Q. For what value of $R$ in the circuit as shown in figure, current passing through 4$\Omega\ $resistance will be zero.
A) $1\ \Omega$
B) $2\ \Omega$
C) $3\ \Omega$
D) $4\ \Omega$
Explanation
(b) Since no current is to flow in the $4\Omega$ resistance, hence resistance $4\Omega$ becomes ineffective in current.
Current through resistances $2\ \Omega$ is
$i = \frac{9 - 6}{2} = \frac{3}{2}A$
In circuit $ABCDEFA,\ 9 - 3 = (2 + R) \times \frac{3}{2}$
or $12 = 6 + 3R\ or\ R = 2\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.