The shunt required for 10% of main current to be sent through the moving coil galvanometer of resistance `99 Omega` will be-

Options

A. `9.9Omega`
B. `10Omega`
C. `11Omega`
D. `9Omega`

Answer

To find the shunt resistance required for 10% of the main current to be sent through a moving coil galvanometer with a resistance of 99 ohms, we can follow these steps:

### Step 1: Understand the Current Distribution
Given that 10% of the main current (I) flows through the galvanometer, we can express this mathematically:
- Current through galvanometer, \( I_g = \frac{I}{10} \)
- Current through the shunt, \( I_s = I - I_g = I - \frac{I}{10} = \frac{9I}{10} \)

### Step 2: Apply Ohm's Law
The potential difference (V) across the galvanometer and the shunt must be the same when they are connected in parallel. According to Ohm's law, we can express the potential difference across both components:
- For the galvanometer:
\[ V = I_g \cdot R_g = \left(\frac{I}{10}\right) \cdot 99 \]
- For the shunt:
\[ V = I_s \cdot R_s = \left(\frac{9I}{10}\right) \cdot R_s \]

### Step 3: Set the Equations Equal
Since the potential differences are equal, we can set the two equations from Step 2 equal to each other:
\[ \left(\frac{I}{10}\right) \cdot 99 = \left(\frac{9I}{10}\right) \cdot R_s \]

### Step 4: Simplify the Equation
We can cancel \( \frac{I}{10} \) from both sides (assuming \( I \neq 0 \)):
\[ 99 = 9R_s \]

### Step 5: Solve for Shunt Resistance
Now, we can solve for \( R_s \):
\[ R_s = \frac{99}{9} = 11 \, \Omega \]

### Conclusion
The shunt resistance required is \( 11 \, \Omega \).

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