Given, galvanometer resistance,
RG=50Ω; battery, V = 3 V; resistance, R = 2950 Ω; deflection = 30 divisions. Now,
Thus, current when deflection is 30 divisions
=10−3A ⇒ Current when deflection is 1 division
= ⇒ Current when deflection is 20 division
=×20 Therefore, in order to have a deflection of 20 division the resistance should be
×10−3= ⇒(50+R)= 50+R=4.5×1000 R=4500−50=4450 Hence, additional resistance required
=4450−2950=1500