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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Electric circuits are fundamental to modern life, powering everything from the lights in our homes to the cell phones we use every day. Understanding how electric current flows and how resistance affects that flow is crucial. In South Africa, with its ongoing efforts to improve electricity access and manage energy consumption, a firm grasp of these principles helps us to become informed citizens who can make responsible choices about energy use. This week, we'll delve into the relationship between resistance and current in electric circuits. Understanding this relationship is vital for understanding energy saving strategies that can help your families manage their household expenses.
Electric Current: Electric current is the rate of flow of electric charge through a circuit. Think of it as the number of electrons that pass a given point in a circuit every second.
Symbol: I Unit: Ampere (A)
Analogy: Imagine water flowing through a pipe. The electric current is like the amount of water flowing past a certain point in the pipe per second.
How to Measure: Ammeters are used to measure current. They are always connected in series in the circuit.
Potential Difference (Voltage): Potential difference (voltage) is the electrical potential energy difference between two points in a circuit. It's the "push" that drives the electrons through the circuit.
Symbol: V Unit: Volt (V)
Analogy: In our water pipe analogy, voltage is like the pressure pushing the water through the pipe.
How to Measure: Voltmeters are used to measure voltage. They are always connected in parallel across the component you want to measure the voltage of.
Resistance: Resistance is the opposition to the flow of electric current in a circuit. It's like a bottleneck in the water pipe, making it harder for the water to flow.
Symbol: R Unit: Ohm (Ω)
Analogy: In the water pipe analogy, resistance is like a narrow section of the pipe that restricts the water flow.
Factors Affecting Resistance: Material: Different materials offer different resistance to current flow. Conductors (like copper and silver) have low resistance, while insulators (like rubber and plastic) have high resistance.
Length: Longer wires have higher resistance. Imagine a longer pipe offering more friction to the water flow.
Thickness (Cross-sectional Area): Thicker wires have lower resistance. A wider pipe allows water to flow more easily.
Temperature: For most materials, resistance increases with temperature.
Ohm's Law: This is a fundamental relationship in electrical circuits. 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 (Voltage = Current x Resistance)
Explanation: If you increase the voltage (V), the current (I) will increase proportionally, assuming the resistance (R) stays the same. If you increase the resistance (R), the current (I) will decrease proportionally, assuming the voltage (V) stays the same.
Example 1: A light bulb in a shack in Khayelitsha has a resistance of 10 Ω. If the voltage supplied is 220 V, what is the current flowing through the bulb?
Solution:
Given: R = 10 Ω, V = 220 V
Formula: V = IR
Rearrange to solve for I: I = V/R
Substitute the values: I = 220 V / 10 Ω = 22 A
Answer: The current flowing through the bulb is 22
A. Why: We used Ohm's Law (V=IR) because we knew the voltage and resistance and wanted to find the current.
Example 2: A heater in a home in Soweto draws a current of 5 A when connected to a 220 V power supply. What is the resistance of the heater?
Solution:
Given: I = 5 A, V = 220 V
Formula: V = IR
Rearrange to solve for R: R = V/I
Substitute the values: R = 220 V / 5 A = 44 Ω
Answer: The resistance of the heater is 44 Ω.
Why: We used Ohm's Law (V=IR) because we knew the voltage and current and wanted to find the resistance.
Example 3: A long extension cord used in a garden in Durban has a resistance of 2 Ω. If a lawnmower connected to the extension cord draws a current of 8 A, what is the voltage drop across the extension cord?
Solution:
Given: R = 2 Ω, I = 8 A
Formula: V = IR
Substitute the values: V = 8 A x 2 Ω = 16 V
Answer: The voltage drop across the extension cord is 16
V. Why: We used Ohm's Law (V=IR) because we knew the current and resistance and wanted to find the voltage drop. A voltage drop means that the lawnmower is only getting 204V of the 220V available. This means that some of the energy is being wasted by the extension cord.
Guided Practice (With Solutions)
Question 1: A cellphone charger has a resistance of 20 ohms and requires a current of 0.5 Amperes to charge a phone. What voltage is required to power the charger?
Solution:
Given: R = 20 Ω, I = 0.5 A
Formula: V = IR
Substitute the values: V = 0.5 A x 20 Ω = 10 V
Answer: A voltage of 10 V is required.
Commentary: This is a straightforward application of Ohm's Law. It's important to identify the known values (R and I) and the unknown value (V) before applying the formula.