240 × I (4 × 60 × 60) = 28.8 × 10⁶

Example 14

A hoist motor powered by a 240V mans

supply requires a current of 30A to lift a

load of mass 3 tonnes at the rate of 5m per

minute. Calculate:

28 8000 000

퐼 =

240 × 4 × 60 × 60

퐼 = 8.3퐴

The current through the filament lamp is

8.3A

(a) The power input

Solution

Example 12

Power input = IV

= 30 × 240

= 7200W

In the circuit shown in the figure below,

which bulb, A or B, is bright? Give reason

for your answer.

(b) The power output

F = mg

F = 3 × 1000 × 10 N

F = 30,000 N

Solution

distance

Speed =

Since the bulbs are in series, the current, I

passing through each bulb is the same.

time

5m

Speed =

60 s

Power output = I²R

P_A= I²× 576 and

P_B= I²× 1 440.

푆푝푒푒푑 = 0.8 푚/푠

Solution

To find the power out put

Since the resistance of bulb B is more,

power output from B is more. Bulb B is

brighter.

Force × distance

P =

time

distance

P = Force ×

time

Example 13

What is the maximum number of 100W

bulbs which can be connected from a 240V

source supplying a current of 5A?

distance

P = mg ×

time

solution

P = F × V

P = 30, 000 N × 0.8 m/s

P = 2 500 W

Power input = IV

100n = 5 × 240

n = 12 bulbs

푅

ꢀ

퐵

Case III: To find the ratio =

(c) The overall efficiency

Solution

ꢁ_ꢂ

ꢁ_ꢃ

360

900

power output

efficiency =

=

× 100%

power input

ꢁ_ꢂ

4

2 500

=

× 100%

ꢁ_ꢃ

10

7 200

ꢁ_ꢂ

ꢁ_ꢃ

2

= 34.72%

=

5

ꢁ_ꢂ: ꢁ_ꢃ= 2: 5

Example 15

Example 16

Two heaters A and B are connected in

parallel across a 10 volts supply. Heater A

produces 1000J of heat in one hour while

heater B produces 200J in half an hour.

A battery of e.m.f 12V and unknown

internal resistance is connected in series

with a load resistance, R reads 11.4V and

the power dissipated in the battery is

0.653W. Find the:

푅

ꢀ

Calculate the ratio =

푅

퐵

Solution

(i)

Current flowing

To find the resistance of heaters A and B

Solution

From

V²

Energy =

t

R

V²

R =

t

Eergy

Given that:

V = 11.4V, E = 12V and P =

0.653W

Case I: For heater A

Case I: Voltage loss de to internal

resistance , r

t = 3600s, energy = 1000J, V= 10V

(10)²

ꢁ_ꢂ=

× 3600

1000

퐸 = 퐼(ꢁ + 푟)

퐼푟 = 퐸 − 푉

ꢁ_ꢂ= 360 Ω

퐼푟 = 12.0 − 11.4

퐼푟 = 0.6 ………….. (i)

Case II: For heater B

t = 1800s, energy = 200J,

Case II: Power dissipated in the

(10)²

battery

loss

de

to

internal

ꢁ_ꢃ=

× 1800

resistance , r

200

푃 = 퐼²푟

ꢁ_ꢃ= 900Ω

퐼²푟 = 0.653 …………….. (ii)

marked on the body of an appliance.

Example light bulb marked : 240V, 10W

Power rating: - is the amount of power that

an appliance uses at a specific value of

voltage

Dividing eqn (i) ÷ eqn (i)

퐼²푟

퐼푟

0.653

0.6

=

퐼 = 1.088퐴

Example 01:

A heater marked 240V,

1000W, What does the statement mean?

(ii) Internal resistance of the battery

Given that

Answers

The statement means that the heater

consumes 1000J of electrical energy every

second when connected to 240V main

supply.

P = 0.653W and 퐼 = 1.088퐴

From eqn (ii)

퐼²푟 = 0.653

(1.088)²푟 = 0.653

푟 = 0.55Ω

Example 02: An appliance marked 3000W,

240V, What does the statement mean?

Answers

(iii) Load resistance, R

The statement means that the appliance

dissipates energy at the rate of 300 joules

per second when connected to a 240V

supply.

푉 = 퐼 ꢁ

푉

ꢁ =

퐼

11.4

ꢁ =

Example 03

1.088

Explain the effect on power rating if the

main supply is 220V (i.e less voltage than

rated voltage)

ꢁ = 10.48Ω

(iv) Efficiency of the circuit

Answers

It causes overheating of the appliance.

Solution

power output

efficiency =

× 100%

A low voltage forces a device to draw extra

current to deliver the power expected of it

thus overheating the motor windings

power input

퐼²ꢁ

efficiency =

× 100%

퐼퐸

1.088 (10.48)

2

(

)

=

× 100%

Example 04

(

)

1.088 (12)

= 95%

Why power rating of most appliances does

not exceeds 240V of the main supply?

Because the devices are designed to be

compatible with the standard mains voltage.

Power rating:

Power rating: - is the rate at which an

appliance dissipates energy and is usually

If rated higher, they would not function

properly on the supply available.

= 1000 × 60 × 60 × 60 joules

1 kWh = 3, 600,000J

1 kWh = 3.6MJ

Example 05

What are the advantages of 240V?

1푘푊ℎ = 3.6 × 10⁶퐽

Answers

(i) It delivers low current for the same power

transmitted; this reduces the heating of the

device.P = IV

COST OF ELECTRIC ENERGY

(ii) Reduces poer loss in transmission since

it delivers low current for the same power

transmission over long distance. P = I²Rt

(iii) Give higher efficiency for heavy

appliances

퐍퐮퐦퐛퐞퐫 퐨퐟 퐮퐧퐢퐭퐬 퐮퐬퐞퐝

(Energy consumed )

[

]

Power

(in kW)

Time

(in hours)

= [

] × [

]

퐍퐮퐦퐛퐞퐫 퐨퐟 퐮퐧퐢퐭퐬 퐮퐬퐞퐝

(Energy consumed )

Time

[

]

퐓퐨퐭퐚퐥 퐜퐨퐬퐭 = [

]

× [

]

(in hours)

The rate per unit is the cost per unit of

electrical energy consumed. Thus the cost of

using an appliance is given by;

COMERCIAL ELECTRIC ENERGY

퐏퐨퐰퐞퐫

(in kW)

퐓퐢퐦퐞

(in hours)

Total cost = [

] × [

]

Instead of using the Joule, electricity supply

companies use the killowatt – hours as the

unit of measuring electrical energy.

퐜퐨퐬퐭 퐩퐞퐫

퐮퐧퐢퐭

× ⌈

⌉

Simply

Cost = price × total units

The commercial unit of electrical energy is

kilowatt – hour, (kWh) since a watt second

is very small.

Example 01

Four bulbs each rated at 75W operate for

120 hours. If the cost of electricity is sh. 100

per unit, find the total cost of electricity

used.

1 kilowatt hour (kWh):- is the energy

supplied when an electric appliance which is

rated 1kW is used for one hour.

Solution

A kilowatt hour:- is the electrical energy

used by a rate of working of 1000 watts for

1 hour.

Step 1: To find the total power in kW

Given that

Number of bulbs = 4

Power rating in each bulb = 75W

1 watt = 1 joule per second

Total power rating for 4 bulbs

= 75푊 × 4

1 kWh = 1000 × 1hr

= 300푊