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A galvanometer having a resistance of 10...

A galvanometer having a resistance of `100 Omega` has 25 divisions. A current of` 0.4 mA` deflect the pointer of the galvanometer into a voltmeter of 0 to 25 V, it should be connected with a resistance of

A

`2000 Omega` as shunt

B

`2400 Omega` as shunt

C

`2400 Omega` in series

D

`2500 Omega` in series

Text Solution

Verified by Experts

The correct Answer is:
C
`i_g=25 xx 0.4=10mA=0.01 A` The resistance (X) that sholud be joined in series is`X=V/I_g-G=(25)/(0.01)=2400Omega`.
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