Question

Answer the following :
(a) Why are the connections between the resistors in a meter bridge made of thick copper strips?
(b) Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire?
(c) Which material is used for the meter bridge wire and why?
OR
A resistance of R $\Omega$ draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance $R_0\Omega$. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Answer 1

(a) The resistivity of copper wire is very low. The connections are made of thick copper wires to minimize the resistance of connecting wires. Because the connection resistance have not been accounted in the formula, (v Resistance is inversely proportional to cross-sectional area, so thick wire has low resistance).

(b) The balance point is obtained in the middle of the meter bridge wire so as to increase the sensitivity of the meter bridge and there by no deflection in the galvanometer.

(c) Alloys, such as manganin or constantan are used for making meter bridge wire due to their low temperature coefficient of resistance and high resistivity. Due to which the resistance of the wire does not change with increase in temperature of the wire due to flow of current.

Given: A resistance of$R\Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_0\Omega$. A voltage $V$ is supplied to the potentiometer.

To find an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

Solution:

Total resistance is given by,

$R_{tot}=\frac{R_0}{2}+\frac{\frac{R_0}{2}\times R}{\frac{R_0}{2}+R}=\frac{R(R_0+4R)}{2(R_0+2R)}$

Total current through the device is given by,

$I_{tot}=\frac{V}{R_{tot}}=\frac{2V(R_0+2R)}{R(R_0+4R)}$

Current through resistance $R$ is given by,

$I_2=I_{tot}\times \frac{\frac{R_0}{2}}{\frac{R_0}{2}+R}⟹I_2=I_{tot}\times \frac{R_0}{R_0+2R}⟹I_2=\frac{2V(R_0+2R)}{R(R_0+4R)}\times \frac{R_0}{R_0+2R}⟹I_2=\frac{2V{R}_0}{R(R_0+4R)}$

Voltage across resistance is given by

$I_{2}R=\frac{2V{R}_0}{(R_0+4R)}$

is the required expression for the voltage when sliding contact is in the middle of the potentiometer.