The resistance of an ammeter is `13 Omega` and its scale is graduated fro a current upto `100 A`. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto `750 A` by this meter. The value of shunt resistance is

Options

A. `20 Omega`
B. `2 Omega`
C. `0.2 Omega`
D. `2 k Omega`

Explanation

To solve the problem, we need to find the value of the shunt resistance (S) that allows the ammeter to measure currents up to 750 A when it has an internal resistance of 13 Ω and can measure up to 100 A without the shunt.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Internal resistance of ammeter (R) = 13 Ω
- Maximum current without shunt (I_max) = 100 A
- Maximum current with shunt (I_total) = 750 A

2. **Calculate the Current through the Ammeter:**
- The current through the ammeter when measuring 750 A is 100 A.
- Therefore, the current through the shunt (I_shunt) can be calculated as:
\[ I_{shunt} = I_{total} - I_{max} = 750 A - 100 A = 650 A \]

3. **Apply the Principle of Parallel Circuits:**
- Since the ammeter and shunt are in parallel, the voltage across both must be the same.
- The voltage across the ammeter (V) can be expressed as:
\[ V = I_{max} \times R = 100 A \times 13 Ω = 1300 V \]

4. **Set Up the Equation for the Shunt:**
- The voltage across the shunt can also be expressed as:
\[ V = I_{shunt} \times S \]
- Substituting the values we have:
\[ 1300 V = 650 A \times S \]

5. **Solve for the Shunt Resistance (S):**
- Rearranging the equation gives:
\[ S = \frac{1300 V}{650 A} = 2 Ω \]

6. **Conclusion:**
- The value of the shunt resistance is \( S = 2 Ω \).

### Final Answer:
The value of the shunt resistance is \( 2 Ω \).

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