With a resistance of ‘X’ in the left gap and a resistance of 9 Ω in the right gap of meter bridge, the balance point is obtained at 40 cm from the left end. In what way and to which resistance 3 Ω res

Question

With a resistance of ‘X’ in the left gap and a resistance of 9 Ω in the right gap of meter bridge, the balance point is obtained at 40 cm from the left end. In what way and to which resistance 3 Ω resistance be connected to obtained the balance at 50 cm from the left end?

TestHub · Accepted Answer

The balance condition of bridge is shown below in the first case

Then, $\frac{X}{9} = \frac{40}{60} \Rightarrow X = \frac{2}{3} \times 9 = 6 Ω$
To get the balance point at 50 cm from left end, the balanced, condition becomes,
$\frac{X^{'}}{9} = \frac{50}{50} = 1 \Rightarrow X^{'} = 9 Ω$
As, in series combinations resistances are added, so to get $9 Ω$ at left gap the $3 Ω$ resistance should be added in series with $X Ω (6 + 3 = 9 Ω)$ .

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