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A moving coil galvanometer of resistance...

A moving coil galvanometer of resistance `100Omega` is used as an ammeter using a resistance `0.1 Omega`. The maximum diflection current in the galvanometer is `100 muA`. Find the minimum current in the circuit so that the ammeter shows maximum deflection

A

`100.1 mA`

B

`1000.1mA`

C

`10. 01mA`

D

`1.01mA`

Text Solution

Verified by Experts

The correct Answer is:
A
a. `I_(g)G = (I-I_g) S. Here, I_g = 100 xx 10^(-6)A`
`G= 100Omega, S = 0.1Omega`

`:. I = I_g (G/S+1) = 100 xx 10^(-6) (100/0.1 + 1)`
`=100 xx 10^(-6) xx 1000.1 = 100.01mA`.
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