A galvanometer of resistance 50 is connected to a battery 3 ~V along with a resistance 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

Resistance of galvanometer, R_g=50 Emf of battery, V=3 ~V Resistance connected in series, R_s=2950 Total resistance, R^ =R_g+R_s=50+2950=3000 Current, I= V R^ = 3 3000 =10⁻³ ~A If the deflection has to be reduced to 20 divisions, then current, I^ = I 30 20= 2 3 10⁻³ ~A Let R_E be the effective resistance of the circuit, hence aligned 3 & =R_E I^ R_E & = 3 I^ = 3 2 3 10⁻³ =4.5 10^3=4500 aligned aligned Resistance to be added & =R_E-R_g & =4500-50=4450 aligned