You are Here: Home >< Physics

Power and energy supply to a resistor help please! watch

1. So part f) for the efficiency would be (59/60) x 100 = 98.3%?
Right so for 2) d) I worked out the p.d. lost as 2000V, and then did 120000V - 2000V = 118000 V = 118 kV as the p.d. supplied to the town.

Then for 2) e) I used Ohm's law and did 118000/500 = 236 ohms, both right?

And for the efficiency what would it be?
That looks fine.

The efficiency I gave you earlier way back in post #13.
3. (Original post by Stonebridge)
That looks fine.

The efficiency I gave you earlier way back in post #13.
So (59/60) x 100 = 98.3%?
So (59/60) x 100 = 98.3%?
That's it.
The station starts with 60MW and the town gets 59MW because 1MW is lost in the cables.

Question done.
You got most of it right without our help. Just a couple of hints needed.

When you have problems with questions, always show what you have attempted, even if you think it's all wrong.
If you do, there are plenty of people on here who will be willing to help.
5. (Original post by Stonebridge)
That's it.
The station starts with 60MW and the town gets 59MW because 1MW is lost in the cables.

Question done.
You got most of it right without our help. Just a couple of hints needed.

When you have problems with questions, always show what you have attempted, even if you think it's all wrong.
If you do, there are plenty of people on here who will be willing to help.
Thanks for the help!

And right I will in future, thanks for the advice And yes I can see that now, thanks again! Repped

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: September 29, 2010
Today on TSR

Get the low down

University open days

• University of Exeter
Wed, 24 Oct '18
Wed, 24 Oct '18
• Northumbria University