Electric power is a widely used term in physics as you must be aware of. Moreover, it is a very important thing that allows us to see the rate at which electrical energy transfers from an electric circuit. Thus, it helps us in measuring the efficiency of the same. Similarly, you will learn about the electric power formula in detail over here. In addition, to its application so as to get a better understanding of the whole thing.

**Definition**

Similar to mechanical power, we see that electric power is the rate at which work is done. However, over here, we measure it in the units of watts. Moreover, people use the term wattage in informally to refer to the electric power in units of watts.

You must have often seen electric power gets produced from electric generators and even other sources like electric batteries. Moreover, businesses and home usually utilize this supply through electric power industry via an electric power grid. After that, electric utilities measure power with the use of an electricity meter, that keeps a running total of the electric energy distributed to a consumer.

Electrical power offers quite a low entropy kind of energy. Further, it can be carried through long distances. In addition, it can also convert to other types of energy like light, motion or heat.

**Get the huge list of Physics Formulas here**

## Electric Power Formula

Electric power can be said to be the rate at which energy gets transported to or from a part of an electric circuit. Energy can be delivered by a battery or a circuit element, for instance, a resistor which releases energy as heat. For any element of the circuit, the power equals to the difference in voltage across the element which multiplies by the current. If we look at it from the Ohmâ€™s Law, we see that V = IR here, thus there are other forms of the electric power formula for resistors. We measure power in units of Watts. Moreover, it is important to know that a Watt = Joule per second. (1 W = 1 J/s). Therefore:

P = VI

Over here,

P refers to the electric power

V is the voltage difference

I is the electric current

Then we have the formula for resistors which means, it combines Ohmâ€™s law with Joules Law. Therefore, we have:

P = I^{2 }R = \(\frac{V^{2}}{R}\)

Over here:

P is the electric power (W)

V refers to the difference in voltage (V= J/C)

I is the electric current (A = C/s)

R refers to the resistance (â„¦ = V/A)

**Solved Example onÂ Electric Power**

**Question- **The battery of mobile phone functions at 12.0 V, further, it needs to provide a current of 0.9 A when music plays, how much power will it require for it?

**Answer- **We can find out the power which we require from the battery by applying the electric power formula. Thus, we will have:

P = VI

P = (12.0 V) (0.9 A)

P = (12.0 J/C) (0.9 C/s)

P = 10.8 J/s

P = 10.8 W

Therefore, the power which the battery requires of the mobile phone is 10.8 W.

**Question- **A resistor has a potential difference of 24.0 V and radiates heat. The rate at which thermal energy produces is 16.0 W. Find out the resistance value.

**Answer- **We can easily find out the resistance value upon rearranging one of the forms of electric power formula. Therefore, we will have:

P = \(\frac{V^{2}}{R}\) \(\rightarrow\) R = \(\frac{V^{2}}{P}\)

R= \(\frac{V^{2}}{P}\)

R = \(\frac{24.0V^{2}}{16.0W}\)

R = \(\frac{576V^{2}}{16.0W}\)

R = 36.0 \(\frac{V^{2}}{W}\)

R = 36.0 \(\frac{V^{2}}{J/s}\)

R = 36.0 \(\frac{V^{2}}{(J/C)(C/s)}\)

R = 36.0 \(\frac{V^{2}}{(V)(A)}\)

R = 36.0 V/A

R = 36.0 â„¦

Therefore, we have the resistance value as 36.0 â„¦.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…