Question

A moving coil galvanometer has 24 turns and 12 ohm resistance

Answer 1

$\huge \underline \green{\sf{ Answer :- }}$

Current sensitivity of the galvanometer is given by

    Cs=kNAB

where k= resistance couple per unit test

Case (1)            

 C=kNAB→(1)  

Case(2)

   C′=(C+10025∗C)

             =1.25C

      and C′=kN′AB→(2) 

Taking ratio of (1) and (2)

  we get,

C′C=N′N

   N′=N∗C′C=N∗1.25

Hence N′=24∗1.25=30 turns

So we need to increase 6 turns as voltage sensitivity is given by

   Sv = RkNAB

 as Sv ∝ R1 where R=resisatnce

$\small \underline {\sf{ ↬ \ Hence \ increase \ in \ resistance \ will \ decrease}}$

$\small \underline {\sf{voltage \ sensitivity. }}$

Answer 2

Answer:

Current sensitivity of the galvanometer is given by

Cs=kNAB

where k= resistance couple per unit test

Case (1)

C=kNAB→(1)

Case(2)

C′=(C+10025∗C)

=1.25C

and C′=kN′AB→(2)

Taking ratio of (1) and (2)

we get,

C′C=N′N

N′=N∗C′C=N∗1.25

Hence N′=24∗1.25=30 turns

So we need to increase 6 turns as voltage sensitivity is given by

Sv = RkNAB

as Sv ∝ R1 where R=resisatnce

\small \underline {\sf{ ↬ \ Hence \ increase \ in \ resistance \ will \ decrease}}

↬ Hence increase in resistance will decrease

\small \underline {\sf{voltage \ sensitivity. }}

voltage sensitivity.

Explanation:

31♥️+flw≈ib