In the given circuit, current $I$ is independent of the resistance $R_6$. Then

In the given circuit, current $I$ is independent of the resistance $R_6$. Then
  1. $R_1 R_2 R_5=R_3 R_4 R_6$
  2. $\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_1+R_2}+\frac{1}{R_3+R_4}$
  3. $R_1 R_4=R_2 R_3$
  4. $R_1 R_3=R_2 R_4$

Solution

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, $\frac{R_1}{R_2}=\frac{R_3}{R_4} \Rightarrow R_1 R_4=R_2 R_3$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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