Question

A moving coil galvanometer when shunted with \(2\,\Omega\) resistance gives a full scale deflection for a current of \(500\,\text{mA}\). When a resistance of \(470\,\Omega\) is connected in series it gives a full scale deflection for \(10\,\text{V}\) potential applied on it. The value of resistance of galvanometer coil is _____ \(\Omega\).

Answer 1

Correct Answer: 50

Step 1: Understanding the Question:
We have two circuit configurations for a galvanometer (Ammeter and Voltmeter) and need to find its internal coil resistance \(G\).
Step 2: Key Formula or Approach:
1. Ammeter case: \(I_g G = (I - I_g) S\).
2. Voltmeter case: \(V = I_g (G + R)\).
Step 3: Detailed Explanation:
Let \(I_g\) be the full-scale current of the galvanometer and \(G\) be its resistance.
1. From Ammeter case: \(S = 2 \text{ }\Omega, I = 0.5 \text{ A}\).
\[ I_g G = (0.5 - I_g) \times 2 = 1 - 2I_g \Rightarrow I_g (G + 2) = 1 \dots (i) \]
2. From Voltmeter case: \(V = 10 \text{ V}, R = 470 \text{ }\Omega\).
\[ 10 = I_g (G + 470) \dots (ii) \]
3. Dividing equation (ii) by (i):
\[ \frac{10}{1} = \frac{I_g (G + 470)}{I_g (G + 2)} \]
\[ 10(G + 2) = G + 470 \]
\[ 10G + 20 = G + 470 \Rightarrow 9G = 450 \Rightarrow G = 50 \text{ }\Omega \]
Step 4: Final Answer:
The resistance of the galvanometer coil is 50 \(\Omega\).