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Electricity - Ohm's Law

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Ohm's Law states that the current (II) flowing through a conductor is directly proportional to the potential difference (VV) across its ends, provided physical conditions such as temperature remain constant.

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A conductor that obeys Ohm's law is called an Ohmic conductor, characterized by a linear I−VI-V graph that passes through the origin.

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Resistance (RR) is the property of a material to oppose the flow of electric current. It is measured in Ohms (Ω\Omega).

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Current (II) is the rate of flow of charge, measured in Amperes (AA).

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Potential Difference (VV), also known as voltage, is the work done per unit charge to move it between two points, measured in Volts (VV).

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Non-ohmic components, such as filament lamps or diodes, do not show a constant ratio of VI\frac{V}{I} as their resistance changes with temperature or voltage polarity.

📐Formulae

V=I×RV = I \times R

I=VRI = \frac{V}{R}

R=VIR = \frac{V}{I}

1Ω=1V1A1\Omega = \frac{1V}{1A}

💡Examples

Problem 1:

A circuit contains a resistor with a resistance of 20Ω20\Omega. If a 12V12V battery is connected to the circuit, what is the current flowing through the resistor?

Solution:

I=12V20Ω=0.6AI = \frac{12V}{20\Omega} = 0.6A

Explanation:

Using Ohm's Law in the form I=VRI = \frac{V}{R}, we substitute the known values for potential difference (12V12V) and resistance (20Ω20\Omega) to find the current in Amperes.

Problem 2:

An electric heater draws a current of 5A5A when connected to a 230V230V supply. Calculate the resistance of the heating element.

Solution:

R=230V5A=46ΩR = \frac{230V}{5A} = 46\Omega

Explanation:

To find the resistance, we rearrange the formula to R=VIR = \frac{V}{I}. Dividing the voltage by the current gives the resistance in Ohms.

Problem 3:

If the current passing through a 1.5kΩ1.5k\Omega resistor is 10mA10mA, calculate the potential difference across it.

Solution:

V=(10×10−3A)×(1.5×103Ω)=15VV = (10 \times 10^{-3}A) \times (1.5 \times 10^{3}\Omega) = 15V

Explanation:

First, convert units to standard SI units: 10mA=0.01A10mA = 0.01A and 1.5kΩ=1500Ω1.5k\Omega = 1500\Omega. Then apply V=IRV = IR to find the voltage.

Ohm's Law - Revision Notes & Key Formulas | IB Grade 9 Science