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In the given circuit, it is observed that the current $I$ is independent of the value of the resistance $R_{6}$. Then the resistance values must satisfy 
A$R_{1}R_{2}R_{3} = R_{3}R_{4}R_{6}$
B$\frac{1}{R_5} +\frac{1}{R_6} = \frac{1}{R_1 + R_2} +\frac{1}{R_3 + R_4}$
C$R_{1}R_{4} = R_{2}R_{3}$
D$R_{1}R_{3} = R_{2}R_{4} = R_{3}R_{6}$
Answer & Solution
Correct answer: C. $R_{1}R_{4} = R_{2}R_{3}$
Let the network after $R_5$ have input terminals at the top and bottom nodes. Since $R_5$ is in series with the rest, for current $I$ to be independent of $R_6$, the equivalent resistance of the bridge network must be independent of $R_6$.
This happens when no current flows through $R_6$, so the bridge is balanced. Then the potentials at the midpoints of the two vertical branches are equal.
Hence the potential division on the left and right branches must satisfy
$$\frac{R_1}{R_2}=\frac{R_3}{R_4}$$
Therefore
$$R_1R_4=R_2R_3$$
Now compare with the options. The matching condition is option $C$.
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