Lesson Notes By Weeks and Term v5 - Grade 8
Electrical systems: more complex circuits and switches – Week 6 focus
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Electrical systems: more complex circuits and switches – Week 6 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Technology
Class: Grade 8
Term: 2nd Term
Week: 6
Theme: General lesson support
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Electrical systems are the backbone of modern technology, powering everything from our homes and schools to industries and transportation. Understanding how circuits work, especially more complex circuits and the role of switches, is crucial for anyone interested in technology and essential for future careers in fields like electrical engineering, mechatronics, and even plumbing (think geysers and electrical pumps). In South Africa, where access to reliable electricity can be a challenge in some areas, a practical understanding of electrical systems can empower individuals to troubleshoot basic problems and potentially contribute to innovative solutions for energy access.
2.1 Series Circuits In a series circuit, components are connected one after another along a single path. The same current flows through each component in the circuit.
Imagine a single lane road: all cars (electrons) have to travel the same route.
Current (I): The current is the same at every point in a series circuit. I total = I 1 = I 2 = I 3 ...
Voltage (V): The total voltage supplied by the battery is divided among the components. V total = V 1 + V 2 + V 3 ...
Resistance (R): The total resistance is the sum of the individual resistances. R total = R 1 + R 2 + R 3 ...
Example: Three resistors, R 1 = 10 ohms, R 2 = 20 ohms, and R 3 = 30 ohms, are connected in series to a 12V battery.
Total Resistance: R total = 10 + 20 + 30 = 60 ohms Total Current: Using Ohm's Law (V = IR), I total = V total / R total = 12V / 60 ohms = 0.2 Amperes Voltage Drop across each resistor: V 1 = I total R 1 = 0.2A * 10 ohms = 2V V 2 = I total R 2 = 0.2A * 20 ohms = 4V V 3 = I total R 3 = 0.2A * 30 ohms = 6V Notice that 2V + 4V + 6V = 12V (the total voltage).
Why this matters: If one component fails in a series circuit (e.g., a light bulb burns out), the entire circuit breaks, and nothing works.
Think of old-fashioned Christmas lights: if one bulb went out, the whole string went dark. 2.2 Parallel Circuits In a parallel circuit, components are connected across each other, providing multiple paths for the current to flow.
Imagine a highway with multiple lanes: cars (electrons) can choose different routes.
Current (I): The total current supplied by the battery is divided among the different branches. I total = I 1 + I 2 + I 3 ...
Voltage (V): The voltage across each component is the same and equal to the voltage of the battery. V total = V 1 = V 2 = V 3 ...
Resistance (R): The total resistance is calculated as: 1/R total = 1/R 1 + 1/R 2 + 1/R 3 ... or R total = 1 / (1/R 1 + 1/R 2 + 1/R 3 ...)
Example: Three resistors, R 1 = 10 ohms, R 2 = 20 ohms, and R 3 = 30 ohms, are connected in parallel to a 12V battery.
Total Resistance: 1/R total = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60 Therefore, R total = 60/11 = approximately 5.45 ohms.
Total Current: I total = V total / R total = 12V / 5.45 ohms = approximately 2.2 Amperes.
Current through each resistor: I 1 = V total / R 1 = 12V / 10 ohms = 1.2A I 2 = V total / R 2 = 12V / 20 ohms = 0.6A I 3 = V total / R 3 = 12V / 30 ohms = 0.4A Notice that 1.2A + 0.6A + 0.4A = 2.2A (the total current).
Why this matters: If one component fails in a parallel circuit, the other components continue to function. This is why household wiring is in parallel: if one light bulb burns out, the others stay on. 2.3 Combined Series-Parallel Circuits These circuits combine both series and parallel arrangements. To analyze them, you need to simplify the circuit step-by-step.
Example: Consider a circuit with a 12V battery. R 1 (5 ohms) is in series with a parallel combination of R 2 (10 ohms) and R 3 (15 ohms). Find the equivalent resistance of the parallel section (R 2 and R 3 ): 1/R parallel = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 R parallel = 30/5 = 6 ohms Find the total resistance of the circuit: R total = R 1 + R parallel = 5 ohms + 6 ohms = 11 ohms Find the total current: I total = V total / R total = 12V / 11 ohms = approximately 1.09 Amperes Find the voltage drop across R 1 : V 1 = I total R 1 = 1.09A 5 ohms = approximately 5.45V The voltage across the parallel section (R 2 and R 3 ) is the same: V parallel = V total - V 1 = 12V - 5.45V = approximately 6.55V Find the current through R 2 and R 3 : I 2 = V parallel / R 2 = 6.55V / 10 ohms = approximately 0.66A I 3 = V parallel / R 3 = 6.55V / 15 ohms = approximately 0.44A Notice that 0.66A + 0.44A = 1.1A, which is roughly equal to the total current (allowing for rounding errors). 2.4 Switches Switches control the flow of electricity in a circuit. They act as either an open (off) or closed (on) gate.
Here are some common types: SPST (Single-Pole, Single-Throw): The simplest type. It has one input and one output. It either connects the circuit (on) or disconnects it (off). Think of a simple light switch. SPDT (Single-Pole, Double-Throw): Has one input and two outputs. It can connect the input to either one output or the other. Think of a switch that selects between two different lights or speakers. DPST (Double-Pole, Single-Throw): Two SPST switches controlled by a single mechanism. It simultaneously switches two separate circuits on or off. Used for heavier loads or isolating both the live and neutral wires. DPDT (Double-Pole, Double-Throw): Two SPDT switches controlled by a single mechanism. It can switch two separate circuits between two different sets of connections. More complex applications, such as reversing the polarity of a motor.
Push-button Switch: Momentary contact switch. It only closes the circuit when pressed. Used for doorbells or some types of security systems.