A galvanometer of resistance `95 Omega`, shunted resistance of `5 Omega`, gives a deflection of `50` divisions when joined in series with a resistance of `20 k Omega` and a `2 V` accumulator. What is the current sensitivity of the galvanomter (in div`//mu A`)?

Explanation

In accordance with the given problem, the situation is
depicted by the circuit diagram in Fig. `6.23`. As `20 k Omega` is much
greater than the resistance of the shunted galvanometer `(lt 5 Omega)`,
the current in the circuit will be
`l = (2)/(20 xx 10^(3)) = 10^(-4) A = 100 mu A`
Since this current produces a deflection of `50` divisions in the galvanometer,
`CS = (theta)/(I) = (50 "div")/(100 mu A) = (1)/(2) ("div")/(mu A)`