Question

In the adjacent circuit a resistance R is used. Initially with 'wire AB' not in the circuit, the galvanometer shows a deflection of d divisions. Now, the 'wire AB' is connected parallel to the galvanometer and the galvanometer shows a deflection nearly $d/2$ division. Therefore current sensitivity of the galvanometer is about:

Answer 1

Given: a circuit with a resistance R is used. Initially with 'wire AB' not in the circuit, the galvanometer shows a deflection of d divisions. Now, the 'wire AB' is connected parallel to the galvanometer and the galvanometer shows a deflection nearly $\frac{d}{2}$ division

To find current sensitivity of the galvanometer

Solution:

When wire AB is not in the circuit (fig(i))

Where R>>G

Current through galvanometer G is case I,

$I_0=\frac{E}{R+G}\approx \frac{E}{R}$

When wire AB is connected in the circuit (fig(ii)) as given in the question,

Where R>>G

Current through galvanometer G is case II,

$I_0=\frac{E}{R+\frac{G}{2}}\approx \frac{E}{R}$

above condition is possible when resistance of wire AB is equal to resistance of galvanometers. and current $I_0$ remain constant in circuit, it is possible when R>>G.