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

Options

A. 100.1 mA
B. 1000.1 mA
C. 10.01 mA
D. 1.01 mA

Answer

A

Explanation

(a) KEY CONCEPT :
`I_g G = (I-I_g)S`
Here, `I_g = 100 xx 10^-6 A , G = 100 Omega , S = 0.1 Omega`

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