A galvanometer has resistance `G` and Current `I_(g)` produces full scale deflection. `S_(1)` is the value of the shunt which converts it into an ammeter of range `0-I` and `S_(2)` is the value of shunt for the range `0-2I`. The ratio of `S_(1)` and `S_(2)` is

Options

A. 1
B. 2
C. `(1)/(2)((i-i_(a))/(2iu-i_(a)))`
D. ` ((2i-i_(a))/(i-i_(a)))`

Answer

To find the ratio of the shunt resistances \( S_1 \) and \( S_2 \) that convert a galvanometer into an ammeter for different ranges, we can follow these steps:

### Step 1: Understand the setup
A galvanometer has a resistance \( G \) and a current \( I_g \) that produces full-scale deflection. We need to find the shunt resistances \( S_1 \) for a range of \( 0 \) to \( I \) and \( S_2 \) for a range of \( 0 \) to \( 2I \).

### Step 2: Write the equations for shunt resistances
For the first shunt \( S_1 \):
- The total current \( I \) splits into the galvanometer and the shunt. Therefore, the current through the galvanometer is \( I_g \) and the current through the shunt is \( I - I_g \).
- Using Ohm's law, we can write the equation for the galvanometer:
\[ I_g \cdot G = (I - I_g) \cdot S_1 \tag{1} \]

For the second shunt \( S_2 \):
- The total current is \( 2I \), with \( I_g \) through the galvanometer and \( 2I - I_g \) through the shunt.
- The equation for the galvanometer becomes:
\[ I_g \cdot G = (2I - I_g) \cdot S_2 \tag{2} \]

### Step 3: Solve for \( S_1 \) and \( S_2 \)
From equation (1):
\[ S_1 = \frac{I_g \cdot G}{I - I_g} \]

From equation (2):
\[ S_2 = \frac{I_g \cdot G}{2I - I_g} \]

### Step 4: Find the ratio \( \frac{S_1}{S_2} \)
Now, we can find the ratio of \( S_1 \) to \( S_2 \):
\[ \frac{S_1}{S_2} = \frac{\frac{I_g \cdot G}{I - I_g}}{\frac{I_g \cdot G}{2I - I_g}} = \frac{2I - I_g}{I - I_g} \]

### Step 5: Simplify the ratio
Thus, the ratio of the shunt resistances is:
\[ \frac{S_1}{S_2} = \frac{2I - I_g}{I - I_g} \]

### Conclusion
The final answer for the ratio of \( S_1 \) to \( S_2 \) is:
\[ \frac{S_1}{S_2} = \frac{2I - I_g}{I - I_g} \]