An ammeter gives full scale deflection when current of 1.0 A is passed in it. To convert it into 10 A range ammeter, the ratio of its resistance and the shunt resistance will be

Options

A. `1 :9`
B. `1:10`
C. `1:11`
D. `9:1`

Answer

To solve the problem of converting an ammeter that gives full-scale deflection at 1.0 A into one that can measure up to 10 A, we need to determine the ratio of the resistance of the ammeter (R_g) to the shunt resistance (R_s). Here’s how we can derive that step by step:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The ammeter (or galvanometer) gives full-scale deflection at a current \( I_g = 1 \, \text{A} \).
- We want to convert it to measure up to \( I = 10 \, \text{A} \).
- A shunt resistor \( R_s \) is used in parallel with the ammeter to allow most of the current to bypass the ammeter.

2. **Current Distribution**:
- The total current \( I \) flowing through the circuit is the sum of the current through the ammeter \( I_g \) and the current through the shunt \( I_s \):
\[ I = I_g + I_s \]
- Rearranging gives:
\[ I_s = I - I_g \]

3. **Using Ohm's Law**:
- The voltage across the ammeter (galvanometer) and the shunt resistor must be the same since they are in parallel:
\[ V_g = V_s \]
- Using Ohm's law, we can express the voltages:
\[ V_g = I_g \cdot R_g \quad \text{and} \quad V_s = I_s \cdot R_s \]
- Setting these equal gives:
\[ I_g \cdot R_g = I_s \cdot R_s \]

4. **Substituting for \( I_s \)**:
- From the previous step, we substitute \( I_s \):
\[ I_g \cdot R_g = (I - I_g) \cdot R_s \]

5. **Rearranging to Find the Ratio**:
- Rearranging the equation to find the ratio \( \frac{R_g}{R_s} \):
\[ \frac{R_g}{R_s} = \frac{I - I_g}{I_g} \]

6. **Substituting Known Values**:
- We know \( I = 10 \, \text{A} \) and \( I_g = 1 \, \text{A} \):
\[ \frac{R_g}{R_s} = \frac{10 - 1}{1} = \frac{9}{1} \]

7. **Final Result**:
- Therefore, the ratio of the resistance of the ammeter to the shunt resistance is:
\[ \frac{R_g}{R_s} = 9:1 \]