A galvanometer of resistance `50Omega` gives full deflection for a current of `0*05A`. Calculate the length of shunt wire required to convert the galvanometer into an ammeter of range 0 to `5A`. The diameter of the shunt wire is `2mm` and its resistivity is `5xx10^-7Omegam`.

Answer

`3*174m` in parallel

Explanation

`S=(I_gG)/(I-I_g)=(0*50xx50)/(5-0*05)=(0*05xx50)/(4*95)=50/99Omega`
Now, `l=(Spir^2)/(rho)=50/99xx22/7xx((10^-3)^2)/(5xx10^-7)`
`=3*174m`