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Explain how value of unknown resistor can be obtained by using meter bridge.

Solution
In one gap of meter bridge unknown resistor $\mathrm{R}$ and in second gas resistance $\mathrm{S}$ from resistance box is connected.
Jockey key is slided (moved) on wire$ AC$.
Assume that when jockey key is on position $D$, galvanometer shows zero deflection.
Let length from $A$ to $D$ is $l$.
Resistance of AD wire $=\mathrm{R}_{\mathrm{cm}} l$ where $\mathrm{R}_{\mathrm{cm}}$ is resistance of wire per unit length $(1 \mathrm{~cm})$.
$\therefore$ Resistance of$DC$ wire $=\mathrm{R}_{\mathrm{cm}}(100-l)$
Wire $\mathrm{AB}, \mathrm{BC}, \mathrm{DA}$ and $\mathrm{CD}$ form $\mathrm{Wheatstonebridge}$,
where, $\mathrm{AB}=\mathrm{R}, \mathrm{BC}=\mathrm{S}, \mathrm{DA}=\mathrm{R}_{\mathrm{cm}} l$
$\mathrm{CD}=\mathrm{R}_{\mathrm{cm}}(100-l)$
From principle of Wheatstone bridge for balancing of Wheatstone bridge,
$\quad \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DA}}{\mathrm{CD}}$
$\therefore \frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{R}_{\mathrm{cm}}^{l}}{\mathrm{R}_{\mathrm{cm}}(100-l)}$
$\therefore \frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l}$
Unknown resistance can be obtained by using equation $\mathrm{R}=\mathrm{S} \frac{l}{100-l} .$
For different value of S, value of unknown resistance $R$ can be obtained by taking average value error can be eliminated.
To eliminate end correction, position of resistance $R$ and resistance $S$ from box can be interchanged and null point can be obtained and error can be reduced.
For balancing bridge, null point should be at $50 \mathrm{~cm}$ (between $40 \mathrm{~cm}$ and $60 \mathrm{~cm}$ ) and percentage error can be minimized.
From value of resistance obtained by meter bridge temperature of unknown resistance can be obtained.
By using meter bridge value of small resistance can be known.