QJEEAdv15Ph5fa8: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph5fa8 Solution: In the following circuit, the current through the resistor R(=2 …

Consider the following figure:

ACGA constitutes a Wheatstone bridge; hence, 8 $\Omega$ is redundant and hence can be removed. Therefore,

${R_{AG}} = {{3 \times 6} \over 9} = \,2\,\Omega$

AGDFA again constitutes a Wheatstone bridge 10 $\Omega$ which is redundant and hence can be removed.

${R_{AB}} = {{6 \times 18} \over {24}} = 4.5\,\,\Omega$

$I = {{6.5} \over {6.5}} = 1\,A$