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A glavanometer of 50 Omega resistance...

A glavanometer of `50 Omega` resistance has 25 divisions. A current of `4xx10^(-4)` A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of `25 V`, it should be connected with a resistance of

A

245 `Omega` as a shunt

B

2550 `Omega` in series

C

2450 `Omega` in series

D

2500 `Omega` as a shunt

Text Solution

Verified by Experts

The correct Answer is:
C

According to question 25=I(R+Rg) =`(4xx10^(-4)xx25)(R+50)`
`rArr R+50=2500 rArr R=2450 Omega`
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