A moving coil galvanometer has a resistance of `990Omega.` in order to send only `10%` of the main currect through this galvanometer, the resistance of the required shunt is

Options

A. `0.9Omega`
B. `100Omega`
C. `405Omega`
D. `90Omega`

Explanation

To find the resistance of the required shunt for the moving coil galvanometer, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Values**:
- Resistance of the galvanometer, \( R_g = 990 \, \Omega \)
- The galvanometer should carry only \( 10\% \) of the main current.

2. **Define Variables**:
- Let the total current flowing through the circuit be \( I \).
- The current through the galvanometer, \( I_g = 0.1I \) (which is \( 10\% \) of \( I \)).
- The current through the shunt, \( I_s = 0.9I \) (which is the remaining \( 90\% \) of \( I \)).

3. **Use Ohm's Law**:
- The potential drop across the galvanometer and the shunt is the same since they are in parallel.
- According to Ohm's law, we have:
\[ V = I_g R_g = I_s R_s \]
- Substituting the values of \( I_g \) and \( I_s \):
\[ 0.1I \cdot 990 = 0.9I \cdot R_s \]

4. **Simplify the Equation**:
- Cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ 0.1 \cdot 990 = 0.9 R_s \]
- This simplifies to:
\[ 99 = 0.9 R_s \]

5. **Solve for \( R_s \)**:
- Rearranging gives:
\[ R_s = \frac{99}{0.9} = 110 \, \Omega \]

### Final Answer
The resistance of the required shunt is \( R_s = 110 \, \Omega \).