# Electric Energy and Power – Definition, Formula, Sample Problems

Almost all the electrical devices around us in daily life have a rating written on them. That rating is expressed in Watts and intuitively explains the amount of electricity the device will consume. Bigger devices like AC, refrigerators, etc consume more electricity and thus come with a higher rating. Since these ratings are written almost everywhere, it becomes essential to study what these numbers mean and how do they work. Let’s look at the concepts of electric energy and power in detail.

### Electric Energy

Consider a conductor with endpoints A and B, assume that a current I is flowing through the conductor. Let us denote the potential at ends by V(A) and V(B). Since the current is flowing from A to B, it means that potential is decreasing from point A to point B.

V = V(A) – V(B) > 0

In time “t”, a charge Q travels from point A to point B. It is known that the potential is the energy spent to bring a charge from infinity to the current point. So, the above equation can be modified to find the potential energy at each point. Let’s say potential energy at point A is denoted by U(A) while the potential energy at point B is denoted by U(B).

U(A) = QV(A)

U(B) = QV(B)

Let the change in potential energy be denoted by U_{net}

U_{net} = Final Potential Energy – Initial Potential Energy

U_{net} = U(B) – U(A)

⇒ U_{net} = ∆ QV(B) – QV(A)

⇒ U_{net} = – ∆ Q.V

⇒ U_{net} = -I × V × ∆ t < 0

If the charges inside the conductor moved freely, this potential energy would have gotten converted into kinetic energy, so that the total energy remains unchanged.

∆K = -∆U

Thus, in case the charges could move freely inside the conductor under the action of the electric field, their velocity would have increased as they move. However due to the collisions between electrons and different ions inside the conductor. The charge carriers due to not move with acceleration but with a steady velocity. During the collisions, energy is transferred from these electrons to the ions which then vibrate more vigorously, and that in turn increases the temperature. Thus the derivation is given above indirectly calculates the energy dissipated in the conductor in form of heat.

∆ W = I × V × ∆ t

### Electric Power

From the work-energy principles studied in earlier classes, it is known that power is the rate of work done. In this case, it can be thought of the amount of energy dissipated in form of heat when an electric current is passing through the conductor. In all electrical appliances, heat dissipation occurs. This heat dissipation is called power loss or ohmic loss because these losses are due to resistances offered in conductors.

Rewriting the previous equation,

P = ∆W / t

P = I × V

Using the ohm’s law relation V = IR for substituting the values inside the above equation,

P = I × V

⇒ P = I × I × R

⇒ P = I^{2}R

It can also be written as,

P = I × V

⇒ P = (V/R) × V

⇒ P = V^{2}/R

Thus, the power dissipated in a conductor can be written as,

P = VI = V

^{2}/R = I^{2}R

### Sample Problems

**Question 1: Find the power dissipated in a conductor with a 10V potential difference and a current of 5A. **

**Answer: **

It is known that the power dissipated in the conductor is given by,

P = VI

Given:

V = 10

I = 5

P = VI

⇒ P = (10)(5)

⇒ P = 50W

**Question 2: Find the power dissipated in a conductor with a 5V potential difference and a current of 2A. **

**Answer: **

It is known that the power dissipated in the conductor is given by,

P = VI

Given:

V = 5

I = 2

P = VI

⇒ P = (5)(2)

⇒ P = 10W

**Question 3: An electric heater is connected to a battery of 5V potential difference. The heater has a total resistance of 50 ohms. Find the power dissipated by the electric heater. **

**Answer: **

It is known that the power dissipated in the conductor is given by,

P = V

^{2}/RGiven:

V = 5

R = 50

P = V

^{2}/R⇒ P = (5

^{2})/(50)⇒ P = 0.5 W

**Question 4: An electric fan is connected to a battery of 20V potential difference. Assume that the fan has a total resistance of 15 ohms. Find the power dissipated by the electric fan. **

**Answer: **

It is known that the power dissipated in the conductor is given by,

P = V

^{2}/RGiven:

V = 20

R = 15

P = V

^{2}/R⇒ P = (20

^{2})/(15)⇒ P = 400/15

⇒ P = 26.67 W

**Question 5: An electrical appliance is connected to a battery due to which a current of 5A flows through it. The appliance has a total resistance of 10 ohms. Find the power dissipated by the appliance. **

**Answer: **

It is known that the power dissipated in the conductor is given by,

P = I

^{2}RGiven:

I = 5

R = 10

P = I

^{2}R⇒ P = (5

^{2})(10)⇒ P = (25)(10)

⇒ P = 250 W

**Question 6: An electrical appliance is connected to a battery due to which a current of 10A flows through it. The appliance has a total resistance of 20 ohms. Find the power dissipated by the appliance. **

**Answer: **

It is known that the power dissipated in the conductor is given by,

P = I

^{2}RGiven:

I = 10

R = 20

P = I

^{2}R⇒ P = (10

^{2})(20)⇒ P = (100)(20)

⇒ P = 2000 W