Lesson Notes By Weeks and Term v5 - Grade 8

Electrical systems: more complex circuits and switches – Week 6 focus

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Subject: Technology

Class: Grade 8

Term: 2nd Term

Week: 6

Theme: General lesson support

Lesson Video

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Performance objectives

Lesson summary

This week, we're diving into more complex electrical circuits and switches. Building upon our knowledge of basic circuits, we'll explore how multiple components can be arranged to achieve specific functions. Understanding these concepts is crucial because electrical systems are fundamental to modern life in South Africa, from powering our homes and schools to operating essential equipment in industries and healthcare facilities. Knowing how these systems work allows us to troubleshoot simple problems, understand the energy usage of appliances, and even explore potential careers in electrical engineering or related fields, which are vital for South Africa's infrastructure development.

Lesson notes

2.1 Series Circuits: In a series circuit, components are connected along a single path. The electric current has only one route to flow through. Think of it as a single lane road – all the cars (electrons) must travel along the same path.

Current (I): The current is the same at all points in a series circuit. This is because the electrons have no other path to take.

Voltage (V): The total voltage (supplied by the battery) is divided among the components in the series circuit. The voltage drop across each component depends on its resistance.

Resistance (R): The total resistance in a series circuit is the sum of the individual resistances.

Formula: R total = R 1 + R 2 + R 3 + ...

Example: Imagine three resistors, R 1 = 10 Ω, R 2 = 20 Ω, and R 3 = 30 Ω, connected in series. The total resistance is R total = 10 Ω + 20 Ω + 30 Ω = 60 Ω. 2.2 Parallel Circuits: In a parallel circuit, components are connected along multiple paths. The electric current has several routes it can flow through. Think of it as a multi-lane highway – cars (electrons) can choose different lanes.

Current (I): The total current supplied by the battery is divided among the different branches of the parallel circuit. The current through each branch depends on its resistance.

Voltage (V): The voltage is the same across all branches in a parallel circuit. Each component receives the full voltage from the battery.

Resistance (R): The total resistance in a parallel circuit is less than the resistance of the smallest individual resistor.

Formula: 1/R total = 1/R 1 + 1/R 2 + 1/R 3 + ...

Example: Imagine three resistors, R 1 = 10 Ω, R 2 = 20 Ω, and R 3 = 30 Ω, connected in parallel. 1/R total = 1/10 Ω + 1/20 Ω + 1/30 Ω 1/R total = (6 + 3 + 2) / 60 Ω = 11/60 Ω R total = 60/11 Ω ≈ 5.45 Ω

Note: The total resistance (5.45 Ω) is less than the smallest resistor (10 Ω). 2.3 Combination Circuits (Series-Parallel): Most real-world circuits are a combination of series and parallel connections. To analyze these, we need to simplify the circuit step-by-step. First, identify any series or parallel sections. Calculate the equivalent resistance of those sections. Then, redraw the simplified circuit and repeat the process until you are left with a single equivalent resistance.

Example: Consider a circuit with R 1 (10 Ω) in series with a parallel combination of R 2 (20 Ω) and R 3 (30 Ω).

Solve the parallel part first: 1/R parallel = 1/20 Ω + 1/30 Ω = 5/60 Ω.

Therefore, R parallel = 60/5 Ω = 12 Ω. Now we have a series circuit with R 1 (10 Ω) and R parallel (12 Ω).

Total resistance: R total = R 1 + R parallel = 10 Ω + 12 Ω = 22 Ω. 2.4 Switches: Switches are used to control the flow of current in a circuit. They can be open (OFF) or closed (ON).

SPST (Single-Pole Single-Throw): The simplest type of switch. It has one input and one output. It either allows current to flow (closed) or blocks it (open). Think of a simple light switch in your home.

SPDT (Single-Pole Double-Throw): Has one input and two outputs. It can connect the input to either one of the outputs. Think of a switch that controls which of two lights turns on.

Example (SPST): Consider a simple circuit with a battery, a resistor, an LED, and a SPST switch connected in series. When the switch is closed, the circuit is complete, and the LED lights up. When the switch is open, the circuit is broken, and the LED turns off.

Example (SPDT): Consider a circuit with a battery, two LEDs (LED1 and LED2), a resistor, and an SPDT switch. The input of the switch is connected to the battery. One output is connected to LED1, and the other output is connected to LED

2. When the switch is flipped to one position, LED1 lights up, and when it's flipped to the other position, LED2 lights up. Only one LED can be lit at a time. Guided Practice (With Solutions)

Question 1: Two resistors, R 1 = 5 Ω and R 2 = 15 Ω, are connected in series with a 12V battery. Calculate the total resistance of the circuit.

Solution: Since the resistors are in series, the total resistance is the sum of the individual resistances. R total = R 1 + R 2 = 5 Ω + 15 Ω = 20 Ω Answer: The total resistance is 20 Ω.

Question 2: Two resistors, R 1 = 4 Ω and R 2 = 12 Ω, are connected in parallel. Calculate the total resistance of the circuit.

Solution: Since the resistors are in parallel, we use the reciprocal formula: 1/R total = 1/R 1 + 1/R 2 1/R total = 1/4 Ω + 1/12 Ω = (3 + 1) / 12 Ω = 4/12 Ω = 1/3 Ω R total = 3 Ω Answer: The total resistance is 3 Ω.

Question 3: A circuit consists of a 9V battery connected to a 3 Ω resistor in series with a parallel combination of a 6 Ω resistor and a 3 Ω resistor. What is the total resistance of the circuit?

Solution: Find the equivalent resistance of the parallel part: 1/R parallel = 1/6 Ω + 1/3 Ω = (1 + 2) / 6 Ω = 3/6 Ω = 1/2 Ω.

Therefore, R parallel = 2 Ω. Now the circuit consists of a 3 Ω resistor in series with a 2 Ω resistor.

Total resistance: R total = 3 Ω + 2 Ω = 5 Ω.

Answer: The total resistance is 5 Ω.