A galvanometer has a resistance of `98Omega`. If 2% of the main current is to be passed through the meter what should be the value of the shunt?

Explanation

To solve the problem of finding the value of the shunt resistance \( R_s \) for a galvanometer with a resistance of \( R_g = 98 \, \Omega \), where 2% of the main current \( I \) is to be passed through the meter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Distribution**:
- The galvanometer passes 2% of the main current \( I \). This means the current through the galvanometer \( I_g \) is:
\[
I_g = 0.02 I
...