In an ammeter `5%` of the main current is passing through the galvanometer. If the resistance of galvanometer is G, then the resistance of shunt S will be

Options

A. G/19
B. G/5
C. 5G
D. 19 G

Explanation

To find the resistance of the shunt \( S \) in an ammeter where \( 5\% \) of the main current is passing through the galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Current Distribution**:
- Let the total current from the source be \( I \).
- The current passing through the galvanometer is \( \frac{5}{100}I = 0.05I \).
- Therefore, the current passing through the shunt \( S \) is \( I - 0.05I = 0.95I \).

2. **Voltage Drop Across the Galvanometer**:
- The resistance of the galvanometer is \( G \).
- The voltage drop across the galvanometer can be calculated using Ohm's law:
\[ V_G = I_G \cdot G = 0.05I \cdot G \]

3. **Voltage Drop Across the Shunt**:
- Let the resistance of the shunt be \( S \).
- The voltage drop across the shunt can also be expressed as:
\[ V_S = I_S \cdot S = 0.95I \cdot S \]

4. **Equating Voltage Drops**:
- Since the galvanometer and the shunt are in parallel, the voltage drops across both must be equal:
\[ V_G = V_S \]
- Therefore, we can set the equations from steps 2 and 3 equal to each other:
\[ 0.05I \cdot G = 0.95I \cdot S \]

5. **Simplifying the Equation**:
- We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ 0.05G = 0.95S \]

6. **Solving for Shunt Resistance \( S \)**:
- Rearranging the equation to solve for \( S \):
\[ S = \frac{0.05G}{0.95} \]
- Simplifying further:
\[ S = \frac{5G}{95} = \frac{G}{19} \]

### Final Answer:
The resistance of the shunt \( S \) is:
\[ S = \frac{G}{19} \]