The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below: V (volts): 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 4.0, 5.0; I (amperes): 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.8, 1.0. Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.

Electricity

Ohm's Law

Official Answer

To determine the resistance of the resistor, we plot the potential difference ( $V$ ) on the y-axis and the current ( $I$ ) on the x-axis. The resulting graph is a straight line passing through the origin, which verifies Ohm's law ( $V \propto I$ ).

The resistance ( $R$ ) is given by the slope of the $V-I$ graph:

$R = \frac{\Delta V}{\Delta I} = \frac{V_2 - V_1}{I_2 - I_1}$

Using any two points from the data, say $(I_1, V_1) = (0.2\ \text{A}, 1.0\ \text{V})$ and $(I_2, V_2) = (0.8\ \text{A}, 4.0\ \text{V})$ :

$R = \frac{4.0\ \text{V} - 1.0\ \text{V}}{0.8\ \text{A} - 0.2\ \text{A}} = \frac{3.0\ \text{V}}{0.6\ \text{A}} = 5\ \Omega$

Thus, the resistance of the resistor is $5\ \Omega$ .

Ohm's lawV-I graphstraight linesloperesistance5 \Omega

Marking Scheme

11 mark for correctly plotting the V-I graph with labeled axes (Current on x-axis, Potential Difference on y-axis) and drawing a straight line passing through the origin.
21 mark for calculating the resistance using the slope formula ( $R = \frac{\Delta V}{\Delta I}$ ) and obtaining the correct value of $5\ \Omega$ .

Hint

Plot the potential difference $V$ on the y-axis and current $I$ on the x-axis. The slope of this straight-line graph ( $\frac{\Delta V}{\Delta I}$ ) will give the resistance $R$ .

Quick Oral Answer

The resistance of a conductor can be determined from its V-I graph by calculating the slope of the straight line. If potential difference ( $V$ ) is on the y-axis and current ( $I$ ) is on the x-axis, the slope ( $\frac{\Delta V}{\Delta I}$ ) directly gives the resistance $R$ . For this given data, the slope is constant and equals $5\ \Omega$ .

Analysis & Explanation

Ohm's law states that the current ( $I$ ) flowing through a metallic conductor is directly proportional to the potential difference ( $V$ ) across its ends, provided its temperature remains constant ( $V \propto I$ ). Mathematically, this is written as $V = IR$ , where the constant of proportionality $R$ is the electrical resistance of the conductor.

When we plot $V$ on the y-axis and $I$ on the x-axis, the slope of the resulting straight line represents the resistance $R$ . Since the ratio $\frac{V}{I}$ is constant at all points, the graph is a straight line passing through the origin. This constant ratio confirms that the resistor obeys Ohm's law (it is an ohmic conductor). The value of resistance calculated from any pair of coordinates on this line will yield the same result, which is $5\ \Omega$ .

Common Mistakes

1Plotting current on the y-axis and potential difference on the x-axis without adjusting the slope calculation (the slope of an I-V graph is $\frac{1}{R}$ , not $R$ ).
2Not drawing a straight line through the origin, or joining the points in a zig-zag manner.
3Omitting the unit of resistance ( $\Omega$ ) in the final answer.

Interesting Facts

Georg Simon Ohm's work was initially rejected by the German Minister of Education, who declared that 'a physicist who professed such heresies was unworthy to teach science.' It took more than a decade for his genius to be recognized globally.

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2018?

This question carries 2 marks in the CBSE Class 10 Science 2018 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Electricity" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Ohm's Law" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2018 paper?

This is a Short Answer question from Section B in the CBSE Class 10 Science 2018 paper.