What is the value of the shunt that passes `10%` of the main current through a galvenomenter of `99 Omega`?

Explanation

To find the value of the shunt resistance that allows 10% of the main current to pass through a galvanometer with a resistance of 99 ohms, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Distribution**:
- Let the total main current be \( I \).
- According to the problem, 10% of this current passes through the galvanometer. Therefore, the current through the galvanometer \( I_G = 0.1 I \).
- The remaining current, which passes through the shunt resistance \( S \), is \( I_S = 0.9 I \).
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