To solve this problem, we need to calculate the shunt resistance required when 10% of the main current is passed through the moving coil galvanometer. The galvanometer is given to have a resistance of \( 99 \, \Omega \).
Concept: In an electrical circuit, a shunt resistor is used in parallel with the galvanometer to allow a fraction of the total current to pass through the galvanometer itself. The remaining current, which forms the larger part of the main current, bypasses the galvanometer through this shunt.
Step-by-step Solution:
- The current that should pass through the galvanometer is 10% of the total current, denoted as \( I_g \). Hence, the remaining 90% of the current \( (I - I_g) \) should pass through the shunt resistor, denoted as \( I_s \).
- The voltage across the galvanometer and shunt resistor must be the same because they are in parallel, which means: \(I_g \times R_g = I_s \times R_s\) where \( R_g = 99 \, \Omega \) is the resistance of the galvanometer and \( R_s \) is the shunt resistance.
- From the current relationship, \( I_g = \frac{I}{10} \) and \( I_s = I - I_g = \frac{9I}{10} \).
- Substitute these values into the equation: \(\frac{I}{10} \times 99 = \frac{9I}{10} \times R_s\)
- Cancel \( I \) from both sides and simplify the equation: \(99 = 9 \times R_s\)
- Solve for \( R_s \): \(R_s = \frac{99}{9} = 11 \, \Omega\)
Therefore, the required shunt resistance is 11 \( \Omega \).
Conclusion: The correct answer is:
11 \( \Omega \)
, which matches the provided correct option in the choices.