Question

To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained: If $V_0$ is almost zero, identify the correct statement:

• A. The value of the resistance $R$ is $1.5 \,\Omega$
• B. The emf of the battery is $1.5\, V$ and the value of $R$ is $1.5 \, \Omega$
• C. The emf of the battery is $1.5\, V$ and its internal resistance is $1.5$ $\Omega$
• D. The potential difference across the battery is $1.5\, V$ when it sends a current of $1000\, mA$

Answer 1

The Correct Option is C

To solve this problem, we must understand Ohm's Law and how it applies to the circuit described. Ohm’s Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, represented by the formula:

\(V = IR\)

where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance.

If the student’s graph plots voltage as a function of current and the graph is linear going through the origin, it indicates that the system is following Ohm's Law, and there is little or no internal resistance affecting the battery.

1. Analyzing the Options:

The value of the resistance \(R\) is \(1.5 \, \Omega\): 
This option gives a fixed resistance value but does not consider the internal resistance of the battery or provide information about the emf. There’s insufficient data to conclude this solely from a graph.

The emf of the battery is \(1.5 \, V\) and the value of \(R\) is \(1.5 \, \Omega\): 
The emf describes the battery's voltage under no load and does not indicate resistance directly. This needs a correct analysis of both emf and resistance.

The emf of the battery is \(1.5 \, V\) and its internal resistance is \(1.5 \, \Omega\): 
This option considers both emf and internal resistance, which can be validated through proper substitution in Ohm's Law and from hint \(V_0 \approx 0\).

The potential difference across the battery is \(1.5 \, V\) when it sends a current of \(1000\, mA\): 
This just confirms an instance of the measured voltage at a specific current but doesn't ascertain emf or internal resistance.

Conclusion:

The correct answer is:

The emf of the battery is \(1.5\, V\) and its internal resistance is \(1.5\) \(\, \Omega\).

 

This option correctly identifies the emf and internal resistance based on Ohm’s Law applied to a potential difference measured across a current, assuming \(V_0 \approx 0\) signifies low or no starting value inaccuracies.