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In the circuit shown, the galvanometer G...

In the circuit shown, the galvanometer G of resistance `60 Omega` is connected to a supply voltage of 5.0 V.

If the galvanometer G is shunted by a resistance `r = 0.02 Omega`, then the effective resistance of the galvanometer will be nearly,

A

`2 Omega`

B

`0.02 Omega`

C

`0.2 Omega`

D

`20 Omega`

Text Solution

Verified by Experts

The correct Answer is:
b
Here, `R_G = 60 Omega, r = 0.02 Omega`
When the galvanometer is shunted by a resistance r, its effective resistance is
`R. = (R_G r)/(R_G + r)`
`= ((60 Omega)(0.02 Omega))/(60 Omega div 0.02 Omega) = 0.02 Omega`
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