Question

If an ammeter of negligible internal resistance has been inserted in series with circuit, it reads $1A$. If a voltmeter of very high value of resistance is connected across ${{R}_{1}}$​, it reads $3V$. But if the points A and B are short circuited with the use of a conducting wire, then the voltmeter measures $10.5V$ across the battery. Find out the internal resistance of the battery?
$\begin{aligned} & A.\dfrac{3}{7}\Omega \\\ & B.5\Omega \\\ & C.3\Omega \\\ & D.\text{none of these} \\\ \end{aligned}$

Answer 1

The electric potential of the conducting wire will be equivalent to the product of the current passing through it and the resistance across it. This is the ohm’s law. The internal resistance will be the difference between the initial electric potential and the electric potential after short circuiting to the current flowing when short circuited. This all will help you in solving this question.

Complete step by step answer:
the electric potential has been mentioned as,
$E=3V$
The current in the circuit can be mentioned as,
$I=1A$
Using this, the value of the resistance can be written as,
${{R}_{1}}=\dfrac{E}{I}$
Substituting the values in it will give,
${{R}_{1}}=\dfrac{3}{1}=3\Omega$
When the points A and B are connected using a conducting wire, ${{R}_{2}}$ will be short-circuited. Hence we can write that,
$\begin{aligned} & 10.5={I}'{{R}_{1}} \\\ & \Rightarrow 10.5={I}'\times 3 \\\ & {I}'=\dfrac{10.5}{3}=3.5A \\\ \end{aligned}$
From this the internal resistance should be calculated. We can write that,
$10.5=E-{I}'r$
Substituting the values in it will give,
$10.5=12-3.5\times r$
Rearrange the obtained equation as,
$r=\dfrac{1.5}{3.5}=\dfrac{3}{7}\Omega$
Therefore the value of the internal resistance has been calculated.

So, the correct answer is “Option A”.

Note: Internal resistance is the resistance or the opposition provided against the flow of electrons or the current in a battery. This will cause the production of heat in it. Internal resistance is also having the units of ohm. Every battery will be having some internal resistances as the zero internal resistance is ideal.