Question

A moving coil galvanometer is converted into an ammeter of range 0 to 5mA. The galvanometer resistance is 90Ω and the shunt resistance has a value of 10Ω. If there are 50 divisions in the galvanometer-turnedammeter on either sides of zero, its current sensitivity is

• A. 1 × 10⁵ A/div
• B. 2 × 10⁴ A/div
• C. 1 × 10⁵ div/A
• D. 2 × 10⁴ div/A

Answer 1

The Correct Option is C

Given Information: 
Galvanometer resistance, \(G = 90\,\Omega\)
Shunt resistance, \(S = 10\,\Omega\)
Ammeter range, \(I = 5\,mA = 5 \times 10^{-3}\,A\)
Number of divisions on one side = \(50\) divisions

Step-by-Step Explanation:

Step 1: Find the galvanometer current (\(I_g\)) for full-scale deflection.

Using the formula for shunt resistance in galvanometer conversion:

\[ I_g \cdot G = (I - I_g) \cdot S \]

Substitute given values:

\[ I_g \times 90 = (5\times10^{-3} - I_g)\times10 \]

Solving for \(I_g\):

\[ 90\,I_g = 5\times10^{-2} - 10\,I_g \]

Bring terms involving \(I_g\) together:

\[ 90\,I_g + 10\,I_g = 5\times10^{-2} \]

\[ 100\,I_g = 5\times10^{-2} \]

Thus,

\[ I_g = \frac{5\times10^{-2}}{100} = 5\times10^{-4}\,A \]

Step 2: Find current sensitivity (divisions per ampere):

Number of divisions for full-scale deflection = \(50\) divisions

Thus, current sensitivity (\(S_I\)) is:

\[ S_I = \frac{\text{Number of divisions}}{I_g} = \frac{50}{5\times10^{-4}} = 1\times10^{5}\,\text{div/A} \]

Final Conclusion:
Current sensitivity is \(1\times10^{5}\,\text{divisions per ampere}\).

Answer 2

The problem asks for the current sensitivity of a moving coil galvanometer after it has been converted into an ammeter.

Given:

• Galvanometer resistance, \(G = 90 \, \Omega\)
• Shunt resistance, \(S = 10 \, \Omega\)
• Maximum current of the ammeter (range), \(I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A}\)
• Number of divisions on the galvanometer scale for full deflection, \(N = 50 \, \text{div}\)

Current sensitivity is defined as the deflection (in divisions) per unit current flowing through the galvanometer. \[ \text{Current Sensitivity} = \frac{\text{Number of divisions}}{\text{Current through galvanometer for full deflection}} = \frac{N}{I_g} \] where \(I_g\) is the current flowing through the galvanometer that causes full-scale deflection (50 divisions).

When the ammeter measures its maximum current \(I\), the current \(I_g\) flows through the galvanometer, and the remaining current \(I_s = I - I_g\) flows through the shunt resistance \(S\). Since the galvanometer and the shunt are in parallel, the potential difference across them is equal: \[ V_g = V_s \] \[ I_g \times G = I_s \times S \] \[ I_g \times G = (I - I_g) \times S \]

Now, we solve for \(I_g\): \[ I_g G = IS - I_g S \] \[ I_g G + I_g S = IS \] \[ I_g (G + S) = IS \] \[ I_g = \frac{S}{G + S} \times I \]

Substitute the given values: \[ I_g = \frac{10 \, \Omega}{90 \, \Omega + 10 \, \Omega} \times (5 \times 10^{-3} \, \text{A}) \] \[ I_g = \frac{10}{100} \times (5 \times 10^{-3} \, \text{A}) \] \[ I_g = 0.1 \times (5 \times 10^{-3} \, \text{A}) \] \[ I_g = 0.5 \times 10^{-3} \, \text{A} \]

This current \(I_g = 0.5 \times 10^{-3} \, \text{A}\) causes a full-scale deflection of \(N = 50\) divisions.

Now, calculate the current sensitivity: \[ \text{Current Sensitivity} = \frac{N}{I_g} = \frac{50 \, \text{div}}{0.5 \times 10^{-3} \, \text{A}} \] \[ \text{Current Sensitivity} = \frac{50}{0.5} \times 10^3 \, \text{div/A} \] \[ \text{Current Sensitivity} = 100 \times 10^3 \, \text{div/A} \] \[ \text{Current Sensitivity} = 1 \times 10^5 \, \text{div/A} \]

Our calculated value and units (\(1 \times 10^5 \, \text{div/A}\)) match option (C).