If a shunt is to be applied to a galvanometer of resistance `50Omega` so that only `5%` of total current passes through the galvanometer. The resistance of shunt should be

Options

A. `1.63Omega`
B. `4.2Omega`
C. `3.5Omega`
D. `2.63Omega`

Explanation

To solve the problem of finding the resistance of the shunt to be applied to a galvanometer with a resistance of \(50 \, \Omega\) such that only \(5\%\) of the total current passes through the galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Division**:
Let \(I\) be the total current flowing through the circuit, and \(I_G\) be the current passing through the galvanometer. According to the problem, \(I_G\) is \(5\%\) of \(I\):
\[ I_G = 0.05 I \]

2. **Calculate the Ratio of Currents**:
From the above equation, we can express the ratio of the total current to the galvanometer current:
\[ \frac{I_G}{I} = \frac{5}{100} = \frac{1}{20} \]
This implies:
\[ \frac{I}{I_G} = 20 \]

3. **Use the Shunt Resistance Formula**:
The formula for the resistance of the shunt \(S\) is given by:
\[ S = \frac{G}{\frac{I}{I_G} - 1} \]
where \(G\) is the resistance of the galvanometer.

4. **Substitute the Known Values**:
Here, \(G = 50 \, \Omega\) and \(\frac{I}{I_G} = 20\):
\[ S = \frac{50}{20 - 1} = \frac{50}{19} \]

5. **Calculate the Resistance of the Shunt**:
Now, we perform the division:
\[ S \approx 2.63 \, \Omega \]

6. **Final Answer**:
Thus, the resistance of the shunt should be approximately \(2.63 \, \Omega\).

### Summary:
The resistance of the shunt required is \(2.63 \, \Omega\).