• # question_answer The resistance of a galvanometer is 60. and it can measure a maximum current of 2amp, then required shunt resistance to convert it into an ammeter reads upto 6 amp, will be: A)  50                  B)  40C)  30                  D)  20

The resistance of the galvanometer $G=6\Omega$ maximum current through the galvanometer${{i}_{g}}=2$ amp Final current in ammeter $i=6$ amp The current through the galvanometer is given by ${{i}_{g}}=\frac{S}{S+G}\times i$ (where S is the resistance of shunt) $\Rightarrow$               $2=\frac{S}{S+6}\times 6$ $\Rightarrow$               $2S+12=6S.$ $\Rightarrow$               $4S=12$ $\Rightarrow$               $S=3\Omega$.