Single Choice

In a experiment of meter bridge, a null point is obtained at the centre of the bridge wire. When a resistance of $$10 \ ohm$$ is connected in one gap, the value of resistance in other gap is

A$$10 \Omega$$
Correct Answer
B$$5 \Omega$$
C$$\dfrac{1}{5} \Omega$$
D$$500 \Omega$$

Solution

Given that Wheatstone bridge is balanced, therefore
condition of balanced wheatstone bridge is:
$$\dfrac{P}{Q}=\dfrac{R}{S}$$
According to question
$$1=\dfrac{10}{S} \Rightarrow S=10\ ohm$$


SIMILAR QUESTIONS

Current Electricity

In a meter bridge experiment $$S$$ is a standard resistance. $$R$$ is a resistance wire. It is found that balancing length is $$l=25\ cm$$. If $$R$$ is replaced by a wire of half length and half diameter that of $$R$$ of same material, then the balancing distance $$l'$$ (in $$cm$$) will now be--------.

Current Electricity

Let $$\vec{a},\vec{b}$$ and $$\vec{c}$$ be three vectors such that $$|\vec{a}|=\sqrt{3},|\vec{b}|=5,\vec{b}.\vec{c}=10$$ and the angle between $$\vec{b}$$ and $$\vec{c}$$ is $$\dfrac{\pi}{3}$$. If $$\vec{a}$$ is perpendicular to the vector $$\vec{b}\times \vec{c}$$, then $$|\vec{a}\times (\vec{b}\times \vec{c})|$$ is equal to

Current Electricity

A resistance wire connected in the left gap of a metre bridge balances a $$10\Omega$$ resistance in the right gap at a point which divides the bridge wire in the ratio $$3 : 2$$. If the length of the resistance wire is $$1.5 m$$, then the length of $$1 \Omega$$ of the resistance wire is :

Current Electricity

In a meter bridge, the balancing length from the left end (standard resistance of one ohm in the right gap) is found to be $$20 \ cm$$. The value of the unknown resistance is

Current Electricity

If the balance point is obtained at the $$35^{th} cm$$ in a metre bridge the resistances in the left and right gaps are in the ratio of

Current Electricity

In the shown arrangement of the experiment of meterbridge, if AC corresponding to null deflection of galvanometer is x , what would be its value if the radius of the wire AB is doubled

Current Electricity

Two resistances X and Y in the two gaps of a meter-bridge gives a null point dividing the wire in the ratio $$2:3$$. If each resistance is increased by $$3 \Omega$$, the null point divides the wire in the ratio $$5:6$$, calculate the value of X and Y.

Current Electricity

In a meter bridge shown in the figure, the balance point is found to be 40 cm from and A.If a resistance of $$10 \Omega$$ is connected in series with R, balance point is obtained 60 cm from A.Calculate the value of R and S.

Contact Details