Maximizing Power Transfer in Electrical Circuits

[kira506]

Homework Statement

It is required to transfer an electric power of 80. Kilowatts from a powerstation to a factory which is at 2 km from it.If the potential difference (here it means the voltage of the source) at the station is 400 volts and the resistance of 1km of the 2 conducting wires is 0.1 ohm, find the power lost

Homework Equations

Power=VI (where v here is the potential difference between the source ie: power station and the output : factory ) or I^2.R
voltage in secondary coil(factory)/voltage in primary coil (station)=current intensity in primarycoil/current intensity in secondarycoil

The Attempt at a Solution

P= VI
80000=400I ,I=200 ampere
Here's the tricky part for me , In case of connecting the 2 wires in parallel , resistance of one wire = 2*0.1=0.2ohm
Rtotal=0.2/2 (since the 2 wires have the same length and assumingly made of the same material and have the same cross sectional area ,therefore we consider their value)=0.1ohm
The model answer shows that Rtotal =0.4 ohm ,how ,? We won't get this value unless we connect them in series and usually in these kinds of problems the wires are connected in parallel ,besides , nothing in the problem indicates the kind of connection , please help

 

[BvU]

One wire to bring the current to the user, one wire to bring it back to the powerstation. A schoolbook simplest possible kind of connection...

 

[adjacent]

Here's the tricky part for me , In case of connecting the 2 wires in parallel , resistance of one wire = 2*0.1=0.2
Rtotal=

How do you know that the two wires are connected in parallel?
When you connect a bulb to a battery by two wires,are the considered parallel?

 

[kira506]

How do you know that the two wires are connected in parallel?
When you connect a bulb to a battery by two wires,are the considered parallel?

Actually , I've never tried that so I don't know

 

[kira506]

One wire to bring the current to the user, one wire to bring it back to the powerstation. A schoolbook simplest possible kind of connection...

Out of curiosity , why do they need a wire to bring it back ?

 

[BiGyElLoWhAt]

P = IV
also
V = IR
That should be all you need to solve the problem.

 

[BvU]

Otherwise the charge would accumulate at the end and by the (incredibly short) time enough charge is accumulated to generate 400 volt, the charge would stop flowing... In other words:

You need to close the circuit to make the current flow

In reality powerstations transfer three phases over their wires and the fourth ground wire can be thinner.

DId you get an answer for the (considerable) power loss ? And the voltage the plant receives ?

 

[adjacent]

Actually , I've never tried that so I don't know

Oh come on...

Edit: BvU types faster than me.

 

[kira506]

Otherwise the charge would accumulate at the end and by the (incredibly short) time enough charge is accumulated to generate 400 volt, the charge would stop flowing... In other words:

You need to close the circuit to make the current flow

In reality powerstations transfer three phases over their wires and the fourth ground wire can be thinner.

DId you get an answer for the (considerable) power loss ? And the voltage the plant receives ?

I think I get why there should be two wires but I don't get the part where we have to close the circuit to make the current flow , I mean in case this happened , the potential difference would be zero so even if we closed the circuit (isn't already closed?) No current will flow , right ? Uhm and in the figure , I'm lost ;c; please help , now I can't differentiate between connection in series and in parallel. ( so sorry ;^;) and thank you so much c:

 

[kira506]

Oh come on...

Edit: BvU types faster than me.

I feel so dumb for not getting it XD. Of course I have but can I consider the factory a bulb ? (I mean in terms of connection please don't take it as me being sarcastic , I'm seriously trying to understand it ) and Lol at "BvU types faster than me" , thanks a lot c: