Year 12s: what will be the first way you research your future options? >>

A little help with part B?

Thanks in advance

For a) I got the answer as D, which is equal to a resistance of

3.67\Omega

For part b) you have the voltage generated at the power station, which is NOT the voltage in the cables- this electricity is being delivered somewhere and most of the voltage will be going to the city. Remember voltage is different at different points in a series circuit. You also have the power being produced at the power station (again, most of this is being dissipated in the city so this figure doesn't relate to the cables, it's at the source).

The thing that stays constant in a series circuit is the current. The current at the power station can be calculated as I=P/V and we have both voltage and power at the power station.

I=\frac{1 \times10^9}{400 \times10^3}.=2500A

This gives the current in the whole circuit, including the cables which are what you're interested in. You now have resistance and current for the cables, so use

P=I^2 R[br]P= 2500^2 \times 3.68 =23 \times 10^6 =23MW

So I think the answer's D.

Reply 2

RLJCMorrissey

The answer is E according to the mark scheme, so i guess you need to double the last 23MW?

Thanks for the help anyway, my mistake was in assuming it was a parallel circuit.