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

Question

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

TestHub · Accepted Answer

For a meter bridge, the balancing condition is given by $\frac{R _{1}}{R _{2}} = \frac{l}{100 - l}$

Initially, $R_{1} = x$ , $R_{2} = 9 Ω$ , and $l = 40$ cm.

Substituting the values:

$\frac{x}{9} = \frac{40}{100 - 40}$

$\frac{x}{9} = \frac{40}{60} = \frac{2}{3}$

$x = 6 Ω$

To obtain the balance point at $50$ cm, the ratio of resistances in the left and right gaps must be $1 : 1$ .

$\frac{R _{1}^{'}}{R _{2}^{'}} = \frac{50}{100 - 50} = 1$

$R_{1}^{'} = R_{2}^{'}$

Since the right gap has a resistance of $9 Ω$ , the left gap must also have a resistance of $9 Ω$ .

The current resistance in the left gap is $x = 6 Ω$ . To make it $9 Ω$ , a resistance of $3 Ω$ must be connected in series with $x$ .

$R_{1}^{'} = x + 3 = 6 + 3 = 9 Ω$

Answer: In series with $x Ω$

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