Question

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

• A. $0.5\, \Omega$
• B. $1/3\, \Omega$
• C. $1/4\, \Omega$
• D. $1\, \Omega$

Answer 1

Answer: (B)

$1/3\, \Omega$

Answer 2

$I =\frac{ E }{ R + r }$
$2=\frac{ E }{2+ r }\,\,\,\,\,\,\,\,...(1)$
$0.5=\frac{ E }{9+ r }\,\,\,\,\,\,\,\,...(2)$
(1) divided by (2)
$4=\frac{9+ r }{2+ r }$
$8+4 r=9+r$ or $3 r=1$
$\therefore r =\frac{1}{3} \Omega$