A galvanometer of resistance 50 is connected to a battery of 3 ~V along with a resistance of 2950 in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Step-by-step solution

Current through the galvanometer (I= 3 (50+2950) =10⁻³ ~A ) Current for 30 divisions (=10⁻³ ~A ) Current for 20 divisions (= 10⁻³ 30 20 ) (= 2 3 10⁻³ ~A ) For the same deflection to obtain for 20 divisions, let resistance added be R ( aligned & 2 3 10⁻³= 3 (50+1 R ) & or R =4450 aligned )