P=IV & V=IR Question

[pegggu]

noobie question, but wondering if someone can clarify this for me.

i know according to Ohms law V =IR that the voltage and current are directly proportional that if one goes up the other does as well.

but when we use the power equation P=IV it is the opposite.

am i missing something here? I know in distrubtion and transmission they up the voltage so they can use smaller conductors as the current is alot smaller. but doesnt V=IR contradict this?

a bit confused here.

thanks

 

[david luchini]

noobie question, but wondering if someone can clarify this for me.

i know according to Ohms law V =IR that the voltage and current are directly proportional that if one goes up the other does as well.

but when we use the power equation P=IV it is the opposite.

am i missing something here? I know in distrubtion and transmission they up the voltage so they can use smaller conductors as the current is alot smaller. but doesnt V=IR contradict this?

a bit confused here.

thanks

V=IR says that for a CONSTANT resistance, the voltage is proportional to the current. If you applied 120V to 10ohm resistor, you would get a 12A current. If you applied 240V to the same 10ohm resistor, you would get a 24A current.

P=VI says that for the same amount of power, the current is inversely proportional to the voltage. If you had a 10000VA load at 120V, the current would be 83.8A. If you increased the voltage to 240V, the current would be 41.6A, so you could use a smaller conductor.

But, note that the resistance for a simple 10000VA load at 120V and 240V would NOT be the same. For the 120V load, the resistance would be 1.44ohm and for the 240 load, the resistance would be 5.76ohm.

So, V=IR does not contradict P=VI.

 

[charlie b]

What may be causing your confusion is the mistaken belief that V is the same as V, that I is the same as I, that R is the same as R, and the P is the same as P. What I mean is that you need to understand what specific values of voltage, current, resistance, and power are being considered in any given equation. The V you use in one equation might relate to the voltage at the energy source, or the voltage dropped along the circuit wires, or the voltage at the load location. Just showing the letter V does not convey all the information that is needed.

The other thing you need to keep in mind is that for different types of equipment, the thing that might change, the thing that might respond to that change, and the thing that stays constant throughout will be different. For example, in a resistive load, such as a light bulb or a space heater, it is the resistance that always remains the same. (OK, so the resistance changes as the device heats up, but for the sake of the present discussion let's ignore that.) So if you were to double the voltage applied to the heater, you would double the current. On the other hand, for motors, it is generally (again not precisely, but it's good enough an approximation for this discussion) the value of power (P) that is constant. So if you were to double the voltage applied to the motor, you would get half the current flowing.

 

[Sahib]

If you remove the constraint 'power is constant', then current will increase with voltage in P=VI consistent with V=IR.

 

[Smart $]

As David explained, when voltage (E) and current (I) vary proportionally, you are holding the resistance (R) constant. All three values are variables. For example, if you hold the voltage constant, I and R are inversely proportional when their values vary.

The basic power formula only uses two of those variables. Substituting I=E/R you get P=E²/R... or substituting E=IR you get P=I²R.

 

[JDBrown]

noobie question, but wondering if someone can clarify this for me.

i know according to Ohms law V =IR that the voltage and current are directly proportional that if one goes up the other does as well.

but when we use the power equation P=IV it is the opposite.

am i missing something here? I know in distrubtion and transmission they up the voltage so they can use smaller conductors as the current is alot smaller. but doesnt V=IR contradict this?

a bit confused here.

thanks

The bold part is a common misunderstanding, which Charlie touched on in his response. What you have to remember is that V in these equations is the voltage difference between two points in the circuit. When dealing with a device, we just call this the input voltage -- one screw connects to L1 @ 120V, and the other screw connects to Neutral @ 0V, so the voltage difference across the device terminals is 120V. However, when dealing with transmission line losses we're not concerned with the voltage difference between Line and Neutral; we're concerned with the voltage at Point A on Line 1 compared to the voltage at Point B several miles away, also on Line 1. We commonly refer to this as voltage drop.

Once we realize that we're talking about voltage drop, rather than line voltage, it becomes clearer why there is no contradiction. As David pointed out, if you have, say, a step-down transformer that feeds several homes, the amount of power required will remain constant regardless of what voltage is on the transformer's primary winding (as long as the same loads are turned on, anyway). Perhaps an example is in order:

Let's say we have a step-down transformer that is supplying a constant 100A, 240V, single-phase to a building. The feeder on the primary side of the transformer has a resistance of 1 Ohm. We can calculate the power being fed to the building as:
P = IV = 100A x 240V = 24,000VA, or 24kVA

If the primary voltage is 480V, single-phase, then we can calculate the primary current with:
I = P/V = 24kVA / 480V = 50A
The voltage dropped across the feeder to the transformer can then be calculated with:
V = IR = 50A x 1 Ohm = 50V
So, you would actually need 530V at the source of your feeder in order to have 480V at the transformer primary. This is obviously way too high to be an acceptable voltage drop, but let's continue with the example. Our power loss in the feeder is:
P = IV = 50A x 50V = 2500W (notice that V is the voltage drop across the feeder, NOT the nominal system voltage)
This can also be calculated by:
P = I²R = (50A)² x 1 Ohm = 2500W (this is the form of the equation that we would normally use for this calculation)
P = V²/R = (50V)² / 1 Ohm = 2500W

If, however, the primary voltage is 12kV, then our calculations become:
Primary Current: I = P/V = 24kVA / 12kV = 2A
Feeder Voltage Drop: V = IR = 2A x 1 Ohm = 2V
Feeder Power Loss: P = I²R = (2A)² x 1 Ohm = 4W

This is why we use high voltages to traverse long distances -- it lowers the amount of current that must be delivered to the step-down transformer at the far end, which reduces the power lost as heat in the transmission lines. There are other factors that get considered in real life, such as skin effect, AC reactance, etc., but this gives the basic idea.

Sorry for being so long-winded.

 

[mgookin]

Those are some awesome tutorials there guys. It's been said that a smart person is not one with knowledge but one who can explain complex things in simple and logical fashion.

 

[Besoeker]

noobie question, but wondering if someone can clarify this for me.

i know according to Ohms law V =IR that the voltage and current are directly proportional that if one goes up the other does as well.

but when we use the power equation P=IV it is the opposite.

Only if you assume that P is constant. In many/most cases it isn't.

 

[gar]

140313-1712 EDT

pegggu:

Your question has nothing to do with electrical circuit theory, it is simply an algebra question.

In one equation I and V are on opposite sides of the equal sign, and in the other equation they are on the same side. Both equations represent a straight line thru 0,0 .

.

 

[Smart $]

... Both equations represent a straight line thru 0,0.

:?

If P and R are a constant non-zero value, a plot of possible E and I values would not be a straight line and not pass through 0,0.

 

[gar]

140313-1801 EDT

Both the equations in the original post were of the form y = K*x and that plots as a straight line for y vs x.

If you want to hold power constant, then write the equation as i = P/v or v = P/i . And yes that is not a linear curve.

.

 

[GoldDigger]

It is exactly a hyperbola. Where the part of the curve in the third quadrant (both I and V negative) is usually not included.

Tapatalk!

 

[junkhound]

am i missing something here

yep, you are leaving power factor out of the power equation for ac

and R is rarely a simple R, it is likely a complex impedance comprised of L, R, and C.

wont try to explain that unless you say you know how to calculate vectors, or at least heard of them ?

 

[Sahib]

V=IR does not contradict P=VI.

It does if P is constant. Because then I is inversely proportional to V. Per Ohm's law I is directly proportional to V (V=IR). The corollary is in constant power circuits Ohm's law does not apply.

 

[Sahib]

am i missing something here

yep, you are leaving power factor out of the power equation for ac

and R is rarely a simple R, it is likely a complex impedance comprised of L, R, and C.

wont try to explain that unless you say you know how to calculate vectors, or at least heard of them ?

In AC circuits replace R by Z-the circuit impedance. Ohm's law is applicable, no need for any vectors if Z is given to find current for various values of V, provided Z is constant.

 

[GoldDigger]

In fact, if you consider Ohm's law to be in effect the definition of resistance, R, then there is no contradiction at all. The value of R just changes as other circuit parameters change. This is the most common understanding of Ohm's Law.
But if you consider that equation to simply be the definition of resistance (for non-reactive components), then the actual content of Ohm's Law is that for simple components (not including incandescent lights and motors) the value of R is a constant. In that case Ohm's Law is not violated in those cases since it does not apply in the first place.

Tapatalk!

 

[Sahib]

In fact, if you consider Ohm's law to be in effect the definition of resistance, R, then there is no contradiction at all.

Then we need to consider the conditions under which the Ohm's law is derived If those conditions permit the resistance to be considered not as a proportionality factor, no problem. Otherwise..........

 

[david luchini]

It does if P is constant. Because then I is inversely proportional to V. Per Ohm's law I is directly proportional to V (V=IR). The corollary is in constant power circuits Ohm's law does not apply.

That is not a contradiction. You didn't read my whole post. In it I said...

V=IR says that for a CONSTANT resistance, the voltage is proportional to the current...

P=VI says that for the same amount of power, the current is inversely proportional to the voltage.

For a Constant P=10,000VA, at V=240V, I=41.6A and at V=120V, I=83.3A.

The R for V=240V, I=41.6A => 5.76ohm and the R for V=120V, I=83.3A => 1.44ohm.

As you can see, the R is not CONSTANT for the same power value at different voltages. P=VI does not contradict V=IR.

 

[junkhound]

no need for any vectors if Z is given .

Hmm, I know there have been many famous mathematicians from India. However, I have not yet seen the mathematics that show how one can find a Z without using vector sums.

Can you enlighten me?

 

[Sahib]

Hmm, I know there have been many famous mathematicians from India. However, I have not yet seen the mathematics that show how one can find a Z without using vector sums.

Can you enlighten me?

Z is a given value as discussed in previous posts. But if you want to find it in a linear AC circuit, divide circuit V by circuit I. The result is Z. See no use of vectors!