ELECTRICAL ENERGY AND POWER

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

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,

1. Both current and resistance are double.
2. Current is half and resistance double
3. 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