Question

Current sensitivities of two galvanometers \( G_1 \) and \( G_2 \) of resistances 100 Ω and 50 Ω are \( 10^8 \) div/A and \( 0.5 \times 10^5 \) div/A respectively. The galvanometer in which the voltage sensitivity is more is:

• A. Same in both galvanometers
• B. More in \( G_2 \)
• C. Zero
• D. More in \( G_1 \)

Hint:

Voltage sensitivity is inversely proportional to resistance. A galvanometer with higher resistance has greater voltage sensitivity.

Answer 1

The Correct Option is D

To determine which galvanometer has higher voltage sensitivity, we need to calculate the voltage sensitivity for both galvanometers and compare them. Given: 
- Galvanometer \( G_1 \): 
- Resistance, \( R_1 = 100 \, \Omega \) 
- Current sensitivity, \( S_{I1} = 10^8 \, \text{div/A} \) 
- Galvanometer \( G_2 \): 
- Resistance, \( R_2 = 50 \, \Omega \) 
- Current sensitivity, \( S_{I2} = 0.5 \times 10^5 \, \text{div/A} \) 

Step 1: Calculate Voltage Sensitivity Voltage sensitivity \( S_V \) is given by: \[ S_V = \frac{S_I}{R} \] where \( S_I \) is the current sensitivity and \( R \) is the resistance of the galvanometer. 

Step 2: Calculate Voltage Sensitivity for \( G_1 \) \[ S_{V1} = \frac{S_{I1}}{R_1} = \frac{10^8}{100} = 10^6 \, \text{div/V} \] 

Step 3: Calculate Voltage Sensitivity for \( G_2 \) \[ S_{V2} = \frac{S_{I2}}{R_2} = \frac{0.5 \times 10^5}{50} = 10^3 \, \text{div/V} \] 

Step 4: Compare Voltage Sensitivities \[ S_{V1} = 10^6 \, \text{div/V} \quad \text{and} \quad S_{V2} = 10^3 \, \text{div/V} \] Clearly, \( S_{V1}>S_{V2} \). 

Final Answer: \[ \boxed{\text{More in } G_1} \] This corresponds to option (4).