With a resistance of ' X ' in the lef | vedclass

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Question

The balance condition of bridge is shown below in the first case
Then, $\frac{X}{9}=\frac{40}{60} \Rightarrow X=\frac{2}{3} 	imes 9=6 \Omega$
To ge

Vedclass · Accepted Answer

In series with $X \Omega$

Vedclass · Answer

The balance condition of bridge is shown below in the first case
Then, $\frac{X}{9}=\frac{40}{60} \Rightarrow X=\frac{2}{3} 	imes 9=6 \Omega$
To get the balance point at 50 cm from left end, the balanced, condition becomes, $\frac{X^{\prime}}{9}=\frac{50}{50}=1 \Rightarrow X^{\prime}=9 \Omega$
As, in series combinations resistances are added, so to get $9 \Omega$ at left gap the $3 \Omega$ resistance should be added in series with $X \Omega(6+3=9 \Omega)$.

With a resistance of ' X ' in the left gap and a resistance of $9 \Omega$ 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 $3 \Omega$ resistance be connected to obtained the balance at 50 cm from the left end?

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