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Physics Question on Moving charges and magnetism

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

A

1.01 mA

B

1.0001 mA

C

10.01 mA

D

100.1 mA

Answer

100.1 mA

Explanation

Solution

Given, G=100 ΩG = 100 \, \Omega
S=0.1 Ω.Ig=100 μAS = 0.1 \, \Omega . I_g = 100 \, \mu A
S=GIgI−Ig=100×100×10−6(I−100×10−6)S = \frac{G I_g}{I - I_g} = 100 \times \frac{100 \times 10^{-6}}{(I - 100 \times 10^{-6})}
∴ 0.1=10−2I−10−4⇒I=100.1mA\therefore \, 0.1 = \frac{10^{-2}}{I - 10^{-4}} \Rightarrow I = 100.1 {mA}