High voltage transmission

Thread Starter

cheddy

Joined Oct 19, 2007
87
I am reading about Transformers and have a question about the specifics of reducing loss in a transmission line.

The book says because engineers cant do much about the wire resistance or the power consumed by the loads they can adjust the voltage and thereby the current using P=EI. If the voltage was increased by a factor of 10 then the current could be reduced by a factor of 10 resulting in reduced loss.

How is it possible to have for example

100W = 10 V * 10 A

and

100W = 100V * 1 A

If the resistance is the same in both? Doesn't that violate Ohms Law?
 

Thread Starter

cheddy

Joined Oct 19, 2007
87
I thought you could only have 10 Amps with 10 Volts and 1 Ohm?

How can you have 1 Amp with 100 Volts if the resistance is still 1 Ohm?
 

mrmeval

Joined Jun 30, 2006
833
It's a formula based on real world observation.

P=V*I This doesn't care what the resistance is it just tells you what the power is *after* you've gotten an answer for voltage and current.

If P=1000 finding all values of V and I to solve the equation would be near infinite.

If P=1000 and V=0.1 I=10,000
if P=1000 and V=1 i=1000
if P=1000 and V=3.1415929 i=318.310

Simple algebra.
 

Dave

Joined Nov 17, 2003
6,970
I am reading about Transformers and have a question about the specifics of reducing loss in a transmission line.

The book says because engineers cant do much about the wire resistance or the power consumed by the loads they can adjust the voltage and thereby the current using P=EI. If the voltage was increased by a factor of 10 then the current could be reduced by a factor of 10 resulting in reduced loss.

How is it possible to have for example

100W = 10 V * 10 A

and

100W = 100V * 1 A

If the resistance is the same in both? Doesn't that violate Ohms Law?
The losses are due to P = I^2*R losses - you reduce the current by a factor of 10 the power loss reduces by the same fator for a constant TM line resistance, R.

If you are wondering where P = I^2*R comes from, take P = VI and factor in Ohm's Law V= IR.

Please note I use V where you use E.

Dave
 

Ron H

Joined Apr 14, 2005
7,014
The losses are due to P = I^2*R losses - you reduce the current by a factor of 10 the power loss reduces by the same fator for a constant TM line resistance, R.
I think you meant to say that if you reduce the current by a factor of 10, the power loss reduces by a factor of 100. ;)


If you are wondering where P = I^2*R comes from, take P = VI and factor in Ohm's Law V= IR.

Please note I use V where you use E.

Dave
 

Dave

Joined Nov 17, 2003
6,970
I think you meant to say that if you reduce the current by a factor of 10, the power loss reduces by a factor of 100. ;)
I have just got in after a night out (its 00.52 here!) ;) :D

Yes you are correct - the power reduces by 10^2 because of the I^2.

Dave
 

recca02

Joined Apr 2, 2007
1,214
How is it possible to have for example

100W = 10 V * 10 A

and

100W = 100V * 1 A

If the resistance is the same in both? Doesn't that violate Ohms Law?
yes that wud have violated ohm's law, but that is not what the case is.
u never connect a 10V device to a 100 volts device.
all u need is 100 watt power an if u are taking it from 100 volts outlet u'll design it for
drawing 1A current.

now suppose about 500MW power is to be transmitted thru a 100 ohms resistance of transmission what do u think is better idea:
to transmit it at 10000V or 100V thru the transmission line.
(it can will be stepped to lower voltage later).
the secondary will determine how much power is going to be consumed based on which
the amount of current in primary is decided and since the voltage ratio is higher on primary side(in distribution transformer) the current for same power is low.

ur confusion will be removed if u transfer all the resistance from secondary to primary side.
or u can transfer all resistance(transmission resistance) to secondary side u'll see the transmission line resistance gets reduced by a factor of (transformation ratio)^2
as seen by secondary.
 
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