vcaa yr 2006 exam 2 question 10

[Torud]

hey,
I'm confused about question 10 in the year 2006 the physics exam 2. They say that in order to prove that the current is 4.55 A at point R, they divide the voltage by the power (I=P/V). However the voltage they used is 22000V, which is the voltage at the secondary coil of the step-up transformer. Shouldn't the voltage be less because you need to subtract the voltage loss from that value to find the voltage at point R?

Here's the question here!
http://www.vcaa.vic.edu.au/Documents/exams/physics/2006physics2-w.pdf

[rayray17]

Hi,
When i was looking at the question i didnt look at anything after the first transformer. The current in the transmission lines (current at r) is constant throughout the transmision lines so the current is the same as it is at start of transmission. As you have correctly said 22000v is the voltage at the secondary coil of the step up transformer. This transformer here is the one on the left. At this point no power has been lost as the voltage hasn't been transmitted yet. Therefore current is 100,000 watts/250v

[Torud]

Hi,
When i was looking at the question i didnt look at anything after the first transformer. The current in the transmission lines (current at r) is constant throughout the transmision lines so the current is the same as it is at start of transmission. As you have correctly said 22000v is the voltage at the secondary coil of the step up transformer. This transformer here is the one on the left. At this point no power has been lost as the voltage hasn't been transmitted yet. Therefore current is 100,000 watts/250v

wouldn't there be voltage loss from the resistance in the wire? o.o