In the given circuit, it is observed& that the current I is...

In the given circuit, it is observed& that the current $I$ is independent of the value of resistance ${R}_{5}$ . Then, the resistance value must satisfy

In the given circuit, it is observed& that the current I is independent of the value of resistance R _ 5 . Then, the

Options

A $\dfrac{1}{R5} + \dfrac{1}{R6} = \dfrac{1}{{R}_{1} + {R}_{2}} \cdot \dfrac{1}{{R}_{3} + {R}_{4}}$
B ${R}_{1}{R}_{4} = {R}_{2}{R}_{3}$
C ${R}_{1}{R}_{2}{R}_{5} = {R}_{3}{R}_{4}{R}_{6}$
D ${R}_{1}{R}_{3} = {R}_{2}{R}_{4} = {R}_{5}{R}_{6}$

Correct answer

B. ${R}_{1}{R}_{4} = {R}_{2}{R}_{3}$

Step-by-step solution

From the given circuit,& we can say that $I$ is independent of resistance ${R}_{5},$ so no current flows through ${R}_{5}.$ This required that the junction of ${R}_{1}$ and ${R}_{2}$ is at the same potential as the junction of ${R}_{3}$ and ${R}_{4}$ (Wheatstone's bridge balancing condition)

So, $\dfrac{R1}{R2} = \dfrac{R3}{R4}$

or ${R}_{1}{R}_{4} = {R}_{3}{R}_{2}$

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