A galvanometer of resistance G can measure 1A current. if a shunt S in used to convert it into an ammeter to measure 10A current ratio of `(S)/(G)` is

Options

A. `(9)/(1)`
B. `(1)/(9)`
C. 10
D. `(1)/(10)`

Explanation

To solve the problem, we need to find the ratio of the shunt resistance \( S \) to the galvanometer resistance \( G \) when a galvanometer can measure a maximum of 1 A and is converted to measure up to 10 A using the shunt.

### Step-by-Step Solution:

1. **Understand the Current Distribution:**
- Let \( I \) be the total current flowing through the circuit, which is given as \( I = 10 \, \text{A} \).
- Let \( I_G \) be the current flowing through the galvanometer, which can measure a maximum of \( I_G = 1 \, \text{A} \).
- The current flowing through the shunt \( S \) can be expressed as:
\[ I_S = I - I_G = 10 \, \text{A} - 1 \, \text{A} = 9 \, \text{A} \]

2. **Apply the Current Division Rule:**
- The voltage across the galvanometer \( V_G \) can be expressed as:
\[ V_G = I_G \cdot G \]
- The voltage across the shunt \( S \) can be expressed as:
\[ V_S = I_S \cdot S \]
- Since both resistances are in parallel, the voltages across them are equal:
\[ I_G \cdot G = I_S \cdot S \]

3. **Substitute the Known Values:**
- Substitute \( I_G = 1 \, \text{A} \) and \( I_S = 9 \, \text{A} \):
\[ 1 \cdot G = 9 \cdot S \]

4. **Rearranging the Equation:**
- Rearranging gives us:
\[ S = \frac{G}{9} \]

5. **Finding the Ratio \( \frac{S}{G} \):**
- To find the ratio \( \frac{S}{G} \):
\[ \frac{S}{G} = \frac{1}{9} \]

### Final Answer:
The ratio \( \frac{S}{G} \) is \( \frac{1}{9} \).

---