Basic Electricity - Junior Secondary 1 - Electrical circuit

TERM: 3^RD TERM

WEEK TWO

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:

Subject: BASIC ELECTRICITY

Topic: ELECTRICAL CIRCUIT

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

I.) Define electric circuit

II.) Carry out calculations on electric circuit

III.) Discuss electric Current, Potential Difference PD and Electromotive Force EMF

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures,

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

NOTE

ELECTRICAL CIRCUIT

An electric circuit is a closed loop or pathway through which electric current can flow. It consists of interconnected components such as resistors, capacitors, inductors, and power sources, connected by conductive wires.

Example 1: A battery of emf 9.0 V and internal resistance, r, is connected in the circuit shown in the figure below.

The current in the battery is 1.0 A.

I.) Calculate the PD between points A and B in the circuit.

II.) Calculate the internal resistance, r.

III.) Calculate the total energy transformed by the battery in 5.0 minutes.

Solution

I.) To calculate the potential difference (PD) between points A and B, you can use Ohm's Law, V = IR,

 where V = potential difference,

I  = current, and

R  = resistance.

Since the current is given as 1.0 A and the total resistance in the circuit is (2.0 + 2.0  + 6.0 = 10.0) ohm's,

You can find the potential difference

V = (1.0 A x 10.0) = 10.0 V between points A and B.

II.) The internal resistance (r) can be found using the formula for the total resistance in a circuit with internal resistance:

R(total) = R(external) + r

Where R(total) = total resistance

R(external) = the sum of external resistances, and

r =  the internal resistance.

We already know:

R(total) = 10.0 and R(external = 2.0 + 2.0 + 6.0 = 10.0) ohm's,

Thus, r = R(total) - R(external) = 10.0 - 10.0

Hence, r = 0 ohm's

III.) To calculate the total energy transformed by the battery in 5.0 minutes, you can use the formula

E = P x t,

where E is energy,

P is power, and t is time.

First, we need to find the power delivered by the battery. Power can be calculated as (P = IV),

where I is the current and V is the potential difference. We already have (I = 1.0 A) and (V = 9.0 V),

so, P = (1.0 A)(9.0 V) = 9.0W.

Now, multiply the power by the time: E = (9.0W x 5.0 min) = 45 J

Electric Current, Potential Difference PD and Electromotive Force EMF

Electric current

Electric current is the flow of electric charge carriers, usually electrons, through a conductor. It is measured in amperes (A) and represents the rate of flow of charge past a given point in a circuit.

Potential difference (PD)

Potential difference, also known as voltage, is the difference in electric potential between two points in an electrical circuit. It is measured in volts (V) and represents the energy per unit charge required to move a charge between the two points.

Electromotive Force (EMF)

Electromotive force (EMF) is the total energy supplied per unit charge by a source such as a battery or generator to drive electrons around a circuit. It is measured in volts (V) and represents the maximum potential difference that the source can provide when no current is flowing.

EVALUATION: 1. What is electric circuit?

2. The two lamps are connected n the circuit shown in the figure below. The battery has an emf of 24 V and negligible internal resistance. The resistors, R1

 and R2 are chosen so that the lamps are operating at their correct working voltage.

I.) Calculate the pd across R1

II.) Calculate the current in R1

III.) Calculate the resistance of R1

IV.) Calculate the pd across R2

V.) Calculate the resstance of R2

 CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively