transmission lines

[perfectscore]

what is the advantage of using a higher turns ratio in the transformers?

also, how does doubling this ratio affect the power loss in the transmission lines?

thanks!

[mark_alec]

Power loss is , so by increasing the number of turns, you increase the voltage, which decreases the current.

[perfectscore]

what about the boubling ratio

[Mao]

doubling the size of stepping up (1:2 --> 1:4) reduces power loss to 25% of what it was

halving the size of stepping up (1:4 --> 1:2) quadruples the power loss, i.e. 400% of what it was

[that is, assuming constant resistance. in reality, however, there will be a whole range of factors]

[perfectscore]

do we need to know the factors; could u list a few please? thx!

[fredrick]

I can answers dat but

EDIT: what was the question?

[perfectscore]

[that is, assuming constant resistance. in reality, however, there will be a whole range of factors]

what are these factors?

[Mao]

power loss in transistor due to eddy currents and dispersion of heat in the primary/secondary coils

power loss where the connections between cables are made

imperfection in cables ==> "hot spots" where more energy is lost

materials the cable is made out of (they may not be perfectly ohmic)

etc etc

[perfectscore]

what other advantages of using a higher turns ratio (in transformers?)?

[perfectscore]

"Power loss is , so by increasing the number of turns, you increase the voltage, which decreases the current."

is this the only way? how do you explain this further?

[Collin Li]

The other way you could decrease it is by lowering the resistance, since:

However, decreasing current is the most effective way. You will hit a wall in reducing resistance. Using a transformer is easy.

To explain it further, you talk about how a step-up transformer increases the voltage according to the turns ratio:

.

Since (for VCE) and , then:

and

So .

This means that if you are stepping up the voltage (increasing ), then you are decreasing and hence decreasing . This should be also be apparent by taking a constant and looking at what happens if must remain constant, while is increasing.

[perfectscore]

The other way you could decrease it is by lowering the resistance, since:

However, decreasing current is the most effective way. You will hit a wall in reducing resistance. Using a transformer is easy.

To explain it further, you talk about how a step-up transformer increases the voltage according to the turns ratio:

.

Since (for VCE) and , then:

and

So .

This means that if you are stepping up the voltage (increasing ), then you are decreasing and hence decreasing . This should be also be apparent by taking a constant and looking at what happens if must remain constant, while is increasing.

how does this relate to "the advantage of using a higher turns ratio in the transformer?"

[Collin Li]

A higher turns ratio means a larger , and hence a higher . This means you are increasing (the 1 and 2 swap positions because voltage and current have an inverse relationship for a constant given amount of power).

This ultimately means that you are decreasing for any given (so that increases). Where refers to the current in the 2nd coil (after the transformation), then this means now power being transmitted from this transformer will have less power loss following the formula

[perfectscore]

thanks-got it

[Mao]

by the conservation of energy, power (energy/time) is conserved as we can assume it is constant (so energy is transmitted at a constant rate)

since , we can either reduce the current or reduce the resistance

if we want to reduce resistance, the cross-sectional area of the cable must be increased. This is costly as well as unpractical.

instead, we reduce the current using a transformer. Since power is conserved in a transformer, we have , and hence , hence, as we want current in the secondary to go down, voltage in secondary must go up. this is a step-up transformer, hence the turn-ratio is large.

==> increasing turn ratio reduce power loss

[because you requested me to help?]