TERM – 3RD TERM
WEEK NINE
Class: Junior Secondary School 1
Age: 12 years
Duration: 40 minutes of 5 periods each
Date:
Subject: BASIC TECHNOLOGY
Topic: BASIC ELECTRICITY 3
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers,
demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures,
INSTRUCTIONAL PROCEDURES
PERIOD 1-2
PRESENTATION |
TEACHER’S ACTIVITY |
STUDENT’S ACTIVITY |
STEP 1 INTRODUCTION |
The teacher Identify the most important concept involving electrical calculations (current, resistance, voltage, power and charge) |
Students pay attention |
STEP 2 EXPLANATION |
Teacher discuss ohms laws and describe the relationship between current, Voltage and resistance. |
Students pay attention and participate |
STEP 3 DEMONSTRATION |
Teacher guide students to carry out basic calculations of electricity |
Students pay attention and participate |
STEP 4 NOTE TAKING |
The teacher writes a summarized note on the board |
The students copy the note in their books |
NOTE
BASIC ELECTRICITY 3
Basic electrical calculations often involve concepts like:
V = I x R
Where:
- V is the voltage across the circuit,
- I is the current flowing through the circuit,
- R is the resistance in the circuit.
This equation helps in understanding how these three parameters are interrelated. It can also be used in terms of current and resistance.
That is; Current (I) = Voltage (V).
Resistance(R) = V/R
Current is measured in amperes
And Resistance(R) = Voltage (V)
Current(I). =V/I
Resistance is measured in ohms
I. (P = V x I): This is the general power formula, where P is power, V is voltage, and I is current.
II. (P = I² x R ): This formula is applicable when you know the current and resistance in a circuit.
Power is measured in watts (W).
Q = C x V
Where Q is the charge
C is the capacitance
V is the voltage.
Example 1: Calculate the charge across the capacitor 5mF and the voltage applied is 25 V.
Solution
Capacitance(C) = 5 mF,
Voltage applied V = 25 V,
The Charge across the capacitor is given by Q = CV
= 5 mF x 25 V = 125 x 10^² C
= 0.125 C.
Example 2: A wire carrying a current of 4 Amperes is having a resistance of 5. Calculate the potential difference across its ends.
Solution:
Current I = 4 A,
Resistance R = 5 ohm
Formula for calculating Potential difference is V = IR
= 4 x 5 = 20 v.
Example 3: Calculate the amount of current that flows in the circuit below:
Solution
Voltage (V) = 9v
Resistance (R) = 10 ohms
Current (I) = V/R
= 9/10 = 0.9 ampere
EVALUATION: 1. State ohm's law.
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively