A galvanometer of resistance ( 50 ) is connected to a battery of ( 3 ~V ) along with a resistance of ( 2950 ) in series shows full-scale deflection of ( 30 ) divisions. The additional series resistance required to reduce the deflection to ( 20 ) divisions is

Step-by-step solution

Given, galvanometer resistance, ( R_ G =50 ; ) battery, ( V =3 ~V ; ) resistance, ( R =2950 ; ) deflection ( =30 ) divisions. Now, ( I= V (R_ G +R ) = 3 (50+2950) =0.001=1 10⁻³ ~A ) Thus, current when deflection is ( 30 ) divisions ( =10⁻³ ~A ) ( ) Current when deflection is ( 1 ) division ( = 10⁻³ 30 ) ( ) Current when deflection is ( 20 ) division ( = 10⁻³ 30 20 ) Therefore, in order to have a deflection of ( 20 ) division the resistance should be [ array l 2 3 10⁻³= 3 (50+R) (50+R)= 3 3 2 10⁻³ 50+R=4.5 1000 R=45