P=I*R

[Stoneww]

Just a short question, wondering why people tend to use P=I^2*R instead of the standard P=IV

 

[duke37]

P = I*I*R NOT P = I*R
P = I*V
P = V*V/R

 

[Stoneww]

P = I*I*R NOT P = I*R
P = I*V
P = V*V/R

Ye my bad I know it is Im just wondering why P=I^2*R is used more commonly

 

[Electrobrains]

If people are using that formula, it's probably because they find the resistance (R) interesting (or don't know or are not interested in the voltage).

Basically you can express Power in the way you like best: P = IV = I²R = V²/R

Be aware of that above expressions are in fact only "half of the truth".
The general expression for power is the Complex Power:
S= P + jQ (the Complex Power is the sum of the Active power and the Reactive power), which in turn can be expressed in complex terms with Impedance (Z) etc.

 

[Harald Kapp]

Another way to look at it:
P = I*V
V = I*R
Therefore
P = I^2 * R
or
P = V^2/R
Whicherver equation you use, you get the same value. Use the one which is most convenient in your situation.

And of course this is for real values only. With complex elements things get, well, complex, see #4 .

 

[Stoneww]

If people are using that formula, it's probably because they find the resistance (R) interesting (or don't know or are not interested in the voltage).

Basically you can express Power in the way you like best: P = IV = I²R = V²/R

Be aware of that above expressions are in fact only "half of the truth".
The general expression for power is the Complex Power:
S= P + jQ (the Complex Power is the sum of the Active power and the Reactive power), which in turn can be expressed in complex terms with Impedance (Z) etc.

Thanks a lot dude this is the first time im hearing about complex power, I think P=IV will do me for now but I'm definitely going to research complex

 

[davenn]

Just a short question, wondering why people tend to use P=I^2*R

because for power transmission the power losses directly relate to the current through the resistance of the power line
It means that if you increase the transmission line voltage substantially, then the current is reduced and therefore the I² R
losses are reduced

from wiki ....

Transmitting electricity at high voltage reduces the fraction of energy lost to resistance, which varies depending on the specific conductors, the current flowing, and the length of the transmission line. For example, a 100 mi (160 km) span at 765 kV carrying 1000 MW of power can have losses of 1.1% to 0.5%. A 345 kV line carrying the same load across the same distance has losses of 4.2%.[20] For a given amount of power, a higher voltage reduces the current and thus the resistive losses in the conductor. For example, raising the voltage by a factor of 10 reduces the current by a corresponding factor of 10 and therefore the I 2 R {\displaystyle I^{2}R}

losses by a factor of 100, provided the same sized conductors are used in both cases. Even if the conductor size (cross-sectional area) is reduced ten-fold to match the lower current, the I 2 R {\displaystyle I^{2}R}

losses are still reduced ten-fold. Long-distance transmission is typically done with overhead lines at voltages of 115 to 1,200 kV. At extremely high voltages, more than 2,000 kV exists between conductor and ground, corona discharge losses are so large that they can offset the lower resistive losses in the line conductors. Measures to reduce corona losses include conductors having larger diameters; often hollow to save weight,[21] or bundles of two or more conductors.

Dave

 

[(*steve*)]

Just a short question, wondering why people tend to use P=I^2*R instead of the standard P=IV

Well, if they know the current and the resistance, it's the logical choice.

P = V²/R and V = √(PR) are also useful in many cases.

 

[AnalogKid]

One equation is Joule's Law. The other is Watt's Law. Although they were developed independently and for different reasons, one transforms into the other using Ohm's Law. The third permutation is P = E^2/R. You can use which ever one suits the information you have available.

ak

 

[Electrobrains]

because for power transmission the power losses directly relate to the current through the resistance of the power line

Dave is right. A good reason for using the formula P = I²R does not only have to be interest in R or lack of interest in V, but can be an interest in I².

Expanding on that, the other formula P = V²/R shows exactly the same proportionality. So using that formula can be an interest in R, lack of interest in I or an interest in V².

The proportionality is expressed like this:
P ∝ I²
or
P ∝ V² (far back, when I studied, we used this sign as proportionality sign: ~)

As described in the Wiki article above the consequence of a squared proportionality is immense:
Doubling the Current (or Voltage) will quadruple the Power. To triple the Current (or Voltage) will increase the Power 9 times!

The same way downwards: Decreasing the Current (or Voltage) to half, will reduce the Power to a quarter of the original Power...

That gives you a good reason to increase the voltage of the Power Transfer Lines (fixed P-transferred gives reduced I, that gives much reduced P-dissipated) and to decrease the Voltage in ICs like OP-Amps and Micro Processors.

Keyword: Green Power