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In the given circuit, current $I$ is independent of the resistance $R_6$. Then
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The correct answer is:
$R_1 R_4=R_2 R_3$
If resistance $R_1, R_2, R_3, R_4$ and $R_6$ from Wheatstone bridge, then current will independent of resistance $R_6$.
For this, $\frac{R_1}{R_2}=\frac{R_3}{R_4} \Rightarrow R_1 R_4=R_2 R_3$
For this, $\frac{R_1}{R_2}=\frac{R_3}{R_4} \Rightarrow R_1 R_4=R_2 R_3$
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