A galvanometer having a coil resistance of `100 omega` gives a full scale deflection , when a current of `1 mA` is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of `10A` , is :

Options

A. `0.01 Omega`
B. `2 Omega`
C. `0.1 Omega`
D. `3 Omega`

Answer

a

Explanation

To convert it into an ammeter for measuring up to 10 A, we need to connect a low value resistance called shunt in parallel with the coil and let S is the resistance of shunt. The value of hunt needs to be adjusted in such a manner that when total current is 10 A then the current passing through the coil must be only 1mA and remaining current must pass through shunt connected in parallel. For the same potential difference across coil and shunt we can write the following equation.
`i_(g) G = (i - i_(g)) S`
`rArr (1 xx 10^(-3)) (100) = (10 - 1 xx 10^(-3))S`
Hence S is approximately `0.01 Omega` . So, option (a) is correct.