An ammetre is obtained by shunting a `30 Omega` galvanometer with a `30 Omega` resistance. What additional shunt should be connected across it to double the range ?

Options

A. `15 Omega`
B. `10 Omega`
C. `5 Omega`
D. none of these

Explanation

To solve the problem, we need to determine the additional shunt resistance required to double the range of an ammeter that is already shunted with a 30-ohm galvanometer and a 30-ohm shunt resistance.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of galvanometer, \( R_G = 30 \, \Omega \)
- Initial shunt resistance, \( S = 30 \, \Omega \)

2. **Understand the Current Relationships:**
- Let \( I \) be the total current through the ammeter.
- The current through the galvanometer is \( I_G \).
- The current through the shunt is \( I - I_G \).

3. **Apply the Voltage Equality:**
- The voltage across the galvanometer and the shunt must be equal:
\[ I_G \cdot R_G = (I - I_G) \cdot S \]
- Substituting the known values:
\[ I_G \cdot 30 = (I - I_G) \cdot 30 \]
- Simplifying this gives:
\[ I_G = I - I_G \implies 2I_G = I \implies I = 2I_G \]

4. **Determine New Current for Doubling the Range:**
- To double the range of the ammeter, the new total current \( I' \) should be:
\[ I' = 2I = 2(2I_G) = 4I_G \]

5. **Set Up the New Voltage Equality:**
- For the new situation with the new shunt resistance \( S' \):
\[ I_G \cdot R_G = (I' - I_G) \cdot S' \]
- Substituting \( I' = 4I_G \):
\[ I_G \cdot 30 = (4I_G - I_G) \cdot S' \implies 30I_G = 3I_G \cdot S' \]
- Dividing both sides by \( I_G \) (assuming \( I_G \neq 0 \)):
\[ 30 = 3S' \implies S' = 10 \, \Omega \]

6. **Calculate the Additional Shunt Resistance:**
- The initial shunt resistance \( S = 30 \, \Omega \) and the new required shunt resistance \( S' = 10 \, \Omega \).
- The additional shunt resistance \( X \) can be found using the formula for resistances in parallel:
\[ \frac{1}{X} = \frac{1}{S'} - \frac{1}{S} \]
- Substituting the values:
\[ \frac{1}{X} = \frac{1}{10} - \frac{1}{30} \]
- Finding a common denominator (30):
\[ \frac{1}{X} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \]
- Therefore, \( X = 15 \, \Omega \).

### Final Answer:
The additional shunt resistance that should be connected across the galvanometer to double the range is \( \boxed{15 \, \Omega} \).