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A galvanometer of resistance 50 Omega gi...

A galvanometer of resistance 50 `Omega` gives a full scale deflection for a current `5 xx 10^(-4)` A. The resistance that should be connected in series with the galvanometer to read 3 V is

A

`5050 Omega`

B

`5950 Omega`

C

`595 Omega`

D

`5059 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`R_(G)= 50 Omega, I_(g) = 5 xx 10^(-4) A`
`V= I_(g) (R + R_(G))`
`rArr R = V/I_(g) - R_(G) = 3/(5 xx 10^(-4)) - 50`
`= 6000 - 50 = 5950 Omega`
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