Question

Explain with a neat circuit diagram how will you determine unknown resistance 'X' by using meter bridge.

Answer 1

Watch the diagram. You can see a wire AC which is a metre long wire uninsulated and has a considerable resistance and a uniform cross-section.

You can see 2 L shaped copper strips on the sides and a ribbon shaped strip at the centre.

The unknown resistance X is connected on left and a known variable resistance is connected on right. The middle point B is connected to a galvanometer and the other side of the galvanometer$G$ is connected to a jockey, The jockey is a small rod-like conducting device with the other end sharp.

When the jockey is touched to any random point on the metre bridge, the G shows a deflection, meaning that current is flowing through the jockey to the point of touch.

When the jockey is touched to a particular point on the metre bridge, the G shows no deflection, meaning that no current is flowing through the jockey to the point of touch. Here a situation similar to the wheatstone's network's balancing condition is achieved.

If resistance of AD is $R_{AD}$ and similarly $R_{DC}$ then we have

$\frac{X}{R_{AD}}=\frac{R}{R_{DC}}$ $\therefore \frac{R_{AD}}{R_{DC}}=\frac{X}{R}$

$R_{AD}=\rho \frac{l_{{}_X}}{A}$ where A is area of cross section and $\rho$ is resistivity.

Similarly $R_{DC}=\rho \frac{l_{{}_R}}{A}$

$\therefore \frac{R_{AD}}{R_{DC}}=\frac{l_{{}_X}}{l_{{}_R}}$

By substituting we get $\frac{X}{R}=\frac{l_{{}_X}}{l_{{}_R}}$

R is known and $l_{{}_X}$ and $l_{{}_R}$ can be easily measured with the help of the metre scale. Thus X can be calculated.

The same procedure is repeated for different values of R and various values of X is calculated. Unless the experiment system is faulty, the values of X should be approx equal. The mean value of these values is the true value of X.