Question

A current of $2\text{A}$ flows through a $2\Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5\text{A}$ when connected across a $9\Omega$ resistor. The internal resistance of the battery is

• A. $\frac{1}{4}\Omega$
• B. $1\Omega$
• C. $0.5\Omega$
• D. $\frac{1}{3}\Omega$

Answer 1

The correct option is D $\frac{1}{3}\Omega$

Let the internal resistance of the battery be $r$. Then, the current flowing through the circuit is given by
$I=\frac{E}{R+r}$

Case 1:
$2=\frac{E}{2+r}\dots i$

Case 2:
$0.5=\frac{E}{9+r}\dots ii$

Equating $E$ from (i) and (ii)
$4+2r=4.5+0.5r$
$1.5r=0.5$
$r=\frac{1}{3}\Omega$