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${R_1}{R_2}{R_5} = {R_3}{R_4}{R_6}$
  • 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${R_1}{R_3} = {R_2}{R_4} = {R_5}{R_6}$
IIT 2001, Medium
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c
(c) As $I$ is independent of ${R_6},$ no current flows through ${R_6}$ this requires that the junction of ${R_1}$ and ${R_2}$ is at the same potential as the junction of ${R_3}$ and ${R_4}$. This must satisfy the condition $\frac{{{R_1}}}{{{R_2}}} = \frac{{{R_3}}}{{{R_4}}},$ as in the Wheatstone bridge.
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