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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 like temperature remain constant. This is expressed as V∝IV \propto I.

A basic circuit diagram showing a battery, an ammeter in series, a resistor, and a voltmeter in parallel across the resistor to verify Ohm's Law.
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Resistance (RR) is the measure of opposition to the flow of electric current. In a V−IV-I graph for an ohmic conductor, the slope of the line represents the resistance R=ΔVΔIR = \frac{\Delta V}{\Delta I}.

A linear graph of Potential Difference (V) versus Current (I) starting from the origin.
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The SI unit of resistance is the Ohm (Ω\Omega). One Ohm is defined as the resistance of a conductor such that a potential difference of 1V1V causes a current of 1A1A to flow through it.

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Factors affecting resistance include the length of the conductor (LL), the cross-sectional area (AA), and the nature of the material (resistivity ρ\rho). It is given by R=ρLAR = \rho \frac{L}{A}.

📐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.

Problem 4:

A simple circuit is constructed using a 9V9V battery and a resistor. An ammeter connected in series measures a current of 0.45A0.45A. Calculate the value of the unknown resistance RR.

A circuit diagram showing a 9V battery, an ammeter reading 0.45A, and a resistor R in a single loop.

Solution:

Given: Potential Difference V=9VV = 9V Current I=0.45AI = 0.45A

Using Ohm's Law: R=VIR = \frac{V}{I}

Substitute the values: R=90.45R = \frac{9}{0.45} R=20ΩR = 20\Omega

Explanation:

To find the resistance, we rearrange the Ohm's Law formula to isolate RR. By dividing the voltage of the battery by the measured current, we find that the resistor opposes the current with a value of 20Ω20\Omega.

Problem 5:

A circuit is designed with a lamp that has a resistance of 15Ω15\Omega. When the circuit is closed, the ammeter shows a reading of 0.8A0.8A. Determine the potential difference supplied by the battery to the lamp.

A series circuit diagram showing a battery connected to an ammeter reading 0.8 Amperes and a lamp with a resistance of 15 Ohms.

Solution:

V=I×RV = I \times R V=0.8A×15ΩV = 0.8A \times 15\Omega V=12VV = 12V

Explanation:

To find the potential difference (VV), we use the Ohm's Law formula by multiplying the current (II) flowing through the circuit by the resistance (RR) of the lamp. Substituting the given values 0.8A0.8A and 15Ω15\Omega yields a voltage of 12V12V.