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A potential difference of 0.75V applied ...

A potential difference of `0.75V` applied across a galvanometer causes a current of 15 m A to pass through it. If can be converted into ammeter of range of 25 A , the requried shunt should be

A

`0.3Omega`

B

`0.03Omega`

C

`0.003Omega`

D

`0.0003Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`V_Gi_g G therefore I_g=V_g//G = 5xx10^-4A`
`S=((I_g)/(I-I_g))G=((5xx10^-4)/(5-0.0005))xx100=0.01Omega`
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