Question

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

Answer 1

Step 1: Given

Resistance of galvanometer: $G=100\Omega$

Shunt resistance: $S=0.1\Omega$

Current in galvanometer: $I_g=100\mu A=100\times {10}^{-6}A$

Step 2: Formula Used

$I=I_g\left(\frac{G}{S}+1\right)$

Step 3: Find the minimum current in the circuit for the ammeter to show maximum deflection.

Substitute the given values in the formula

$$\begin{gathered} I=100\times {10}^{-6}\left(\frac{100}{0.1}+1\right)A \\ =100\times {10}^{-6}\left(1000+1\right)A \\ =100\times {10}^{-6}\times 1001A \\ =100.1\times {10}^{-3}A \\ =100.1mA \end{gathered}$$

Hence, the minimum current in the circuit so that ammeter shows maximum deflection is $100.1mA$.