The resistance of the shunt required to allow 2% of the main current through the galvanometer of resistance `49Omega` is

Options

A. `1Omega`
B. `2Omega`
C. `0.2Omega`
D. `0.1Omega`

Answer

A

Explanation

Equating potential drop across shunt and galvanometer, we get `(l-l_(g))S=l_(g)G`
where S is the shhunt resisrtance and G is galvanometer resistance.
given `l_(g)=(2)/(100)l`
`implies(l-(2)/(100)l)S=49xx(2)/(100)l`
or `S=1Omega`