An ammeter has a resistance of G ohm and a range of 'i' ampere. The value of resistance used in parallel, to convert into an ammeter of range 'ni' ampere is

Options

A. nG
B. (n-1)G
C. G/n
D. G/n-1

Answer

To solve the problem of converting an ammeter with a resistance of G ohms and a range of 'i' amperes into an ammeter with a range of 'ni' amperes using a parallel resistance, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ammeter Configuration**:
- An ammeter measures current and has an internal resistance (G ohms).
- The original range of the ammeter is 'i' amperes.
- We want to modify this ammeter to measure up to 'ni' amperes by adding a resistance (R) in parallel.

2. **Applying Kirchhoff's Current Law**:
- When we connect a resistance R in parallel with the ammeter, the total current (I_total) flowing into the parallel combination will be 'ni' amperes.
- The current through the ammeter (I) will still be 'i' amperes.
- Therefore, the current flowing through the parallel resistance (I_R) can be expressed as:
\[ I_R = I_{total} - I = ni - i \]

3. **Using Ohm's Law**:
- The potential difference across the ammeter and the parallel resistance is the same.
- The potential difference (V) across the ammeter can be expressed using Ohm's Law:
\[ V = I \cdot G = i \cdot G \]
- The potential difference across the parallel resistance can also be expressed as:
\[ V = I_R \cdot R = (ni - i) \cdot R \]

4. **Setting the Equations Equal**:
- Since both expressions for V are equal, we can set them equal to each other:
\[ i \cdot G = (ni - i) \cdot R \]

5. **Rearranging the Equation**:
- Rearranging the equation to solve for R gives us:
\[ R = \frac{i \cdot G}{ni - i} \]

6. **Simplifying the Expression**:
- We can factor out 'i' from the denominator:
\[ R = \frac{i \cdot G}{i(n - 1)} = \frac{G}{n - 1} \]

### Final Result:
The value of the resistance used in parallel to convert the ammeter into a range of 'ni' amperes is:
\[ R = \frac{G}{n - 1} \]