Question

A galvanometer with resistance \(100 \Omega\) is converted into an ammeter with a resistance of \(0.1\). The galvanometer shows full scale deflection with current of \(100 \mu \mathrm{A}\). Then what will be the minimum current in the circuit for full scale deflection of galvanometer ? (A) \(0.1001 \mathrm{~mA}\) (B) \(100.1 \mathrm{~mA}\) (C) \(1000.1 \mathrm{~mA}\) (D) \(1.001 \mathrm{~mA}\)

Short answer

The minimum current in the circuit for full scale deflection of the galvanometer is \(100.1 \: mA\), which is Option (B).

Step-by-step solution

Identify the given values

The given values are: - Resistance of the galvanometer (\(G\)): 100 Ω - Resistance of the ammeter (\(R\)): 0.1 Ω - Current in the galvanometer for full scale deflection (\(I_{G}\)): 100 µA

Set up the equation

We will use the formula for converting a galvanometer into an ammeter: \(I_{A} = (G + R) \cdot I_{G}\)

Plug in the given values

Let's plug the values into the equation: \(I_{A} = (100 + 0.1) \cdot 100 \mu A\)

Solve for \(I_{A}\)

Now, let's solve the equation to find the minimum current in the circuit for full scale deflection of the galvanometer: \(I_{A} = (100.1) \cdot 100 \mu A = 10010 \mu A = 100.1 \: mA\) The minimum current in the circuit for full scale deflection of the galvanometer is 100.1 mA, which is Option (B).