Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define Voltage (V), Current (I), and Resistance (R) and their units.
• State and apply Ohm's Law to solve for V, I, or R.
• Calculate total resistance in a series circuit.
• Calculate voltage drop across resistors in a series circuit.
• Describe the effects of changing V, I, and R in a circuit.

Lesson summary

This week, we delve into the fundamental building blocks of electrical circuits: basic electrical quantities and Ohm's Law. Understanding these concepts is crucial, not just for this course, but for comprehending how electricity powers our homes, businesses, and industries across South Africa. From powering our lights during loadshedding with inverters to maintaining electrical appliances, a solid grasp of these concepts is essential for any aspiring electrician, technician, or simply an informed citizen.

Lesson notes

2.1 Basic Electrical Quantities Voltage (V): Also known as potential difference or electromotive force (EMF), voltage is the "push" that forces electrons to move through a circuit. Think of it like the pressure in a water pipe. The higher the voltage, the more "push" there is, and the more current will flow (if the resistance allows). Its unit is the Volt (V). A typical AA battery has a voltage of 1.5V, while the standard household voltage in South Africa is 230

V. Current (I): Current is the rate of flow of electric charge (electrons) through a circuit. Imagine it as the amount of water flowing through a pipe. The more electrons flowing, the higher the current. Its unit is the Ampere (A), often shortened to "Amp." A small LED might require a current of only a few milliamps (mA), while a stove might draw several Amps.

Resistance (R): Resistance is the opposition to the flow of current in a circuit. Think of it as the narrowness of a water pipe. The higher the resistance, the less current will flow for a given voltage. Its unit is the Ohm (Ω). A resistor is a component specifically designed to provide resistance. Things like light bulb filaments and heating elements also provide resistance. 2.2 Ohm's Law Ohm's Law is a fundamental relationship between voltage (V), current (I), and resistance (R) in an electrical circuit.

It states: Voltage (V) = Current (I) x Resistance (R)

This can be written as: V = I * R We can also rearrange this formula to solve for current or resistance: Current (I) = Voltage (V) / Resistance (R) or I = V / R Resistance (R) = Voltage (V) / Current (I) or R = V / I 2.3 Series Circuits A series circuit is a circuit where components are connected one after another along a single path.

In a series circuit: The current (I) is the same at every point in the circuit. The total resistance (R T ) is the sum of the individual resistances: R T = R 1 + R 2 + R 3 + ... The total voltage (V T ) is equal to the sum of the voltage drops across each resistor: V T = V 1 + V 2 + V 3 + ... 2.4 Worked Examples Example 1: A light bulb has a resistance of 200 Ω and is connected to a 230V power supply. Calculate the current flowing through the bulb.

Given: R = 200 Ω, V = 230V Formula: I = V / R Solution: I = 230V / 200 Ω = 1.15 A Example 2: A resistor in a radio circuit has a current of 0.05A flowing through it when a voltage of 5V is applied. Calculate the resistance of the resistor.

Given: I = 0.05 A, V = 5V Formula: R = V / I Solution: R = 5V / 0.05 A = 100 Ω Example 3: A series circuit consists of two resistors, R 1 = 10 Ω and R 2 = 20 Ω, connected to a 12V battery.

Calculate: a) The total resistance of the circuit. b) The current flowing through the circuit. c) The voltage drop across each resistor.

Given: R 1 = 10 Ω, R 2 = 20 Ω, V T = 12V Solution: a) R T = R 1 + R 2 = 10 Ω + 20 Ω = 30 Ω b) I = V T / R T = 12V / 30 Ω = 0.4 A c) V 1 = I R 1 = 0.4 A * 10 Ω = 4V V 2 = I R 2 = 0.4 A 20 Ω = 8V Important

Note: Notice that V 1 + V 2 = 4V + 8V = 12V = V T , confirming that the voltage drops in a series circuit add up to the total voltage. Guided Practice (With Solutions)

Question 1: A kettle with a resistance of 25 Ω is connected to a 230V supply. Calculate the current flowing through the kettle.

Solution: Given: R = 25 Ω, V = 230 V Formula: I = V / R Solution: I = 230 V / 25 Ω = 9.2 A

Commentary: This is a direct application of Ohm's Law. Remember to use the correct units (Volts and Ohms) to get the current in Amperes. A kettle draws a significant amount of current!

Question 2: A cellphone charger draws 0.2A from a 5V USB port. What is the charger's resistance?

Solution: Given: I = 0.2 A, V = 5 V Formula: R = V / I Solution: R = 5 V / 0.2 A = 25 Ω

Commentary: Even though chargers are small, they still have internal resistance. This resistance is what allows them to reduce the voltage from the wall socket to the voltage the device needs.

Question 3: A series circuit has three resistors: R 1 = 5 Ω, R 2 = 10 Ω, and R 3 = 15 Ω, connected to a 9V battery. Calculate the current flowing through the circuit.

Solution: Given: R 1 = 5 Ω, R 2 = 10 Ω, R 3 = 15 Ω, V T = 9 V Formula: R T = R 1 + R 2 + R 3 , I = V T / R T Solution: R T = 5 Ω + 10 Ω + 15 Ω = 30 Ω I = 9 V / 30 Ω = 0.3 A

Commentary: First, calculate the total resistance of the series circuit. Then, use Ohm's Law with the total voltage and total resistance to find the current.

Question 4: In the series circuit from Question 3, calculate the voltage drop across resistor R 2 .

Solution: Given: R 2 = 10 Ω, I = 0.3 A (from Question 3)

Formula: V = I R Solution: V 2 = 0.3 A 10 Ω = 3 V

Commentary: Using the current calculated in the previous question and the resistance of R 2 , we can apply Ohm's Law to find the voltage drop across that specific resistor. Independent Practice (Questions Only) A heater element has a resistance of 40 Ω. If it draws a current of 5.75 A, what is the voltage applied to it?