Question

In the given circuit, an ideal voltmeter connected across the $10\,\Omega$ resistance reads $2\,{\text{V}}$. The internal resistance $r$, of each cell is:

A. $1\,\Omega$
B. $0.5\,\Omega$
C. $1.5\,\Omega$
D. $0\,\Omega$

Answer 1

We should use the formulae for the equivalent resistance of the two or more resistors connected in series and parallel arrangement. Also use the expression for Ohm’s law. First calculate the equivalent resistance of the two resistors connected in parallel and then calculate the equivalent resistance of all the resistors connected in the circuit. Calculate the current flowing in the circuit and use Ohm’s law to calculate the internal resistance of each cell.

Formulae used:
The equivalent resistance ${R_{eq}}$ of the two resistors ${R_1}$ and ${R_2}$ connected in parallel is
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ …… (1)
The equivalent resistance ${R_{eq}}$ of the two resistors ${R_1}$ and ${R_2}$ connected in series is
${R_{eq}} = {R_1} + {R_2}$ …… (2)
The expression for Ohm’s law is
$V = IR$ …… (3)
Here, $V$ is the potential difference across the ends of the conductor, $I$ is the electric current and $R$ is the resistance.

Complete step by step answer:
We have given that an ideal voltmeter connected across the $10\,\Omega$ resistance reads $2\,{\text{V}}$.
$V' = 2\,{\text{V}}$
Let us first calculate the equivalent resistance $R'$ of the two resistors connected in parallel.The equation (1) becomes
$\dfrac{1}{{R'}} = \dfrac{1}{{15\,\Omega }} + \dfrac{1}{{10\,\Omega }}$
$\Rightarrow \dfrac{1}{{R'}} = \dfrac{{15 + 10}}{{150}}$
$\Rightarrow R' = 6\,\Omega$
Hence, the equivalent resistance of the two resistors connected in parallel is $6\,\Omega$.

Let us now draw the equivalent circuit diagram of the given circuit diagram.

The equivalent resistance ${R_{eq}}$ in the whole circuit according to equation (2) is
${R_{eq}} = 6\,\Omega + 2\,\Omega + 2r\,\Omega$
$\Rightarrow {R_{eq}} = \left( {8 + 2r} \right)\,\Omega$
The current flowing in the system of two parallel resistors with equivalent resistance is given by Ohm’s law.
$I = \dfrac{{V'}}{{R'}}$
Substitute $2\,{\text{V}}$ for $V'$ and $6\,\Omega$ for $R'$ in the above equation.
$I = \dfrac{{2\,{\text{V}}}}{{6\,\Omega }}$
$\Rightarrow I = \dfrac{1}{3}\,{\text{A}}$
Let now write the expression for Ohm’s law for the whole circuit.
$V = I{R_{eq}}$
Substitute $3\,{\text{V}}$ for $V$, $\dfrac{1}{3}\,{\text{A}}$ for $I$ and $\left( {8 + 2r} \right)\,\Omega$ for ${R_{eq}}$ in the above equation.
$\left( {3\,{\text{V}}} \right) = \left( {\dfrac{1}{3}\,{\text{A}}} \right)\left[ {\left( {8 + 2r} \right)\,\Omega } \right]$
$\Rightarrow 9 = 8 + 2r$
$\Rightarrow 2r = 1$
$\therefore r = 0.5\,\Omega$
Therefore, the internal resistance of each cell is $0.5\,\Omega$.

Hence, the correct option is B.

Note: The students may think that why we have calculated the current through the system of the two resistors connected in parallel and not through the whole circuit. But the students should keep in mind that the system of two resistors in parallel is in series with the whole circuit. Hence, the net current through the system of parallel resistors will be the same through the system as the current in the series network is the same.