Question

A cell supplies a current of 0.9 A through a $2\Omega$ resistor and a current of 0.3 A through a $7\Omega$ resistor. The internal resistance of the cell is :-

• A. $1.0\Omega$
• B. $0.5\Omega$
• C. $2.0\Omega$
• D. $1.2\Omega$

Answer 1

The correct option is B $0.5\Omega$

On cross multiplication, we get $E=I\ast (R+r)$

For the first resistor, we have $I=0.9A$ and $R=2\Omega$

Hence, $E=0.9\ast (2+r)$

For the second resistor, $I=0.3A$ and $R=7\Omega$

Thus, $E=0.3(7+r)$

As the emf of the cell (E) remains constant, we can equate the two equations and hence we get

$0.9\ast (2+r)=0.3(7+r)$

$1.8+0.9r=2.1+0.3r$

Shifting variables to one side and constants to another side, we get

$0.9r-0.3r=2.1-1.8$

$0.6r=0.3$

Multiplying both sides by $10$, we get

$6r=3or$

$r=0.5\Omega$

Hence, the option $B$ is the correct answer.