Question 608: Current Electricity

Question

With a resistance of ' $X$ ' in the left gap and resistance of $9 \Omega$ in the right gap of a 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 obtain the balance at 50 cm from the left end?

Accepted Answer

The balanced 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 combination resistances are added, so to get $9 \Omega$ at left gap the $3 \Omega$ resistance should be added in series with $X \Omega(6+3=9 \Omega)$.

Answer

In series with $9 \Omega$

Answer

Parallel to $X{\Omega}$

Answer

In series with $X \Omega$

Answer

Parallel to $9 \Omega$

Current Electricity

With a resistance of ' X ' in the left gap and resistance of 9 \Omega in the right gap of a 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 obtain the balance at 50 cm from the left end?

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