## Concept of Power

Power is an essential concept in physics. It measures the rate of transfer of energy or how work is done. We can define power as the amount of work done or energy transferred per unit of time. The unit of power is watt (W), and the name was an honour to the inventor of the steam engine, James Watt. In this article, I will walk you through simple steps that will guide you on how to calculate power in physics by solving enough problems.

In physics, power is an important quantity that helps us understand and analyse the behaviour of various physical systems. For example, we can use power to measure:

1. An output of an engine,

2. The rate of heat transfer, or

3. Rate at which an electric circuit consumes or produces energy.

Calculating power involves two key variables: work done or energy transferred, and time taken. The formula for power is P = W/t.

Where:

P is power,

W is the amount of work done or energy transferred, and

t is the time taken.

There are different ways to calculate power depending on the situation. For example, if we know the force applied and the distance moved. We can use the formula P = Fd/t to calculate power. If we know the voltage and current in an electric circuit, we can use the formula P = VI to calculate power.

It is essential to understand power and how to calculate it. This is because it helps us to design, analyse, and optimise various physical systems. Whether it’s designing a car engine, analysing the performance of a power plant, or optimising the efficiency of a solar panel. Power plays a critical role in physics and engineering.

### What is power?

**Definition:** Power is simply the rate of doing work. We can also say that power is the ratio of work to the time taken.

The formula for power is

Power = work / time

The unit of power is in watts, kilowatts, Joule per second, or horsepower

Now we can proceed to the definition of electric power

### What is Electric Power?

**Definition:** Electrical power is the quantity of electrical work done or electrical heat transferred per second.

Since we know that

Power = work/time

and work = force x distance

and work done in an electrical material is

W = QV

but Q = It

which implies that

W = IVt

Also, V = IR

Therefore W = I^{2}Rt

From the formula

Power = work / time

we now have

Power = IVt / t

and W = IVt = I^{2}Rt

Power = I^{2}Rt / t

which implies that

The formula for power in a circuit is Power, P = I^{2}R = IV = V^{2}/R

Where I = current (measured in ampere, A)

R = resistance (measured in ohms, Ω)

V = Potential difference across the conductor (measured in volts)

Q = Quantity of electricity (measured in coloumbs)

W = workdone in transferring heat

P = Power dissipated in watts

## Electrical Energy

**Definition:** Energy is the ability or capacity to do work. In this case, we are dealing with electricity and we are going to relate energy with electricity.

Therefore, we have a potential difference across two points which is known as the dissipation of energy by a charge between two points.

Potential difference (p.d), V = Electrical energy that was dissipated (E) / Charge (Q)

Therefore, we now have

V = E / Q

and current (I) is due to the rate of the charge flow. Now we can write Q = It

Which implies that

V = E / It

and we can see that electrical energy is

E = IVt = I^{2}Rt because V = IR

## Finding the Cost of Electrical Power Consumption

When electrical equipment consumes one kilowatt of power in one hour, we say that electrical energy of one kilowatt-hour (kWh) is dissipated.

1 kWh = 1 x kW x 1 hr = 1 x 10^{3} x 60 x 60 = 3.6 x 10^{6} Joules

## Worked Examples on How to Calculate Power in Physics and Find the Cost of Power Dissipated

Here are examples to acquaint you with the simplest methods of how to calculate power in a circuit. I have also solved examples of how to find the cost of electrical power.

### Example 1

A current of 20A flows through a cable whose resistance is 5Ω. Calculate the electric power dissipated. [UTME 2015]

**Solution:**

**Data:**

current, I = 20A

Resistance, R = 5Ω

Power, P = ?

by applying the formula for power which is P = I^{2}R

we now have

P = 20^{2} x 5 = 400 x 5 = 2000 watts = 2 kW

### Example 2

Two resistors 5Ω and 10Ω are arranged in series and later in parallel to a 24 V source. Calculate the ratio of total power dissipated in the series and parallel arrangements.

**Solution:**

**Data:** For the resistance in series, we have

Total resistance, R = 5 + 10 = 15Ω

and Voltage, V = 24V

Therefore, to calculate power for resistance in series, we apply the formula

P_{series} = V^{2}/R = 24^{2} / 15 = 576 / 15 = 38.4 Watts

Additionally, we now calculate resistance in parallel

1 / R = (1 / 5) + (1 / 10)

and this implies

1 / R = 0.3

Now we make R subject of the formula

R = 1 / 0.3

Which will give us

R = 3.33Ω

We now calculate power in favour of resistance in parallel

P_{parallel} = V^{2}/R = 24^{2}/3.33 = 576 / 3.33 = 172.97 Watts

The ratio of the power is

P_{seriese} **:** P_{parallel}

Which is

38 : 173

and we have the ratio of the total power dissipated in series and in parallel as 38/173

### Example 3

An electric lamp marked 240 volts, 60 Watts is left to operate for an hour. How much energy is generated by the filament?

**Solution:**

**Data:**

We need to first extract our data by going through the question again,

Voltage, V = 240 v

Power, P = 60 W

Time, t = 1 hr = 60 x 60 = 3,600 s

and the formula we need to apply to solve the problem is

Power (P) = Energy of the filament (E) / Time (t)

This is the same thing as P = E / T

by making E the subject of the formula, we now have

E = PT

Now we apply our data into the above formula

E = 60 x 3,600 = 216,000 Joules

Therefore, the energy generated by the filament is 216 Kilojoules or 216 kJ

### Example 4

A heating coil dissipating energy of 9 x 10^{5} Joules is used to heat water for 30 minutes. Calculate the power supplied to the coil.

**Solution**

**Data**

Energy dissipated, E = 9 x 10^{5} Joules

Time, t = 30 minutes = 30 x 60 = 1800 seconds

Applying the formula below

Power (P) = Energy of the filament (E) / Time (t)

This implies that

P = E / t = 9 x 10^{5} / 1800 = 500 Watts

### Example 5

An electric bulb is labelled 240V, and the current taken by the bulb is 0.416A. How much power is dissipated by the bulb?

**Solution:**

**Data:**

Voltage = 240V

Current, I = 0.416A

Power, P =?

and the formula for power is

P = IV

Now we apply our data to the above expression

P = 240 x 0.416 = 99.84 Watts

And the power can be approximated into

P = 100 Watts

### Example 6

If the charge for electricity per kWh is 4 Naira (N4). What is the cost of operating an electrical appliance rated 250V, 2A for 6 hours?

**Solution:**

**Data:**

N4 per kWh was given as the cost

Voltage = 250V

Current, I = 2A

Time, t = 6hrs

Thus,

The formula for power consumed is

P = IV = 2 x 250 = 500 watts

Now we change P into kilowatt-hour

P = (500/1000)kW x 6hrs = 0.5kW x 6hrs = 3kWh = 3 units

since the cost per unit or 1kWh is N4, we can now calculate the cost of consumption is

Cost of consumption = N ( 3 x 4) = N12

### Example 7

Find the cost of consuming eight 60W bulb, for 8hrs if electric energy costs N5 per unit.

**Solution**

**Data**

Power, P = 8 bulbs x 60W = 480W = 0.48kW

Time, t = 8hrs

Energy,E = Power (P) x Time (t) = 0.48kWh x 8hrs = 3.84 kWh

E = 3.84 units

The cost of consumption is

N(3.84 x 5) = N19.2

Thus the cost of consumption is 19.2 Naira.

### Example 8

A 200W bulb is lighted by a 240V a.c main supply. If 1kWh is sold at N40, what is the cost of keeping the bulb lighted for one day.

**Solution:**

**Data:**

We find our data by reading the question again

Power, P = 200W = 0.2kW

Voltage, V = 240V

Time, t = 24hrs

We apply the formula

Cost of consumption will be = Power x time x cost

This will be Cost = 0.2kW x 24hrs x N40 = N192

Therefore, the cost of consumption is N192

### Example 9

An electric bulb is rated 60W for a 240v supply. Calculate the resistance of the bulb and the current it consumes.

**Solutions**

**Data**

Power, P = 60W

Voltage, V = 240v

Resistance = ?

current, I = ?

The formula is

P = V^{2} / R

R = (240)^{2} / 60

R = 57,600 / 60 = 960Ω

and the current (I)

I = P / R

which implies that

I = 240 / 960 = 0.25A

Thus, the current is 0.25A

### Example 10

A lamp is marked 220V, 60W. Calculate the energy it would consume when connected to a 220V source for 1 hour.

**Solution**

**Data:**

Voltage = 220V

Power = 60W

Time, t = 1 hour = 60 x 60 = 3600 seconds

Energy = Power x Time = 60 x 3600 = 216,000 Joule = 216 kJ

You can check our questions and answers page to see more questions on how to calculate power in physics.

*You may also like to read:*

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**Reference:**