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. `1.0001 mA`
C. `10.01 mA`
D. `1.01 mA`

Answer

A

Explanation

The current in the galvanometer `I_g=(S/(S+G))I` For maximum deflection, maximum current of 100muA must pass through the ammeter. The required main current is given by`I=I_g ((S+G)/S)=(100xx10^(-6)(100+0.1)/0.1)=(100.1) xx 10^(-3)A.(100.1)mA`.