An ammeter has resistance `R_(0)` and range l what resistance should be connected in parallel with it to increase its range by nl ?

Options

A. a series resistance of `(G)/(n +1) Omega`
B. a shut of `(G)/(n +1) Omega`
C. a shut of `(G)/(n +1) Omega`
D. a series resistance of `(G)/(n -1) Omega`

Answer

B

Explanation

The range of an ammeter can be increased by connecting a shunt or parallel resistance to it, whose value is given by
`R = ((l_(g))/(l-l_(g))) G Omega`
To increase the range from l to nl, the current passing through it is l` = nl_(g)`,
Using this value in Eq. (i), we get
`R = ((l_(g))/(nl-l_(g))) G Omega`
` =(G)/((n-1)) Omega`