Introduction

Rotation is one of the most important transformations in Geometry. It is used in Mathematics, Engineering, Architecture, Computer Graphics, Robotics, Astronomy, and many other scientific fields.

Whenever an object turns about a fixed point, the transformation is called a rotation.

Examples of rotation in everyday life include:

• Opening and closing a door
• The hands of a clock
• A ceiling fan
• A bicycle wheel
• A spinning top
• A propeller
• A merry-go-round

In each case, the object turns about a fixed point known as the centre of rotation.

---

What is Rotation?

Rotation is a transformation that turns an object about a fixed point called the centre of rotation through a specified angle and direction.

Unlike enlargement, rotation does not change the size of an object.

The shape, size and distances between corresponding points remain unchanged.

Rotation is therefore classified as an isometry.

Key Terms

Term	Meaning
Centre of Rotation	Fixed point about which an object turns
Angle of Rotation	Amount of turning measured in degrees
Direction of Rotation	Clockwise or Anticlockwise
Image	New position of the object after rotation
Object	Original figure before rotation

---

Measuring Rotation

The angle of rotation is measured in degrees.

Common Rotations

Rotation	Angle
Quarter Turn	90°
Half Turn	180°
Three-Quarter Turn	270°
Full Revolution	360°

Illustration

Observe how the arrow changes direction after each turn.

• Quarter Turn = 90°
• Half Turn = 180°
• Three Quarter Turn = 270°
• Full Revolution = 360°

---

Direction of Rotation

There are two possible directions of rotation.

1. Clockwise Rotation

A rotation in the same direction as the hands of a clock.

2. Anticlockwise Rotation

A rotation opposite to the direction of the hands of a clock.

the figure below illustrates rotation of clockwise and anticlockwise rotation of 90^o about the origin

Sign Convention

Mathematicians use signs to indicate the direction.

Direction	Sign
Anticlockwise	Positive (+)
Clockwise	Negative (-)

Examples:

• +90° means 90° anticlockwise
• -90° means 90° clockwise
• +180° means 180° anticlockwise
• -150° means 150° clockwise

---

Rotation Through 90°

A rotation through 90° is called a quarter turn.

Construction Procedure

Suppose triangle ABC in the figure below is rotated through +90° about point P.

1. Join P to A.
2. Measure PA.
3. Draw a 90° angle from PA.
4. Mark point A’ such that:

PA = PA’

5. Repeat for points B and C.
6. Join A’, B’ and C’.

The figure A’B’C’ is the image of triangle ABC.

Important Observation

When a figure is rotated:

• A’B’ = AB
• B’C’ = BC
• A’C’ = AC

Therefore:

Rotation preserves size and shape.

Illustration

---

Rotation Through 180°

A rotation through 180° is called a half turn.

Construction Procedure

Suppose triangle PQR is rotated through 180° about point O.

1. Join P to O.
2. Extend line PO.
3. Mark P’ such that:

PO = OP’

4. Repeat for Q and R.
5. Join P’, Q’ and R’.

The resulting triangle is the image of triangle PQR.

Observation

For a 180° rotation:

• Corresponding sides remain equal.
• Corresponding sides become parallel.

Therefore:

Hence:

Object Size = Image Size

Illustration

---

Rotation Through 360°

A rotation through 360° is called a full revolution.

When an object is rotated through 360° about any point:

The image coincides exactly with the original object.

In other words:

A figure rotated through 360° maps onto itself.

---

Properties of Rotation

Rotation has several important properties.

Property 1: Rotation is an Isometry

The size of the figure remains unchanged.

Property 2: Lengths are Preserved

Corresponding sides remain equal.

Property 3: Angles are Preserved

Corresponding angles remain equal.

Property 4: Shape is Preserved

No distortion occurs.

Property 5: Orientation Changes

The position of the object changes although its size remains unchanged.

---

Worked Example 1

Rotating a Line Through 60°

Line AB is rotated through 60° about point C.

Find the image A’B’ under this rotation

Solution

1. Join B to C.
2. Measure CB.
3. Construct angle BCB’ = 60°.
4. Mark B’ such that:

BC = BC’

5. Repeat the procedure for point A.
6. Join A’ to B’.

The resulting line A’B’ is the image of line AB.

Illustration

---

Worked Example 2

Rotating Triangle PQR Through -150°

The triangle PQR is rotated through -150° about point O.

Solution

1. Join P to O.
2. Construct angle POP’ = 150° clockwise.
3. Mark P’ such that:

OP = OP’

4. Repeat for Q and R.
5. Join P’, Q’ and R’.

The resulting triangle is the required image.

Illustration

---

Finding the Centre of Rotation

Sometimes both the object and its image are given.

The task is to determine the centre of rotation.

Example:

Triangle ABC is mapped onto a triangle A’B’C’ under a certain rotation.

Procedure

Step 1

Join a point and its corresponding image.

For example:

C to C’

Step 2

Construct the perpendicular bisector.

Step 3

Repeat using another pair of corresponding points.

For example:

B to B’

Step 4

The intersection of the two perpendicular bisectors is the centre of rotation.

Illustration

---

Finding the Angle of Rotation

Once the centre of rotation has been found:

1. Join the centre to a point on the object.
2. Join the centre to the corresponding point on the image.
3. Measure the angle between the two lines.

This angle is the angle of rotation.

observations

For triangle ABC and image A’B’C’:

∠COC’ = 110°

Therefore:

• Angle of rotation = -110°
• Equivalent positive angle = 250°

Both answers describe the same rotation.

---

Example: Finding the Centre and Angle of Rotation of WX

Suppose line segment WX is mapped onto W’X’.

Solution

1. Join W to W’.
2. Join X to X’.
3. Construct perpendicular bisectors.
4. The bisectors meet at O.

Therefore:

O is the centre of rotation.

Measure:

∠WOW’ = 60°

Hence:

Angle of rotation = 60°

or

Angle of rotation = -300°

Illustration

---

Rotation Rules in Coordinate Geometry

For rotations about the origin:

90° Anticlockwise rotation

(x, y) → (-y, x)

Example:

(3, 2) → (-2, 3)

---

180° Rotation

(x, y) → (-x, -y)

Example:

(3, 2) → (-3, -2)

---

270° Anticlockwise Rotation

(x, y) → (y, -x)

Example:

(3, 2) → (2, -3)

---

360° Rotation

(x, y) → (x, y)

The point remains unchanged.

---

Interactive Activity

Try the rotation simulator below.

Students should:

• Rotate shapes through 90°, 180°, 270° and 360°.
• Observe how distances remain constant.
• Compare clockwise and anticlockwise rotations.
• Investigate how changing the centre of rotation affects the image.

Rotation Simulator

Explore rotations of a shape about a centre. Observe that distances from the centre remain constant.

Click anywhere on the canvas to change the centre of rotation.

---

Practice Questions

Question 1

A triangle is rotated through +90° about point O. What type of turn has occurred?

Question 2

Describe the image of a figure after a 180° rotation.

Question 3

A point (4, 2) is rotated 90° anticlockwise about the origin. Find its image.

Question 4

A figure is rotated through 360°. Describe the image.

Question 5

Explain how the centre of rotation can be determined when both the object and image are given.

---

Summary

Rotation is a transformation that turns a figure about a fixed point called the centre of rotation.

Key facts:

✓ Rotation preserves size and shape.

✓ Rotation is an isometry.

✓ Positive angles indicate anticlockwise rotation.

✓ Negative angles indicate clockwise rotation.

✓ Common angles are 90°, 180°, 270° and 360°.

✓ The centre of rotation can be found using perpendicular bisectors.

✓ The angle of rotation can be measured from corresponding points and the centre of rotation.

Understanding rotation provides a foundation for advanced studies in coordinate geometry, trigonometry, matrices, computer graphics and engineering.

Related topics

• Key concepts in matrices
• rotational dynamics
• Wave difference
• Exam questions on photoelectric effect
• Energy of radiations

Electricity powers our homes, devices, and industries, yet many students find electrical concepts difficult to understand. Two of the most important ideas in electricity are Electromotive Force (E.M.F) and Potential Difference (P.D). These concepts explain how electrical energy is supplied and how electric current flows in a circuit.

One of the easiest ways to understand E.M.F and P.D is by comparing an electric circuit to the flow of water through pipes. Just as a pump pushes water through a system, a battery pushes electric charges through a circuit. This simple analogy helps explain how voltage is produced, why current flows, and why some energy is lost inside a battery.

In this article, you will learn:

• What Electromotive Force (E.M.F) means
• The meaning of Potential Difference (P.D)
• The difference between E.M.F and P.D
• How batteries supply electrical energy
• The role of internal resistance and lost volts
• Real-life examples of voltage in electric circuits

By the end of this guide, you will have a clear understanding of how electrical energy moves through a circuit and why voltage behaves differently in open and closed circuit

---

What is Potential Difference?

Potential difference is the work done in moving a unit charge from one point to another in a circuit. It is commonly called voltage and is measured in volts (V).

A simple way to understand potential difference is by comparing it to water flowing between two containers.

Water Flow Analogy

Imagine two containers connected by a pipe:

• If one container has a higher water level than the other, water flows from the higher level to the lower level.
• The greater the difference in water levels, the faster the flow.
• When the water levels become equal, the flow stops.

see the figure below:

This difference in water levels is similar to potential difference in electricity. Charges only flow when there is a difference in electrical potential between two points.

---

Potential Energy in Water Flow

Water at a higher position possesses gravitational potential energy.

If water is raised to a height $h_1$​, it can flow down to a lower level $h_0$​. The larger the height difference, the greater the energy available to move the water.

Similarly, electric charges move from a point of higher electrical potential to a point of lower electrical potential.

The figure below shows water falling under gravitational force:

A useful way to understand electric potential difference is by comparing it to the flow of water in a closed circuit. Water raised to a higher level possesses gravitational potential energy. The higher the water is raised, the greater its potential energy and the faster it can flow to a lower level. A pump is needed to lift the water back to the higher level and maintain continuous flow. In the same way, a battery supplies energy to electric charges, creating a potential difference that causes them to move through a circuit from a region of higher potential to a region of lower potential.

Let us study the flow of water in figure below, which can be referred to as a water circuit because water flows round a complete ring.

Water at a height h₁ from the ground level has potential energy because of its position. The greater the height, the higher the potential energy. The rate of flow will depend on the height at which the water had initially been raised. A higher water level results in a faster rate of flow.

The potential energy can be calculated as:

Potential energy = mgh, where m is the mass of water falling and g the gravitational pull on water.

At a height h₀, the water has no potential energy.

If the water is to be raised to h₁, a pump has to be used. So long as the pump in the water circuit is working, the water will move round the complete path, from a point of higher potential energy to a point of lower potential energy.

The pump creates a difference in potential.

---

The Role of a Pump and a Battery in e.m.f

In a water circuit, a pump raises water to a higher level so that it continues flowing around the system. In an electric circuit, the battery performs a similar role:

• The battery pumps charges to a higher electrical potential.
• These charges then move through the conductor and electrical devices such as bulbs or lamps.
• The movement of charges forms an electric current.

The battery therefore creates the potential difference needed for current to flow.

consider the setup below:

For the charges to move through the conductor, there must be a battery which produces an electrical potential difference at the ends of the conductor. The battery does the work of pumping charges to a high potential so that they can flow. The higher the potential difference (p.d.), the stronger the current in the circuit, if other factors like opposition to flow of current (resistance) are kept constant. The model of the circuit shown in the figure above can help suggest that the function of a battery is to cause a potential difference across a conductor.

Not all the energy supplied by the pump is used to drive the water round the circuit. Some of the energy is lost in moving or raising parts of the pump. Similarly, for the battery, some energy is lost in moving charges through the battery itself. The total energy supplied by the battery is called its electromotive force (e.m.f.).

Potential difference is measured in volts, by an instrument called voltmeter.

Although both e.m.f. and p.d. are measured in volts, the potential difference of a cell is different from its e.m.f. The e.m.f. of a cell is the voltage across its terminals when it is supplying no current in the circuit (an open circuit), while the p.d. of a cell is the voltage across the cell in a closed circuit. in the Figure below: (a) and (b) shows the e.m.f. of the cell as 1.5 V and the p.d. as 1.45 V respectively.

illustrating e.m.f and p.d in a battery

When the switch is open No current flows in the circuit (a), therefore the voltmeter reads the full e.m.f of the cell.

When the switch is closed Current flows through the circuit. Some energy is lost inside the cell because of internal resistance. The terminal voltage becomes slightly lower.That is, 1.45V.

The difference between the readings is known as the lost volts, in this case 0.05 V. This voltage is lost because of the opposition to the flow of charges within the cell (internal resistance).”

---

Electromotive Force (E.M.F)

The total energy supplied by a battery to move charges around a complete circuit is called the electromotive force (e.m.f).

Although both e.m.f and potential difference are measured in volts, they are not exactly the same.

by definition; E.m.f is the voltage across the terminals of a cell when the circuit is open, and no current is flowing.

---

Difference Between E.M.F and Potential Difference

Electromotive Force (E.M.F)	Potential Difference (P.D)
Energy supplied by the cell	Energy used between two points
Measured when no current flows	Measured when current flows
Occurs in an open circuit	Occurs in a closed circuit
Represents total supplied energy	Represents useful energy delivered

---

---

e.m.f and the Lost Volts

The difference between the e.m.f and the terminal potential difference is known as the lost volts.

Example:$$\text{Lost volts} = 1.50V - 1.45V = 0.05V$$

This loss in voltage occurs because of the opposition to the flow of charges within the cell. This resistance is referred to as the internal resistance.

---

Key Points to Remember

• Charges flow only when there is a potential difference.
• A battery creates the potential difference in a circuit.
• E.m.f is the total energy supplied by a cell.
• Potential difference is the energy used between two points in a circuit.
• Internal resistance causes some voltage to be lost inside the cell.

---

Conclusion

The concepts of e.m.f and potential difference are fundamental in electricity. Using the water-flow analogy helps simplify these ideas:

• Water level difference corresponds to electrical potential difference.
• A pump corresponds to a battery.
• Water flow corresponds to electric current.

Understanding these concepts provides a strong foundation for studying electric circuits and electrical energy transfer.

Revision exercise

---

Prepared for physics learners and teachers as a simple guide to understanding electromotive force and potential difference.

Related topics

• Electric current and potential difference
• Simple electric circuits
• The ohm’s law
• High energy transmission
• Domestic wiring
• Electromotive force and potential difference

The function of a cell in a circuit is to supply electrical energy. By definition, the electromotive force (e.m.f.) of a cell is the potential difference between its terminals when no charge is flowing out of the cell (cell in open circuit).

Electric cells and batteries are essential sources of electrical energy in everyday life. From powering flashlights and radios to running vehicles and electronic devices, cells convert chemical energy into electrical energy. However, no cell is perfect. Every cell possesses a small internal resistance that affects the voltage supplied to an external circuit.

Figure below shows a circuit that may be used to demonstrate the difference between e.m.f. of a cell and terminal voltage.

The reading of the voltmeter when the switch is open is the e.m.f. of the cell.

Once a cell supplies current to an external circuit, the potential difference across it drops by a value referred to as ‘lost voltage’. This loss in voltage is due to the internal resistance of the cell.

The potential difference across the cell when the circuit is closed is referred to as the terminal voltage of the cell.

---

Relationship Between electromotive force and Internal Resistance

If a resistor $R$ is connected in series with a cell as shown in figure below, the internal resistance of the cell $r$, is considered to be connected in series with the external resistor $R$.

The current flowing in the circuit is therefore given by the equation;

$I = \frac{E}{R+r}$

where $E$ is the e.m.f. of the cell.

Thus,

$E = I(R+r)$

$E = IR + Ir$

=V+Ir

$IR$ is the voltage drop across the external resistor $R$R while Ir is the voltage drop across the internal resistance.

The voltage across the external resistor is called the terminal voltage while the p.d. drop across the internal resistance is called the lost voltage.

Battery Internal Resistance Animation

---

Experiment to determine the internal resistance and electromotive force of a cell

Method 1

Apparatus

Voltmeter, ammeter, rheostat, cells, connecting wires.

Procedure

• Connect the apparatus as shown in figure below

• Switch on the circuit and set the current to the minimum value possible.
• Increase the current in steps and record the corresponding terminal voltage $V$ in table below.

Table of current against voltage

current I(A)
Voltage (V)

• Plot a graph of voltage against current.

Determining Internal Resistance of a Cell

---

Results and Conclusion

The graph of voltage against current is as shown below.

Using the equation $E = V + Ir$ and hence V=E-Ir; the gradient of the graph gives the internal resistance of the cell.

If the graph is extrapolated so as to cut the voltage axis, the point at which it does so gives the electromotive force(e.m.f) of the cell.

---

Method 2

Apparatus

• Ammeter
• voltmeter
• variable resistor
• cells
• connecting wires.

Procedure

• Set the apparatus as shown:

• Switch on the circuit and increase the current in step from a minimum value.
• Record the corresponding voltage $V$.

---

Complete table

Current I (A)
Voltage(V)
R=V/I
1/I

Plot a graph of $\frac{1}{I}$ against $R$

---

Results and Observation

The graph is a straight line with a positive gradient. see the diagram below

The gradient of the graph gives $\frac{1}{E}$

Internal resistance can be obtained in two ways:

(i) Extrapolating the graph to cut $R$ axis gives $r$ as can be observed on the diagram above.

(ii) If the intercept on axis is $A$, then,

$A = \frac{r}{E}$

So,$r = A \times E$

But: $E = \frac{1}{\text{Gradient}}$

Therefore,

$r = A \times \frac{1}{\text{Gradient}}$

---

Example 20

A battery consisting of four cells in series, each of e.m.f. 2.0 V and internal resistance 0.6 Ω, is used to pass a current through a 1.6 Ω resistor. Calculate the current through the battery.

Solution

Current through battery = $\frac{\text{e.m.f. of battery}}{\text{total resistance}}$

The electromotive force(e.m.f) of the battery is the sum of the e.m.f. of all the cells while the internal resistance of the battery is the sum of all internal resistances of the cells.

Therefore, current through the batter

$I = \frac{2.0 \times 4}{(0.6 \times 4)+1.6}$

$I = \frac{8.0}{4.0}$

$I = 2.0\ A$

---

Example 22

A cell drives a current of 2.0 A through a 0.6 Ω resistor. When the same cell is connected to a 0.9 Ω resistor, the current that flows is 1.5 A. Find the internal resistance and the electromotive force(e.m.f) of the cell.

Solution

The first connection is as shown with it's internal resistance.

when connected to 0.9 ohm resistor, the circuit is as shown:

Taking E as the electromotive force(e.m.f) of the cell and r the internal resistance.

E = IR+Ir

from the below figure:

E = (2.0 x 0.6)+2.0r = 1.2+2r --------(i)

using the figure below:

E = (1.5 x 0.9)+1.5r

E = 1.35 + 1.5r ---------(ii)

since e.m.f is the same in both circuits:

1.2+2r = 1.35+1.5r

2r-1.5r = 1.35-1.2

=.5r = 0.15

r = 0.3Ω

substituting for r in the first equation:

E = 1.2+2r = 1.2+2(0.3)

E= 1.2+0.6 = 1.8V

Example problem

A battery consists of two identical cells, each of e.m.f. $1.5\,V$1.5V and internal resistance $0.6\,\Omega$0.6Ω, connected in parallel. Calculate the current the battery drives through a $0.7\,\Omega$0.7Ω resistor.

Solution

When identical cells are connected in parallel,the equivalent e.m.f. is equal to that of only one cell.

The figure below represents the arrangement:

The equivalent internal resistance is equal to that of two such resistors connected in parallel as shown in the diagram above. Figure (a) is simplified to figure (b).

Equivalent e.m.f. $= 1.5\,V$

Equivalent internal resistance will be given by:

$$r_T = \frac{r_1 r_2}{r_1 + r_2}$$

​​substituting:$$= \frac{0.6 \times 0.6}{0.6 + 0.6}$$$$= 0.3\,\Omega$$$$I = \frac{E}{R + r}$$

Current through the $0.7\,\Omega$ will be given by:

$$= \frac{1.5}{0.7 + 0.3}$$$$= 1.5\,A$$

---

Example 23

In an experiment to determine the electromotive force $(E)$ and internal resistance of a cell, the following results were obtained.

I (A)	0.5	1.0	1.5	2.0	2.5
V (V)	1.25	1.0	0.75	0.5	0.25

Plot a graph of $\frac{1}{I}$ against $R$. From the graph, determine the values of $E$ and $r$.

Solution

A table for $\frac{1}{I}$​ and $R$ is generated from the values given as follows:

$\frac{1}{I}$	2.0	1.0	0.67	0.5	0.4
$R = \frac{V}{I}$​	2.5	1.0	0.5	0.25	0.1

A plot of $\frac{1}{I}$ against $R$ is as follows:

The graph is a straight line whose gradient is $\frac{1}{E}$

Thus,$$\frac{1}{E} = \frac{2.0 - 0.75}{2.5 - 0.65}$$$$= \frac{1.25}{1.85}$$

Hence,$$E = \frac{1.85}{1.25} = 1.48\,V$$

The value for $r$ is found by extrapolating the graph until it cuts the R-axis and reading off $r$ as indicated on the graph.

---

Thus, r=0.48Ω

Alternatively

$$y = \frac{r}{E}$$

Thus,$$r = y \times E$$$$= 0.33 \times 1.48$$$$= 0.488\,\Omega$$

---

Practice Questions

1. State the physical quantities whose units are;
   • ampere,
   • ohm,
   • volt,
   • coulomb and watt.
2. State Ohm’s law and describe an experiment to verify it.
3. For the resistor network given, determine
   (a) the total resistance
   (b) the voltage drop across each resistor.
   (c) the current through each resistor.

4. The figure below shows four resistors and a source of voltage of $6\,V$ with internal resistance $0.2\,\Omega$

(a) Find the effective resistance of the circuit.
(b) Calculate the current through $r$.

5. Six resistors are connected in a circuit as shown in the figure below.

Calculate the:
(a) total resistance of the circuit.
(b) total current in the circuit.
(c) current through the $3\,\Omega$ resistor.
(d) current through the $8\,\Omega$ resistor.

6. (a) You are provided with two resistors of values $4\,\Omega$ and $8\,\Omega$.
   (i) Draw a circuit diagram showing the resistors in series with each other and with a battery.
   (ii) Calculate total resistance of the circuit (assume negligible internal resistance).

(b) Given that the battery has an e.m.f of 6V and an internal resistance of 1.33Ω:

calculate the current through:

(i) 8Ω

(ii) 4Ω resistor when the two are in parallel.

practice questions

Related topics

• Current and potential difference
• Simple electric circuits
• The ohm’s law
• High energy transmission
• Domestic wiring
• Internal resistance explained

Trigonometry becomes much easier when you understand the unit circle. The unit circle helps us define trigonometric ratios for all angles, including positive, negative, and angles greater than 90°.

What Is the Unit Circle?

A unit circle is a circle with:

• Centre at O(0,0)
• Radius equal to 1

The circle is drawn on the Cartesian plane with the x-axis and y-axis crossing at the centre. see the figure below

The Four Quadrants

The unit circle is divided into four sections called quadrants:

• First Quadrant (Quadrant I) → top right
• Second Quadrant (Quadrant II) → top left
• Third Quadrant (Quadrant III) → bottom left
• Fourth Quadrant (Quadrant IV) → bottom right

Positive and Negative Angles

Angles are measured from the positive x-axis.

• An angle measured anticlockwise is positive.
• An angle measured clockwise is negative.

Examples:

• 120° is a positive angle and lies in the second quadrant.
• -50° is a negative angle and lies in the fourth quadrant.

Determining Quadrants of Angles

To know where an angle lies:

First Quadrant

Angles between 0° and 90°

Example:
30° lies in Quadrant I

Second Quadrant

Angles between 90° and 180°

Example:
140° lies in Quadrant II

Third Quadrant

Angles between 180° and 270°

Example:
240° lies in Quadrant III

Fourth Quadrant

Angles between 270° and 360°

Example:
330° lies in Quadrant IV

Negative Angles

Negative angles move clockwise.

Example:
-70° lies in Quadrant IV
-120° lies in Quadrant III

The figure below shows angles of 120^o and -50^o marked on the unit circle. They are in the second and fourth quadrants respectively.

Determine which quadrants where 35^o, 45^o, 190^o, 280^o, 330^o,235^o are found.

---

Coordinates on the Unit-Circle

One important idea about the unit circle is that every point on the circle represents:

(x, y) = (cos θ, sin θ)

This means:

• x-coordinate = cos θ
• y-coordinate = sin θ

Figure below is a unit-circle and angle PON=30°. Determine the values of x and y at point P.

Angle AON is a right-angled at N. Therefore:

A right-angled triangle is formed inside the circle.

Since the radius of the unit circle is 1:

OP = 1

Using trigonometric ratios:

Now for cosine:

Therefore, the coordinates of point P are:

P(0.86, 0.5)

Therefore, for a unit circle:

sinθ = y co-ordinates of P

cosθ = x co-ordinates of P.

---

Key Ideas to Remember

• The unit circle has radius 1.
• Positive angles move anticlockwise.
• Negative angles move clockwise.
• Every point on the unit circle represents: (cos θ, sin θ)
• The x-coordinate gives cosine.
• The y-coordinate gives sine.

The unit-circle is the foundation for understanding trigonometric functions, graphing, and solving advanced trigonometry problems.

Related topics

• factorization of quadratic equation
• Introduction to quadratic equation and expression
• Unit circle

Ohm’s Law and Electrical Resistance are fundamental concepts in electricity explaining the relationship among voltage, current, and resistance. Electrical resistance describes how strongly a material opposes the flow of electric current. Understanding these concepts is essential for analyzing circuits, designing electrical systems, and explaining how electronic devices operate in everyday life.

One of the most important principles that helps us understand how electricity behaves in a circuit is Ohm’s Law. This law explains the relationship between voltage, current, and resistance in an electrical conductor.

Electricity powers almost every device we use today, from mobile phones and televisions to electric cars and industrial machines.

In this lesson, we shall explore Ohm’s Law and the electrical resistance, how it is verified experimentally, the meaning of electrical resistance, and the factors that affect resistance in conductors.

---

What is Ohm’s Law?

Ohm’s Law states that:

The current flowing through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant.

This means that when the voltage across a conductor increases, the current flowing through it also increases proportionally.

The mathematical expression of Ohm’s Law is:

V=IR

some of the physical conditions includes pressure and tensional forces on the conductor.

Where:

• V = Voltage (Volts)
• I = Current (Amperes)
• R = Resistance (Ohms)

---

Investigating the Relationship Between Current and Voltage in ohm’s law

To verify Ohm’s Law, a simple experiment can be carried out using a nichrome wire.

Apparatus Required

• Two-metre nichrome wire
• Two dry cells
• Ammeter
• Voltmeter
• Rheostat
• Switch
• Connecting wires

Experimental Setup to study ohm’s law and electrical resistance

• Using a nichrome wire, make a coil of as many turns as possible
• Set up the circuit as shown in figure below

• Set the current flowing in the circuit to the least possible value
• with help of the rheostat, vary in steps the current flowing in the circuit and not e the corresponding voltage drop across the coil.
• Record the results in the table below

Current (A)
Voltage (V)

Observation of ohm’s and electrical resistance

As the current flowing through the nichrome wire increases, the voltage across the wire also increases. When voltage is plotted against current, the graph obtained is a straight line passing through the origin.

The table below represents a sample data from such an experiment.

Current (A)	0.1	0.2	0.3	0.4
Voltage (V)	1.2	2.4	3.6	4.8

when this data is plotted on a grid, a graph as shown is obtained.

Points to note:

The circuit is connected such that:

• The ammeter measures the current flowing through the wire.
• The voltmeter measures the voltage across the nichrome wire.
• The rheostat is used to vary the current in the circuit.

Data Analysis and conclusions

The gradient of the graph will be given by:

---

from the graph, the following observations can be obtained:

As the current increases, the voltage across the coil also increases.

The graph obtained when voltage is plotted against current is a straight line passing through the origin.

Therefore:

• voltage is directly proportional to current
• gradient of the graph is a constant. This constant gives the resistance of the conductor used.

---

This straight-line graph confirms that voltage is directly proportional to current.

---

Understanding Resistance

From the experiment, the ratio of voltage to current remains constant. We can verify the ohm’s law with the same procedure described above when we replace a coil with a standard resistor. The graph of current against voltage is a straight line through the origin.

From the graph above:

From V ∝ I:

V=constant(K)x I

The constant which we represent with K is the resistance of the conductor.

hence;

V=IR where V is the potential difference across the conductor.

Resistance can therefore be calculated using:

The SI unit of resistance is the ohm (Ω).

An ohm is defined as the resistance of a conductor when a current of 1A flowing through it produces a voltage drop of 1 V across it’s ends.

A conductor is said to have a resistance of 1 ohm if a current of 1 ampere flows through it when a potential difference of 1 volt is applied across it.

an ohm have some other units like:

1 kilo ohm(KΩ) = 1000Ω

1 mega ohm(MΩ) = 1 000 000Ω

Worked Examples

Example 1

A current of 2 mA flows through a conductor of resistance 2 kΩ. Calculate the voltage across the conductor.

solution:

Using Ohm’s Law:

V=IR

---

Example 2

Calculate the current flowing through a 50 Ω resistor connected to a 10 V battery.

solution

from ohm’s law:

---

Example 3

A starter motor requires a current of 50 A from a 12 V battery. Determine the resistance of the motor.

solution

---

Ohmic and Non-Ohmic Conductors

Ohmic Conductors

Conductors that obey Ohm’s Law are called Ohmic conductors.

Examples include:

• Nichrome wire
• Metallic resistors

For these conductors, the graph of voltage against current is a straight line.

---

Non-Ohmic Conductors

Some conductors do not obey Ohm’s Law. These are known as Non-Ohmic conductors.

Examples include:

• Filament lamps
• Thermistors
• Semiconductor diodes
• Electrolytes

Their voltage-current graphs are curved instead of straight.

---

Electrical Resistance

Electrical resistance is the opposition offered by a conductor to the flow of electric current.

Resistance occurs because electrons moving through a conductor collide with atoms and impurities inside the material. These collisions reduce the flow of charge and produce heat energy.

An instrument used to measure resistance is called an ohmmeter.

5

---

Factors Affecting Resistance

Several factors determine the resistance of a conductor.

---

1. Length of the Conductor

The resistance of a conductor increases with its length.

that is: R ∝ l

hence; resistance = constant x length

i.e = R = Kl ————————(i)

for a given conductor:

As the length of the conductor increases, the resistance increases because of the increased number of atoms that are available to hinder the flow of electrons.

A longer wire contains more atoms that obstruct the movement of electrons, leading to greater resistance.

---

2. Cross-Sectional Area

Resistance decreases when the cross-sectional area increases A.

That is: resistance is inversely proportional to cross section area(A) of the conductor.

A conductor with a larger cross-section area(A) has many free electrons for conduction, hence better conductivity.

​

Thicker wires allow more electrons to flow easily and therefore have lower resistance.

Combining (i) and (ii) for a conductor with uniform cross-section area;

The constant value in the equation above is referred to as the resistivity(ρ)of a material. It is practically the resistance of sample of a material of unit length and unit cross-section area at a given temperature. The unit of measurement for resistivity(ρ) known as ohmmeter(Ωm).

The table below shows resistivity of some common materials

Material	Resistivity (Ωm)	Common Uses
Silver	1.6 × 10⁻⁸	Contacts on some switches
Copper	1.7 × 10⁻⁸	Connecting wires
Aluminium	2.8 × 10⁻⁸	Power cables
Tungsten	5.5 × 10⁻⁸	Lamp filaments
Constantan	49 × 10⁻⁸	Resistance boxes, variable resistors
Nichrome	100 × 10⁻⁸	Heating elements
Carbon	3,000 × 10⁻⁸	Radio resistors
Glass	10¹⁰ – 10¹⁴	Electrical insulators
Polystyrene	10¹⁵	Electrical insulators

Example problem in resistance

Two wires of A and B are such that the radius of A is twice that of B and the length of B is twice that of A. if the two are of the same material, determine the ratio:

solution:

Therefore, after rearranging the expression:

since l_B = 2l_A ; 2R_B = R_A

---

3. Temperature

For metallic conductors, resistance increases with temperature.

Heating causes atoms in the conductor to vibrate more vigorously. This increases collisions between electrons and atoms, making it more difficult for current to flow.

---

---

Resistivity of Materials

Resistivity is a property that shows how strongly a material opposes electric current.

Materials such as:

• Silver and copper have low resistivity and are good conductors.
• Glass and polystyrene have high resistivity and act as insulators.

Temperature also affects resistance:

• In metals, resistance increases with temperature.
• In semiconductors, resistance decreases with temperature.

---

Resistors

A resistor is an electrical component designed to provide resistance in a circuit.

Resistors are used to:

• Control electric current
• Reduce voltage
• Protect circuit components
• Produce heat in appliances

Most wire-wound resistors are made using materials such as:

• Manganin
• Constantan

These materials are preferred because their resistance changes very little with temperature.

---

Fixed Resistors

Fixed resistors have a constant resistance value.

Types of Fixed Resistors

1. Wire-Wound Resistor

This resistor is made by winding resistance wire around an insulating core.

Features:

• High durability
• Can handle large currents
• Common in power circuits

2. Carbon Resistor

Made using carbon material.

Features:

• Cheap and widely used
• Small in size
• Used in electronic circuits

Resistor Symbol

The electrical symbol of a resistor is represented by a zigzag or rectangular shape depending on the standard used.

---

Variation of Resistance with Temperature

Different materials respond differently to temperature changes.

Metals

Resistance increases as temperature rises.

Thermistors

Resistance decreases as temperature rises.

Constantan

Resistance remains nearly constant despite temperature changes.

This behavior is important in designing temperature-sensitive circuits.

---

Variable Resistors

A variable resistor allows resistance to be adjusted manually.

The resistance changes when a sliding contact moves along the resistance track.

Applications include:

• Volume controls in radios
• Light dimmers
• Fan speed regulators

---

Rheostat

A rheostat is a variable resistor with two terminals.

It is used to control current in a circuit.

As the slider moves:

• The effective length of the resistance wire changes
• Resistance changes accordingly

Increasing the resistance reduces current flow.

---

Potentiometer

A potentiometer is a variable resistor with three terminals.

It is used to:

• Divide voltage
• Control signal levels
• Adjust volume in audio systems

How It Works

A sliding contact moves along the resistor track, selecting different voltage levels.

Potentiometers are commonly used in:

• Audio amplifiers
• Electronic control systems

---

Non-Linear Resistors

These resistors do not obey Ohm’s Law strictly because their resistance changes non-linearly with voltage, temperature, or light.

Examples include:

• Thermistors
• Light-dependent resistors (LDRs)

---

Thermistor

A thermistor is a temperature-dependent resistor.

Characteristics

• Resistance decreases as temperature increases.
• Used in heat-sensitive circuits.

Applications

• Temperature sensors
• Fire alarms
• Electronic thermometers

---

Light-Dependent Resistor (LDR)

An LDR changes resistance according to the amount of light falling on it.

Characteristics

• High resistance in darkness
• Low resistance in bright light

Applications

• Automatic street lights
• Camera light sensors
• Burglar alarms

---

---

Importance of Resistors in Daily Life

Resistors are found in almost every electrical and electronic device.

They help to:

• Protect circuits from excessive current
• Control electrical energy
• Improve device performance
• Enable automatic sensing systems

Without resistors, modern electronics would not function safely or efficiently.

---

Resistors play a vital role in electrical and electronic circuits. From fixed resistors to thermistors and LDRs, these components help control current, voltage, temperature, and light sensitivity in devices we use every day.

Understanding how resistors work gives students a strong foundation in physics and electronics, preparing them for more advanced studies and practical applications in technology.

---

Conclusion

Ohm’s Law is one of the most fundamental principles in electricity and electronics. It helps us understand how voltage, current, and resistance are related in electrical circuits. Through experiments and graphical analysis, students can clearly observe the direct relationship between voltage and current in Ohmic conductors.

Understanding electrical resistance and the factors affecting it is essential in designing safe and efficient electrical systems used in homes, schools, laboratories, and industries.

Whether you are studying basic physics or advanced electronics, mastering Ohm’s Law provides the foundation for understanding the behavior of electric circuits.

Related Topics

• Current and potential difference
• Resistor
• Introduction to electronics
• Intrinsic and extrinsic semiconductors
• Simple electric circuits

Pressure in liquids is a fascinating concept that explains why divers feel greater force underwater and why dams are built thicker at the bottom than at the top. As depth increases, the pressure exerted by a liquid also increases because more liquid presses down from above.

This principle plays an important role in everyday life, engineering, and natural water systems. Understanding how depth affects pressure helps us explain many real-world phenomena, from submarine design to the flow of water in oceans and rivers.

At the same depth in a given liquid, differences in levels obtained is the same regardless of the direction which the funnel faces.

To investigate the variation of liquid pressure with depth and density

Apparatus

A tall jar, liquids of different densities, thistle funnel, U-tube, rubber tubing.

---

Procedure

• Using the nail, make three holes, A, B and C, of the same diameter along a vertical line on one side of the tin.
• Fill the tin with water as shown in figure 4.3.
• With the tin full of water, observe the jets of water from the holes A, B and C.

---

Observation

The lower hole, A, throws water farthest, followed by B

Conclusion

Pressure in liquids increases with density and depth.

In summary:

1. Pressure in a liquid increases with depth below its surface.
2. Pressure in a liquid at a particular depth is the same in all directions.
3. Pressure in a liquid increases with the density of the liquid.

---

Fluid Pressure Formula

Consider a liquid in a container, as shown in figure 4.8.

If A is the cross-sectional area of the column, h the height of the column and ρ the density of the liquid, then:

Volume of the liquid= cross-sectional area × height
= Ah

Mass of the liquid=volume of the liquid × density
= Ahρ

Therefore, weight of the liquid
= mass of the liquid × gravitational force per unit mass
= Ahρg

From the definition of pressure:

So fluid pressure becomes:

$P=h\rho g$P=hρg

---

EXPERIMENT To show the distribution of pressure at a point in a liquid

Apparatus

A tall jar, water, thistle funnels, U-tube, rubber tubing.

Procedure

• Fill the glass vessel G with water.
• Connect one of the thistle funnels to a U-tube filled to some level with water.
• Lower the funnel to a depth from the surface of water and notice the difference in levels, h, of the water in the U-tube.
• Replace the funnel with others, in turn, whose mouths are pointing in different directions.
• Lower the funnel into the water so that the mouth of the funnel is at the same point as the straight one. Observe the difference in levels of the water in the U-tube.

---

Figure 4.6: Pressure variation in a liquid

Procedure

• Fill the glass vessel G with water.
• Connect the thistle funnel to a U-tube filled to some level with water.
• Lower the funnel to different depths from the surface and notice the difference in levels, h, of water in the U-tube.
• Replace water in G with a denser liquid, such as sodium chloride solution (brine).
• Lower the funnels to the same depths as above and compare the heights obtained.

Observations

1. The deeper the funnel goes below the surface, the greater the difference in levels, h.
2. The differences in levels, h, obtained with brine at a particular depth is greater than that obtained with water at that depth.

Liquid Levels in a U-tube

When water is poured into a U-tube, it will flow into the other arm. The water will settle in the tube with the levels on both arms being the same, see figure 4.5(a).

When one arm of the U-tube is blown into with the mouth, the level moves downwards, while on the other arm it rises, see figure 4.5(b). This is caused by the pressure difference between the two arms. The pressure increases on the arm that is blown into and causes water to rise on the other arm.

---

Effect of pressure on liquid levels

---

Pressure of water at A is greater than pressure at B and pressure at B is greater than at C. Hence, pressure increases with depth.

For this reason, a diver at the bottom of the dam experiences pressure due to the weight of water above him. The deeper the diver goes, the greater the pressure.

---

Liquid Levels

When a liquid is poured into a set of connected tubes with different shapes, it flows until the levels are the same in all the tubes, as shown in figure 4.4.

Related topic

• Introduction to pressure
• Pressure in liquids | Matter | Physics | FuseSchool

An inclined plane is one of the simplest yet most useful machines in everyday life. It is a flat surface set at an angle to help move heavy objects from a lower level to a higher level with less effort. Instead of lifting a load straight upward, an inclined plane allows force to be applied over a longer distance, making work easier and more efficient. Common examples include ramps, staircases, sloping roads, and playground slides. Inclined planes have been used since ancient times in construction, transportation, and engineering, and they continue to play an important role in modern technology and daily activities.

solving problems involving inclined plane

Consider an inclined sheet of metal placed to form a slope to ease the process of loading heavy luggage onto a truck as in the diagram below:

The velocity ratio of the inclined plane will be given by:

From trigonometric ratios:

substituting for h in the denominator:

Experiment To find the mechanical advantage of an inclined plane

Apparatus

• A pulley
• string
• two metre rules
• weighing balance
• flat plane
• five blocks of wood of different masses
• pan
• sand

---

Procedure of find the mechanical advantage of an inclined plane

• Fix the flat plane at an angle of inclination of about ${40}^{\circ \; to\; the\; horizontal.}$
• Place the load L in on an inclined plane and tie it with a string running over a pulley and attached to a pan, as in figure below:

• Add sand to the pan until the load moves steadily up the incline.
• Measure the load and the effort (pan + sand) using a weighing balance. Record the results on the table shown in table.

Table 2.1: Finding M.A. of an inclined plane

Load L(N)	Effort E (N)	M.A = E/L

• Repeat the experiment for other values of $L$ and $E$.
• Calculate the M.A. for each pair of values.

---

---

Observation

The ratio of load to effort is found to be a constant, i.e.

---

Experiment To find the velocity ratio of an inclined plane

Apparatus

• A pulley
• two metre rules
• two blocks of wood
• one small and the other big
• pan
• sand.

---

Procedure of finding the velocity ratio of an inclined plane

• Fix a flat plane at an angle of inclination of 30∘.
• Place a block of wood on the plane and tie it with a string running over the pulley and attached to the pan, as shown in figure below:

• Add sand to the pan until the load moves steadily up the incline.
• When the load stops moving, record lengths $D_L$​ and $D_E$.
• Add more sand so that the effort (pan with sand) moves further down.
• Measure the $D_L$DL​ and $D_E$DE​ for the new positions.
• For each pair of values, calculate the velocity ratio, as shown below;

Distance moved by effort DE(cm)	Distance moved by load DL(cm)	V.R = DE\DL

Example 20

A man uses the inclined plane to lift a 50 kg load through a vertical height of 4.0 m. The inclined plane makes an angle of 30° with the horizontal. If the efficiency of the inclined plane is 72%, calculate:

(a)The effort needed to move the load up the inclined plane at a constant velocity.

(b) The work done against friction in raising the load through the height of 4.0 m.

Take: g = 10Nkg^-1

solution

(a)

picture the setup as in the figure below:

/

(b)

Work done against friction = work input – work output

work output = mgh = 50 x 10 x 4 = 2000J

work input = effort x distance moved by effort

= 327.2 x AC

Therefore, the work done against friction = 2777.6 – 2000 = 777.6J

Related Topics

• Introduction to pressure
• floating and sinking
• Introduction to Inclined Planes

Electric resistance is one of the most important concepts in current electricity. Resistance determines how easily electric current flows through a conductor. In physics laboratories, several methods are used to measure resistance accurately, including the voltmeter-ammeter method, the Wheatstone bridge method, and the metre bridge method. These techniques form the foundation for understanding electrical circuits and practical electrical measurements.

---

The Voltmeter-Ammeter Method of Measuring Resistance in Electric Circuits

Aim of the Experiment

To determine the resistance of a resistor using a voltmeter and an ammeter.

---

Apparatus Required

• Two cells
• Switch
• Voltmeter
• Ammeter
• Variable resistor
• Resistor (R)

---

Circuit Arrangement for Measuring Resistance in Electric Circuits

In this method:

• The ammeter is connected in series with the resistor to measure current.
• The voltmeter is connected across the resistor to measure potential difference.
• A variable resistor is used to vary the current flowing through the circuit.

---

Procedure

1. Set up the circuit as shown in the diagram.
2. Keep the switch open and note the voltmeter reading (V) and the corresponding ammeter reading (I).
3. Close the switch.
4. Adjust the variable resistor and record several values of voltage and current.
5. Calculate the value of (\frac{V}{I}) for each reading.
6. Plot a graph of voltage (V) against current (I).
7. Determine the slope (gradient) of the graph.

---

Observation

When the switch is open, no current flows through the resistor. Therefore:

• Ammeter reading = 0
• Voltmeter reading = 0

As current increases, the voltage across the resistor also increases.

According to Ohm’s law:

in other words; The resistance is obtained by dividing the voltage across the resistor by the current flowing through it.

---

Graph of Voltage Against Current

The graph of (V) against (I) is a straight line passing through the origin.

V=IR

The slope (gradient) of the graph gives the resistance of the resistor.

---

Limitation of the Method

This method is not perfectly accurate because the voltmeter draws a small amount of current. Therefore, not all the current measured by the ammeter passes through the resistor.

---

The Wheatstone Bridge Method

The Wheatstone bridge is a more accurate method of measuring resistance.

It consists of:

• Four resistors
• A galvanometer
• A cell

operation of the Wheatstone bridge

Wheatstone bridge operations involves making adjustments to one or two of the resistors until there is no deflection in the galvanometer. The four resisters K,L,M and N are joined as shown.

If K is the unknown resistance, the value of L, M and N must be known. Alternatively, the ratio of M to N must be known. A galvanometer G and a cell are connected as in figure above.

Variable resistor L(resistance box) is adjusted till there is no deflection in the galvanometer G. The bridge is then said to be balanced. At the state of balance, no current is flowing through G, hence the potential difference across BD is zero. When this happens, the potential difference across AB = potential difference across AD. The same current I₁ flows through K and L and current I₂ flows through M and N. then:

I₁ = I₃ and I₂ = I₄

I₁k = I₂M and I₃L = I₄N

As one can see, the currents are cancelling each other. hence when the bridge is balanced:

---

points to note

The four resistors are connected in a bridge arrangement.

When the galvanometer shows no deflection:

• The bridge is said to be balanced.
• No current flows through the galvanometer.
• This equation is used to determine an unknown resistance.

Wheatstone bridge is more accurate in measuring resistance compared to voltmeter-ammeter method because the value obtained does not depend on the accuracy of the current measuring instrument.

---

Advantages of the Wheatstone Bridge

• It gives more accurate measurements.
• The result does not depend greatly on the accuracy of the galvanometer.
• It is suitable for measuring small resistances precisely.

---

The Metre Bridge method of Measuring Resistance in Electric Circuits

The metre bridge is a practical form of the Wheatstone bridge.

It uses:

• A uniform wire one metre long
• Two resistors
• A galvanometer
• A jockey (movable contact)

---

Working Principle

A typical setup of the metre bridge is shown in the figure below:

The wire AC of uniform cross-section area and length 1 m with a resistance of several Ohm’s and made of an alloy such as constantan. The length AD represents resistor M while the length CD represents resistor N. The ratio of M to N is altered by changing the position D on the wire of the movable contact D called ‘jockey’.

The other arm of the bridge contains an unknown resistor K and a known resistor L. The copper strips of low resistance connect the various parts. The position of D is adjusted until there is no deflection in G. Then;

---

Since the wire is uniform cross-section, its resistance will be proportional to its length hence:

The resistor L should be chosen to give balance points near the centre of the wire. This gives a more accurate result. After obtaining the balance, K and L should be interchanged and a second pair of values for $X_1$​ and $X_2$​ obtained. This average of the value eliminates errors due to non-uniformity of the wire and end corrections. In finding the balance point, the cell key or switch should be closed before the jockey makes contact with the wire. This is necessary because of the effect known as ‘self-induction’ in which the currents in the circuit take a short time to grow to their steady values. A high resistance should always be joined in series with the galvanometer to protect it from damage whilst the balance is being sought.

Precautions in Using the Metre Bridge

• Choose balance points near the centre of the wire for greater accuracy.
• Close the switch before the jockey touches the wire.
• Use a high resistance in series with the galvanometer to protect it from damage.

---

Worked Example

In an experiment to determine the resistance of a nichrome wire using the metre bridge, the balance point was found to be at 38 cm mark. If the value of the resistance in the right hand gap needed to balance the bridge was 25 Ω, calculate the value of the unknown resistor.

solution

picture the setup to be as shown in the diagram below:

Since AB = 100 cm and AC = 38 cm, BC = 100 − 38 = 62 cm;

---

Resistors Connected in Series

When resistors are connected end to end, they form a series circuit.

Characteristics of series connection:

• The same current flows through all resistors.
• The total voltage equals the sum of individual voltages.

Using Ohm’s law:

[
V_T = V_1 + V_2 + V_3
]

The total resistance in series is:

genui{“math_block_widget_always_prefetch_v2”:{“content”:”R_T=R_1+R_2+R_3″}}

---

Conclusion

The measurement of resistance is an important practical skill in electricity. The voltmeter-ammeter method provides a simple way to determine resistance using Ohm’s law, while the Wheatstone bridge and metre bridge offer greater precision. Understanding these methods helps students appreciate how electrical measurements are carried out in laboratories and real-world electrical systems.

Related topics

• power consumption
• https://www.youtube.com/watch?v=IjZlPMUnUts

Energy is measured in joules (J). Power is measured in watts (W). You will recall that power is the rate of energy transfer or power = energy ÷ time and that 1 W is the power when 1 J of energy is being transferred every second.

The amount of energy used by a consumer depends on:

(i) Power rating of appliances.
(ii) Time for which they have been used.

Therefore, it follows that the amount of energy consumed by a device (an electric lamp, for example) is equal to its power multiplied by the time for which it has been operating:$$\text{Energy} = \text{power} \times \text{time}$$

An alternative unit for energy is thus the watt-second (Ws) since 1 J = 1 W × 1 s.

In 10 seconds, a 100 watt light bulb consumes:$$100 W \times 10 s = 1\,000 Ws$$

But watt-second is too small a unit for practical use. The unit used in practice is the kilowatt-hour (kWh). This is not an SI unit but is used all over the world.

One kWh is the amount of electrical energy used in one hour (3 600 s) at a power of 1 kW (1 000 W):$$1 kWh = 1\,000 W \times 3\,600 s = 3.6 \times 10^6 J$$

If we know the power rating of an appliance and the time for which it is used, the consumption in kWh is easily calculated.

Electricity companies charge by the kilowatt-hour for the use of the power they produce — they call 1 kWh a unit. So the cost of using an appliance for a certain length of time is worked out using total cost = number of kWh × cost per unit.

An electricity meter records the total number of units consumed on the premises where it is located.

An electricity meter reading units (kWh) is as illustrated below:

---

Example 5

A 1 200 W hair dryer is used for 15 mins. What is the energy used by the dryer?

Answer

---

Example 6

Two 40 W bulbs are left on for 6 hours. How much energy will they consume?

Answer

time in seconds = 6hours x 60min x 60secs = 21600 seconds

E = power x times

power = 40W x 2 = 80W

Energy(E) = 80 x 21600 = 2028000J

---

Example 7

An electric heater is used for 45 minutes. If it consumes 2 250 kJ of energy within this time, calculate its power rating.

Answer

$$E = pt \therefore P = \frac{E}{t}$$$$P = \frac{2\,250 J}{45/60}$$$$P = \frac{2\,250 \times 60}{45}$$​ $$= 3\,000 W$$$$= 3 kW$$

---

Problems on mains electricity

Example 8

If one unit of electricity costs sh 3.5, calculate the cost of using:

a) a 60 W light bulb
b) a 1 kW heater

for 30 minutes each.

Answer

a)

Power = 60 W$$= \frac{60}{1\,000} kW$$

Time = 30 min$$= \frac{30}{60} h$$

Energy consumed$$= \frac{60}{1\,000} kW \times \frac{30}{60} h$$$$= 0.03 kWh$$

Cost = number of kWh × cost per unit$$= 0.03 \times 3.5 = sh 0.105$$

b)

Power = 1 kW

Time = 30 min$$= \frac{30}{60} = 0.5 h$$

Energy consumed:$$= 1 kW \times 0.5 = 0.5 kWh$$

Cost = number of kWh × cost per unit$$= 0.5 \times 3.5 = sh 1.75$$

---

Example 9

A home has the following appliances:

a) A heater (5 kW)
b) Iron box (2.5 kW)
c) Home heater system (1 kW)

What fuse would be required for each appliance if the appliance draws power from 240 V mains supply?

Answers

a) 5 kW heater

$$P = VI \therefore P = \frac{P}{V}$$$$I = \frac{5\,000}{240} = 20.83 A$$

Fuse should be 25 A.

b) 2.5 kW iron box

$$I = \frac{P}{V}$$ $$= \frac{2\,500}{240}$$$$= 10.42 A$$

Fuse should be 15 A.

c) 1 kW home heater

$$I = \frac{1\,000}{240}$$ $$= 4.2 A$$

Fuse should be 5A.

Example 4

A consumer has the following appliances operating in his house for the times indicated in one day:

Appliance	Time
Two 40 W bulbs	30 min
One 500 W fridge	10 hrs
Four 75 W bulbs	3 hrs
One 3 kW electrical heater	45 min
One 100 W television	5 hrs

Calculate:

(a) the total power of the appliances used.
(b) the total electrical power consumed in kWh in 30 days, assuming that the power consumption per day is the same.

---

Solution

(a)

Total power$$= (2 \times 40) + (1 \times 500) + (4 \times 75) + (1 \times 3000) + (1 \times 100)$$=(2×40)+(1×500)+(4×75)+(1×3000)+(1×100) $$= 3\,980 W$$=3980W

---

(b)

Total energy consumed in one day equals the sum of energy consumed by each appliance.

Two 40 W bulbs for 30 minutes:

$$2 \times \frac{40}{1000} \times \frac{30}{60}$$2×100040​×6030​ $$= 0.04 \text{ kWh}$$=0.04 kWh

One 500 W fridge for 10 hours:

$$1 \times \frac{500}{1000} \times 10$$ $$= 5 \text{ kWh}$$=5 kWh

Four 75 W bulbs for 3 hours:

$$4 \times \frac{75}{1000} \times 3$$ $$= 0.90 \text{ kWh}$$

One 3 kW heater for 45 minutes:

$$1 \times 3 \times \frac{45}{60}$$$$= 2.25 \text{ kWh}$$

One 100 W television for 5 hours:

$$1 \times \frac{100}{1000} \times 5$$ $$= 0.50 \text{ kWh}$$

Total energy consumed in one day:$$= 8.69 \text{ kWh}$$

Hence, the electrical energy used in 30 day will be given by:

8.69 x 30= shs 260.7

Revision exercise

“`html

Related topics

• current and electricity
• Physics: Photo electric effects
• The cathode rays
• Properties of waves
• Electromagnetic waves
• American Electric Power