Question

For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :

• A. 1 \(\Omega\)
• B. 2 \(\Omega\)
• C. 4 \(\Omega\)
• D. 5 \(\Omega\)

Hint:

Pay close attention to the units. Current sensitivity is given in div/mA, and voltage is in mV. Converting everything to base SI units (Amperes, Volts) before applying Ohm's law helps prevent calculation errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the voltage for full-scale deflection, the total number of divisions, and the current sensitivity of a galvanometer. We need to find the resistance of the galvanometer.
Step 2: Key Formula or Approach:
1. Current for Full-Scale Deflection (\(I_g\)): We can find this using the current sensitivity and the total number of divisions.
Current Sensitivity \( = \frac{\text{Deflection (div)}}{\text{Current (mA)}} \)
2. Galvanometer Resistance (\(R_g\)): We can find this using Ohm's law, \( R_g = \frac{V_g}{I_g} \), where \(V_g\) is the voltage for full-scale deflection.
Step 3: Detailed Explanation:
Given values:
- Total divisions for full-scale deflection, \(\theta_{\text{max}} = 50\) divisions.
- Voltage for full-scale deflection, \(V_g = 50 \, \text{mV} = 50 \times 10^{-3} \, \text{V}\).
- Current sensitivity, \(S_i = 2 \, \text{div/mA}\).
First, calculate the current required for full-scale deflection (\(I_g\)).
The sensitivity tells us that a current of 1 mA produces a deflection of 2 divisions.
\[ I_g = \frac{\text{Total Divisions}}{\text{Current Sensitivity}} = \frac{50 \, \text{div}}{2 \, \text{div/mA}} = 25 \, \text{mA} \] Convert this current to Amperes:
\[ I_g = 25 \times 10^{-3} \, \text{A} \] Now, use Ohm's law to find the galvanometer resistance (\(R_g\)).
\[ R_g = \frac{V_g}{I_g} \] \[ R_g = \frac{50 \times 10^{-3} \, \text{V}}{25 \times 10^{-3} \, \text{A}} \] \[ R_g = 2 \, \Omega \] Step 4: Final Answer:
The resistance of the galvanometer is 2 \(\Omega\).