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A galvanometer of 50 Omega resistance...

A galvanometer 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

`2500 Omega` as a shunt

B

`245 Omega` as a shunt

C

`2550 Omega` in series

D

`2450 Omega` in series

Text Solution

Verified by Experts

The correct Answer is:
D

To convert a galvanometer into voltmeter, high resistance should be connected in series with it. Let `R` is the resistance connected in series with the galvanometer.

`i_(g)=V/(G+R)` or `R=V/(i_(g))=-G`
Given, `G=50 Omega`,
`i_(g)=25xx4xx10^(-4) =10^(-2) A, V=25 V`
`:. R=25/(10^(-2))-50=2500-50=2450 Omega`
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