JEE Advance - Physics (2015 - Paper 2 Offline - No. 9)
In the following circuit, the current through the resistor R(=2$$\Omega$$) is I amperes. The value of I is :
Answer
1
Explanation
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.
$${R_{AB}} = {{6 \times 18} \over {24}} = 4.5\,\,\Omega $$
$$I = {{6.5} \over {6.5}} = 1\,A$$
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