Q. Three resistances of $4\Omega,\ 6\Omega\ and\ 12\Omega$ are connected in parallel and the combination is connected in series with 4 V battery with internal resistance of $2\ \Omega$. The battery current is
A) 1 A
B) 10 A
C) 2 A
D) 0.5 A
Explanation
(a) Effective resistance $R_{P}$ of 4$\ \Omega,\ 6\Omega$ and $12\ \Omega$ in parallel will be
$\frac{1}{R_{p}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2}\ or\ R_{p} = 2\Omega$
Total resistance of circuit $= 2 + 2 = 4\ \Omega$
The battery current, $i = 4/4 = 1A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.