(i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l₁ with resistors R and S as shown in the figure.

When an unknown resistor X is connected in parallel with the resistor S, the balance point shift to a distance l₂. Find the expression for X in terms of l₁ l₂ and S.

Solution

(i) A slide wire bridge is known as meter bridge. It is constructed on the principle based on balanced Wheatstone bridge and is used to find the resistance of an unknown conductor or, to compare the value of two resistances.

When the jockey is moved along the wire, there is one position where the galvanometer will show no current. Then, a metre bridge is balanced and the balanced condition is,

\frac{R}{S} = \frac{l}{100 - l}

where R is the unknown resistance and S is the standard resistance.

(ii) When, R and S are connected, balance point is found at a distance l₁from the zero and

∴

\frac{R}{S} = \frac{l_{1}}{100 - l_{1}}

...(1)
When unknown resistance X is connected in parallel to S.
∴ total resistance in the right hand gap is

S_{1} = \frac{SX}{S + X}

[∵ \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \Rightarrow R = \frac{R_{1} R_{2}}{R_{1} + R_{2}}]

Since the balance point is obtained at a distance l₂ from the zero end

∴

\frac{R}{S_{1}} = \frac{l_{2}}{100 - l_{2}}

putting the value of S₁, we get

\frac{R}{\frac{SX}{S + X}} = \frac{l_{2}}{100 - l_{2}} \frac{R (S + X)}{SX} = \frac{l_{2}}{100 - l_{2}} . . . (2)

Dividing equation (2) by (1), we get

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