In an ammeter `5%` of the main current is passing through the galvanometer. If the resistance of galvanometer is G, then the resistance of shunt S will be
A. G/19
B. G/5
C. 5G
D. 19 G

4 views

1 Answers

Correct Answer - A
`I_(g)=((10)/(100))I=0.1I`
`S=((I_g)/(I-I_g))g=((0.1I)/(I-0.1I))99=11Omega`

4 views

Related Questions