Lesson Notes By Weeks and Term v5 - Grade 9
Electric circuits: resistance and current – Week 5 focus
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Electric circuits: resistance and current – Week 5 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Natural Sciences
Class: Grade 9
Term: 2nd Term
Week: 5
Theme: General lesson support
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This week, we delve into the fundamental concepts of resistance and current in electric circuits. Understanding resistance and current is crucial for comprehending how electricity powers our homes, schools, and communities. In South Africa, where access to reliable electricity is a significant concern, understanding how electricity works allows us to be more responsible consumers, troubleshoot basic electrical issues, and appreciate the technologies that improve our lives. Moreover, with the increasing adoption of renewable energy sources, such as solar power, understanding electrical circuits becomes even more vital for managing and utilizing these resources effectively.
Electric Current (I): Electric current is the rate of flow of electric charge through a circuit. Think of it like water flowing through a pipe. The amount of water flowing past a certain point per second is similar to the amount of charge flowing past a point in a circuit per second. This flow of charge is caused by the movement of electrons.
Definition: The amount of charge (measured in Coulombs, C) passing a point in a circuit per unit of time (measured in seconds, s).
Formula: I = Q/t (where I = current, Q = charge, and t = time)
Unit: Ampere (A), which is equal to Coulombs per second (C/s).
Analogy: Imagine a busy street in Johannesburg during rush hour. The number of cars passing a specific point every minute is analogous to the current – the higher the number of cars, the higher the current.
Voltage (V): Voltage, also known as potential difference, is the electrical "pressure" that drives the current through a circuit. It's the force that pushes the electrons along the circuit.
Definition: The amount of electrical potential energy per unit charge. In simpler terms, it's the amount of work needed to move a unit of electric charge between two points in an electric circuit.
Unit: Volt (V), which is equal to Joules per Coulomb (J/C).
Analogy: Consider a water pump connected to a pipe. The pump provides the pressure that pushes the water through the pipe. The voltage in a circuit is like the water pump.
Battery as a Source: A battery provides this "pressure" or voltage. A typical AA battery provides 1.5
V. Resistance (R): Resistance is the opposition to the flow of electric current in a circuit. Different materials offer different levels of resistance. For example, copper wires have low resistance, allowing electricity to flow easily, while materials like rubber have high resistance, preventing electricity from flowing easily.
Definition: The opposition to the flow of electric current.
Unit: Ohm (Ω).
Analogy: Imagine a narrow section in a water pipe. This narrow section makes it harder for the water to flow through, creating resistance.
Factors affecting resistance: Length: The longer the wire, the higher the resistance. Think of it like a longer pipe offering more friction to the water flow.
Cross-sectional area: The thinner the wire, the higher the resistance. A thinner pipe restricts water flow more than a wider pipe.
Material: Different materials have different inherent resistances. Copper has low resistance, while nichrome (used in heaters) has higher resistance.
Temperature: For most conductors, resistance increases with temperature.
Ohm's Law: Ohm's Law describes the relationship between voltage (V), current (I), and resistance (R). It states that the voltage across a conductor is directly proportional to the current flowing through it, provided the temperature remains constant.
Formula: V = IR Meaning: If you increase the voltage (V), the current (I) will also increase, assuming resistance (R) stays constant. If you increase the resistance (R), the current (I) will decrease, assuming voltage (V) stays constant.
Analogy: Using our water pipe analogy: Increasing the water pressure (voltage) will increase the water flow (current). Increasing the narrowness of the pipe (resistance) will decrease the water flow (current).
Example 1: A light bulb is connected to a 220V power source. The current flowing through the bulb is 0.5A. Calculate the resistance of the light bulb.
Given:
Voltage (V) = 220V
Current (I) = 0.5A
Formula: V = IR => R = V/I
Solution: R = 220V / 0.5A = 440 Ω
Answer: The resistance of the light bulb is 440 ohms.
Example 2: A heater element has a resistance of 20 ohms. If a current of 10A flows through the element, what is the voltage across the element?
Given:
Resistance (R) = 20 Ω
Current (I) = 10 A
Formula: V = IR
Solution: V = 10A 20 Ω = 200 V
Answer: The voltage across the heater element is 200 volts.
Example 3: A cellphone charger delivers a voltage of 5V. If the internal resistance of the charging circuit is 2.5 ohms, what is the current flowing into the cellphone?
Given:
Voltage (V) = 5V
Resistance (R) = 2.5 Ω
Formula: V = IR => I = V/R
Solution: I = 5V / 2.5 Ω = 2 A
Answer: The current flowing into the cellphone is 2 amps.
Guided Practice (With Solutions)
Question 1: A resistor is connected to a 12V battery, and a current of 0.2A flows through it. What is the resistance of the resistor?
Solution:
Given: V = 12V, I = 0.2A
Ohm's Law: V = IR, therefore R = V/I
R = 12V / 0.2A = 60 Ω
Answer: The resistance of the resistor is 60 ohms.
Commentary: This is a straightforward application of Ohm's Law. We are given the voltage and current and asked to find the resistance.