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The resisitance of a galvanometer is 5...

The resisitance of a galvanometer is `50 Omega ` and it shows full scale diflection for current of 1 mA. . To convert it into a voltmeter to measure 1V and as well as 10 V ( refer circuit diagram ) the resistance `R_(1) ` and `R_(2) ` respecivtively

A

`950 Omega` and `9150 Omega`

B

`900 Omega` and `9950 Omega`

C

`900 Omega` and `9900 Omega`

D

`950 Omega` and `9000 Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
`1V = 1 xx 10^(-3) (R_(G) + R_(1))`
`rArr 10^(3) = 50 + R_(1) rArr R_(1)= 950 Omega`
For 10 V
`10 = 1 xx 10^(-3)(R_(G) + R_(1) + R_(2))`
`rArr 10^(4) = 1000 + R_(2) rArr R_(2) = 9000 Omega`
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