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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 :
JEE Advanced 2015 Paper 2 Offline Physics - Current Electricity Question 21 English
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Consider the following figure:

JEE Advanced 2015 Paper 2 Offline Physics - Current Electricity Question 21 English Explanation 1

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

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

JEE Advanced 2015 Paper 2 Offline Physics - Current Electricity Question 21 English Explanation 2

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

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

JEE Advanced 2015 Paper 2 Offline Physics - Current Electricity Question 21 English Explanation 3

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

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