Electrical Instruments

Let's calculate the current for the full-scale deflection:

Now, let's calculate the current required for a 20-division deflection:

To find the resistance needed, we can use Ohm's law:

For the 20-division deflection:

Solving this equation for , we get:

Calculating this expression gives:

Rounding this value, we get: 4,450 Ω.

A meter bridge works on the principle of the Wheatstone bridge. Let the resistance in the left gap be (the metal conductor) and the resistance in the right gap be .

Let the balancing point be found at a distance from the left end (usually denoted as A or 0 cm mark) on the uniform meter bridge wire of total length (usually 100 cm). The length of the wire segment to the right of the balancing point is .

For a uniform wire, resistance is proportional to length. Therefore, the Wheatstone bridge condition at balance is:

Now, consider the effect of heating the metal conductor connected in the left gap.

For a metal conductor, its resistance increases with an increase in temperature. This is because the increased thermal agitation of the atoms in the conductor hinders the flow of electrons.

So, when the metal conductor is heated, its resistance increases.

The resistance in the right gap, , remains unchanged. The meter bridge wire properties also remain unchanged.

Let the initial balancing length be and the final balancing length (after heating) be . Let the initial resistance be and the final resistance be .

Initially:

After heating:

Since (due to heating) and is constant, the ratio on the left side has increased:

This implies that the ratio on the right side must also increase:

The function is an increasing function of (for ). This means that if the value of the ratio increases, the value of must also increase.

Therefore, .

An increase in the balancing length means the balancing point moves further away from the left end (0 cm) and closer to the right end (100 cm).

Conclusion: The balancing point Shifts towards right.