A moving coil galvanometer converted into an ammeter reads up to 0.03 A by connecting a shunt of resistance 4r across it and ammeter reads up to 0.06 A, when a shunt of resistance r is used. What is the maximum current which can be sent through this galvanometer if no shunt is used?

detailed solution

Correct option is C

Let G be the resistance of galvanometer and ig the  current which on passing through the galvanometer produces full-scale deflection.Since G and S are in parallel, the potential difference across them will beig ×  G=i−ig  ×  Sigi= SS+GHence,   ig= 4r4r+G  ×  0.03                                          ....1ig= rr+G  ×  0.06                                    ....2Equating Eqs. (1) and (2), we get4r + G = 2(r + G)⇒    G=2r∴      ig= 4r4r+2r  ×  0.03=0.02  A     from Eq.  1