Contents

The study of Physics Topics has helped humanity develop technologies like electricity, computers, and space travel.

## Calculating Electric Power Systems

When an electric current flows through a conductor, electrical energy is used up and we say that the current is doing work. We know that the rate of doing work is called power, so electric power is the electrical work done per unit time. That is,

Power = \(\frac{\text { Work done }}{\text { Time taken }}\)

or P = \(\frac{W}{t}\)

### Unit of Power

We have calculated the power by dividing work done by time taken. Now, the unit of work is “joule” and that of time is “second”. So, the unit of power is “joules per second”. This unit of power is called watt. Thus, the SI unit of electric power is watt which is denoted by the letter W. The power of 1 watt is a rate of working of 1 joule per second. That is,

1 watt = \(\frac{1 \text { joule }}{1 \text { second }}\)

Actually, watt is a small unit, therefore, a bigger unit of electric power called kilowatt is used for commercial purposes. it is obvious that :

1 kilowatt = 1000 watts

or 1 kW = 1000 W

It should be noted that the symbol for watt is W and that for kilowatt is kW. When work is done, an equal amount of energy is consumed. So, we can also say that electric power is the rate at which electrical energy is consumed. In other words, electric power is the electrical energy consumed per second. We can now write down another definition of watt based on electrical energy.

When an electrical appliance consumes electrical energy at the rate of 1 joule per second, its power is said to be 1 watt. We have just given two definitions of electric power, one by using the term ‘work’ and another by using the term ‘energy’. We can combine these two definitions and say that : The rate at which electrical work is done or the rate at which electrical energy is consumed, is called electric power.

### Formula for Calculating Electric Power

We know that :

Power = \(\frac{\text { Work done }}{\text { Time taken }}\)

or P = \(\frac{W}{t}\) ………………. (1)

We have already studied that the work done W by current I when it flows for time t under a potential difference V is given by :

W = V × I × t joules

Putting this value of W in equation (1), we get :

P = \(\frac{V \times I \times t}{t}\) joules per second

or P = V × I joules per second

or Power, P = V × I watts,

where V = Potential difference (or Voltage) in volts

and I = Current in amperes

Thus, the power in watts is found by multiplying the potential difference in volts by the current in amperes. We can write down the above formula for electric power in words as follows :

Electric power = Potential difference × Current

Since the potential difference is also known by the name of voltage, we can also say that:

Electric power = Voltage × Current

It is clear from the above discussion that in electric circuits, the power expended in heating a resistor or turning a motor depends upon the potential difference between the terminals of the device and the electric current passing through it.

We can also use the formula P = V × I for defining the unit of power called ‘watt’ in another way as described below.

Power, P = V × I watts

Now, if an electrical device is operated at a potential difference of 1 volt and the device carries a current of 1 ampere, then power becomes 1 watt. That is :

1 watt = 1 volt × 1 ampere

or 1 W = 1 V × 1 A

or 1 W = 1 V A

This gives us another definition of the unit of power called ‘watt’. We can now say that: One watt is the power consumed by an electrical device which when operated at a potential difference (or voltage) of 1 volt carries a current of 1 ampere.

### Some Other Formulae for Calculating Electric Power

We have just obtained a formula for calculating electric power, which is :

P = V × I

This formula can be used when both, the potential difference (or voltage) V and the current I are known to us. Sometimes, however, they do not give us V and I. We are given either voltage V and resistance R or current I and resistance R. In that case we have to take the help of Ohm’s law. This will become clear from the following discussion.

(i) Power P in terms of I and R. We have just seen that:

P = V × I ………………. (1)

Now, from Ohm’s law we have, \(\frac{V}{I}\) = R

or V = I × R …………. (2)

Putting this value of V in equation (1), we get :

P = I × R^{2} × I

or Power, P = I^{2} × R

where I = Current

and R = Resistance

This formula is to be used for calculating electric power when only current I and resistance R are known to us.

(ii) Power P in terms of V and R. We know that:

P = V × I ……………. (1)

Also, from Ohm’s law we have, \(\frac{V}{I}\) = R

or V = I × R

or I = \(\frac{V}{R}\)

Putting this value of I in equation (1), we get:

P = V × \(\frac{V}{R}\)

or Power, P = \(\frac{V^2}{R}\) ……………… (3)

where V = Potential difference (or Voltage)

and R = Resistance

This formula is to be used for calculating power when voltage V and resistance R are known to us.

It is clear from equation (3) that power is inversely proportional to the resistance. Thus, the resistance of high power devices is smaller than the low power ones. For example, the resistance of 100 watt (220 volt) bulb is smaller than that of a 60 watt (220 volt) bulb (see Figure). We have now three formulae for calculating electric power. These are :

First formula for power : P = V × I

Second formula for power : P = I^{2} × R

Third formula for power : P = \(\frac{V^2}{R}\)

These three formulae should be memorized because they will be used to solve numerical problems. Before we solve the problems based on electric power, it is very important to know the meaning of ‘power- voltage’ rating of electrical appliances.

### Power-Voltage Rating of Electrical Appliances

Every electrical appliance like an electric bulb, radio or fan has a label or engraved plate on it which tells us the voltage (to be applied) and the electrical power consumed by it. For example, if we look at a particular bulb in our home, it may have the figures 100 W – 220 V written on it. Now, 100 W means that this bulb has a power consumption of 100 watts and 220 V means that it is to be used on a voltage of 220 volts. The power rating of an electrical appliance tells us the rate at which electrical energy is consumed by the appliance. For example, a power rating of 100 watts on the bulb means that it will consume electrical energy at the rate of 100 joules per second. If we know the power P and voltage V of an electrical appliance, then we can very easily find out the current I drawn by it. This can be done by using the formula : P = V × I. The usual power-voltage ratings of some of the common household electrical appliances and the current drawn by them are given below.

Power-Voltage Ratings of Some Electrical Appliances and the Current Drawn by Them

Let us solve some problems now.

**Example Problem 1.**

What will be the current drawn by an electric bulb of 40 W when it is connected to a source of 220 V ?

**Solution:**

In this case we have been given power P and voltage V, so the formula to be used for calculating the current will be :

P = V × I

Here, Power, P = 40 watts

Voltage, V = 220 volts

And, Current, I = ? (To be calculated)

Now, putting these values in the above formula, we get :

40 = 220 × I

I = \(\frac{40}{220}\)

= \(\frac{2}{11}\)

Thus, Current, I = 0.18 ampere

**Example Problem 2.**

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

**Solution:**

In the first case :

Power, P = 100 W

Potential difference, V = 220 V

And, Resistance, R =? (To be calculated)

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

So, 100 = \(\frac{(220)^2}{R}\)

And R = \(\frac{220 \times 220}{100}\) = 484 Ω

This resistance of 484 Ω of the bulb will remain unchanged.

In the second case:

Power, P = ? (To be calculated)

Potential difference, V = 110 V

And, Resistance, R = 484 Ω (Calculated above)

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

P = \(\frac{(110)^2}{484}=\frac{110 \times 110}{484}\) = 25 W

Thus, the correct answer is : (d) 25 W.

**Example Problem 3.**

Which of the following does not represent electrical power in a circuit ?

(a) I^{2}R

(b) IR^{2}

(c) VI

(d) \(\frac{V^2}{R}\)

**Answer:**

(b) IR^{2}