### At the end of this post, you should be able to;

State Joule’s law

Solve problems involving Electrical energy and power

Solve simple problems on Joule’s law

JOULE’S LAW

It state that, the amount of work required to maintain a current of I amperes through a resistance of R ohms, for time t seconds is I²R

E = IVt, joules

Or E = I²Rt, joules

Or E = V²t/R, joules

The above expression are also known as joules Law, which states that; Electrical Energy E is the energy required for V volts of electricity to make an electric current of I amperes to flow through a conductor of resistance R Ohms for a time T seconds.

Example

## Calculate the work done in moving through a distance of 25m by a force of 20N which acts in the opposite direction of the motion of the body.

Solution

Since work done = F × d

D = 20 ×25 = 500j

## A force of 40N is applied to a body to move it at a uniform velocity through a distance of 15m in 10seconds in the direction of the force. Calculate the power produced.

Workdone = force × distance

40 × 15 = 600J

Power = workdone/time taken

= 600/10 = 60W

## An electric motor develops 5Kw at the speed of 100 rev/min. Calculate (a) the work done in 30min.

## (I) in kilowatt hour (II) in mega joules (III) the torque in N-m

Solution

(I) Workdone in KWh = power (in kw) × time(in hours)

= 5 × 30/60 = 2.5KWh

work done in mega joules, but 1Kwh = 3.6MJ

Therefore w.d = 2.5 × 3.6 = 9MJ

Power= 2π × T × n

5000 = 2πT × 1000/60

T = 47.74N-m

## An electric kettle takes 2KW at 240V. Calculate;

## (a) the current and

## (b) the resistance of the heating element

Calculate current, I

Recall power, P = IV

Make I the subject of the formula,

I = P/V = 2000/240

I = 8.33A

Resistance(R) = ?

P = I²R; R = P/I²

R = 2000/8.33²

2000/69.39 = 28.82Ω

## The power expanded in a certain resistor is given by I²R. If the power expended in the resistor is 175W when the current is 5A, calculate the power in the resistor when,

- Both current and resistance are double.
- Current is half and resistance double
- When current is double and resistance half.

Given that: P = 175W, I = 5A

P = I²R, ⇒R = P/I²

= 175/(5)² = 7Ω

When both current and resistance are double, I = 2i; R = 2R

Therefore, P = (2I)² × 2R = (2² × 5²) × (2 × 7)

4 × 25 × 2 × 7 = 1400W

Half current = ½; double resistance = 2R

Therefore, P = (½I)² × 2R = (½² × 5²) × (2 × 7)

0.25 × 25 × 2 × 7 = 87.5W

Double current= 2i, half resistance = ½R

Therefore, P = (2I)² × ½R = (2² × 5²) × (½ × 7)

4 × 25 × ½ × 7 = 350W