmission line with a resistance of 5 ohms when 10,000 kW are transmitted at 10 kV, use the formula P = VI = I^2R where V is the voltage drop, I is the current, and R is the resistance. The current can be found by dividing the power by the voltage using the formula I = P/V.
  • #1
DB
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-A transmission line has a resistance of 5 ohms. Calculate the power loss in the line when 10 000 kW are transmitted at 10 kV...

I don't know where to start, what formula am I supposed to use? I'm sure the second I find out its going to be soo easy, but any help would be apreciated.

Thanks in Advance
 
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  • #2
DB said:
-A transmission line has a resistance of 5 ohms. Calculate the power loss in the line when 10 000 kW are transmitted at 10 kV...

I don't know where to start, what formula am I supposed to use? I'm sure the second I find out its going to be soo easy, but any help would be apreciated.
Use:

[tex]P = VI = I^2R[/tex] where V is the voltage drop over the transmission line and I is the current it conducts, and R is the resistance of the line.
What is the current if the 10 kV line is conducting 10,000 kW (that is just P = VI)?

AM
 
  • #3


To calculate the power loss in a transmission line, we can use the formula P = I^2 * R, where P is the power loss, I is the current, and R is the resistance. In this case, we are given the voltage (10 kV) and power (10,000 kW), but we need to find the current in order to use the formula. We can use Ohm's Law, V = I * R, to solve for the current.

10 kV = I * 5 ohms
I = 10 kV / 5 ohms
I = 2000 A

Now, we can plug in the current into the power loss formula.

P = (2000 A)^2 * 5 ohms
P = 4,000,000 W or 4 MW

Therefore, the power loss in the transmission line is 4 MW. This means that 10,000 kW (or 10 MW) of power is being transmitted, but 4 MW is being lost due to the resistance in the line. This highlights the importance of minimizing resistance in transmission lines to reduce power loss and improve efficiency.
 

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