Question

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

• A. $\frac{1}{3} \, \Omega$
• B. $\frac{1}{4}\Omega$
• C. $5 \, \Omega$
• D. $0.5 \, \Omega$

Answer 1

Answer: (A)

Current $I=\frac{E}{R + r}$
$\Rightarrow 2=\frac{E}{2 + r}$ .... (i)
$\Rightarrow 0.5=\frac{E}{9 + r}$ .... (ii)
On dividing equation (i) by equation (ii), we get
$\frac{2}{\text{0.5}}=\frac{9 + r}{2 + r}$
$\Rightarrow 4=\frac{9 + r}{2 + r}$
$\Rightarrow 3r=1$
$\Rightarrow r=\frac{1}{3}\Omega$