A galvanometer of resistance 200 is to be converted into an ammeter. The value of shunt resistance which allows 3 % of the main current through the galvanometer is equal to (nearly)

Step-by-step solution

Given: Resistance of galvanometer is G=200 . Current through the galvanometer is i_g=3 % of i . To convert into an ammeter, the galvanometer with resistance G has to be shunted with S . The potential drop across the galvanometer and shunt is the same: aligned & i_g G= (i-i_g ) S & i i_g =1+ G S aligned Given, i_g i = 3 100 aligned & 100 3 =1+ 200 S & S= 200 3 97 6 aligned