How Does Resistance Affect Power Loss in Electrical Circuits?

[Tomtom]

Hi. Something which has been bugging me a bit lately, is this:
I've learned at school that from P=V*I and V=IR, we can insert the second equation in the first, and get P=I^2*R. This equation shows (probably amongst more things) that the power emitted in a wire is the square of the current, and is why when transmitting large amounts of energy, high voltage is used, so that the current is small, and hence, power loss, too, is small.

However, by rearranging the second equation to I=V/R, I can insert it to give P=V^2/R. Now, why isn't this correct is the above situation? This should mean that we should have a large current, and small voltage.

 

[reasonableman]

I think you're over complicating this.

Just think of the P=VI equation. At constant voltage the power (delivered by the circuit) is proportional to the current. At constant voltage the power is proportional to the current.

For high power applications what do you mean high voltage is used? I think power cables both are high...

If there is a different approach it would be due to the resistance. In most electrical applications R is small. So in P=I^2.R a small resistance makes the power smaller. In P=V^2/R a small resistance makes P bigger. I should state this is all guess work on my part.

 

[LURCH]

I think that the key to understanding this is contained within your own statement;

...This equation shows (probably amongst more things) that the power emitted in a wire is the square of the current...

I added the emphasis because I believe it points to the specific line of thought you're looking for.

Did it?

 

[cjl]

Hi. Something which has been bugging me a bit lately, is this:
I've learned at school that from P=V*I and V=IR, we can insert the second equation in the first, and get P=I^2*R. This equation shows (probably amongst more things) that the power emitted in a wire is the square of the current, and is why when transmitting large amounts of energy, high voltage is used, so that the current is small, and hence, power loss, too, is small.

However, by rearranging the second equation to I=V/R, I can insert it to give P=V^2/R. Now, why isn't this correct is the above situation? This should mean that we should have a large current, and small voltage.

It is correct, but you cannot think of the current, voltage, and resistance as independent parameters. In P=V²/R, the current is fixed by the values of V and R, and cannot arbitrarily be changed. If you have a fixed resistance value, and you increase the voltage by a factor of 2, the current will also increase by a factor of 2, causing a power increase of 4x. This is why the V term is squared. The reason the R term is on the bottom is that if for a fixed voltage situation, you decrease the resistance to 1/2 its original value, you have just increased the current to twice the original value. Because the voltage is the same, the power is twice the original value as well (due to the increased current).

 

[Lojzek]

However, by rearranging the second equation to I=V/R, I can insert it to give P=V^2/R. Now, why isn't this correct is the above situation? This should mean that we should have a large current, and small voltage.

You must distinguish between the power lost in the transmitting wire and power lost in the whole circuit (wire+load).

we want to minimize power lost in the wire at the fixed power at load P(load):

P(wire)=I*V(wire)=I^2*R(wire)

Obviously we must minimize I. Since the same I also runs through the load, we get:

I=P(load)/V(load),

so we must maximize V(load)

The equation

P(wire)=V(wire)^2/R,

is correct, but useless since we can't calculate V(wire) directly from the
power or voltage on the load (we can only calculate it from current I).