two galvanometers of resistance `100Omega and 50Omega`, are `10^(-8)A//"div and "2xx10^(-5)`A/div respectively. In which case the voltage sensitivity is more ?

Options

A. more in case I
B. more in case II
C. same in both cases
D. can.t say

Answer

To determine which galvanometer has more voltage sensitivity, we will calculate the voltage sensitivity for both galvanometers using the formula:

\[ V_s = I \times R \]

where \( V_s \) is the voltage sensitivity, \( I \) is the current per division, and \( R \) is the resistance of the galvanometer.

### Step 1: Calculate the voltage sensitivity of the first galvanometer

1. **Identify the values for the first galvanometer:**
- Resistance \( R_1 = 100 \, \Omega \)
- Current per division \( I_1 = 10^{-8} \, A/\text{div} \)

2. **Use the formula to calculate voltage sensitivity:**
\[ V_{s1} = I_1 \times R_1 = (10^{-8} \, A) \times (100 \, \Omega) \]
\[ V_{s1} = 10^{-6} \, V/\text{div} \]

### Step 2: Calculate the voltage sensitivity of the second galvanometer

1. **Identify the values for the second galvanometer:**
- Resistance \( R_2 = 50 \, \Omega \)
- Current per division \( I_2 = 2 \times 10^{-5} \, A/\text{div} \)

2. **Use the formula to calculate voltage sensitivity:**
\[ V_{s2} = I_2 \times R_2 = (2 \times 10^{-5} \, A) \times (50 \, \Omega) \]
\[ V_{s2} = 10^{-3} \, V/\text{div} \]

### Step 3: Compare the voltage sensitivities

- Voltage sensitivity of the first galvanometer: \( V_{s1} = 10^{-6} \, V/\text{div} \)
- Voltage sensitivity of the second galvanometer: \( V_{s2} = 10^{-3} \, V/\text{div} \)

### Conclusion

Since \( 10^{-3} \, V/\text{div} \) (the second galvanometer) is greater than \( 10^{-6} \, V/\text{div} \) (the first galvanometer), the second galvanometer has more voltage sensitivity.

### Final Answer

The second galvanometer (50 Ω, \( 2 \times 10^{-5} \, A/\text{div} \)) has more voltage sensitivity.

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