Shunt required in an ammeter of resistance R to decreases its deflection from 30 A to 10 A is

Options

A. R/4
B. R/3
C. R/2
D. R

Answer

To solve the problem of finding the shunt required in an ammeter of resistance \( R \) to decrease its deflection from 30 A to 10 A, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Circuit**:
- We have an ammeter with a maximum current rating of 30 A and a resistance \( R \).
- We want to measure only 10 A when the total current flowing is 30 A.
- A shunt resistor \( R_s \) is connected in parallel with the ammeter.

2. **Applying Kirchhoff's Law**:
- The potential difference across the ammeter \( V_A \) is equal to the potential difference across the shunt \( V_s \).
- Therefore, we can write:
\[ V_A = I_A \cdot R_A \quad \text{and} \quad V_s = I_s \cdot R_s \]
- Since \( V_A = V_s \), we have:
\[ I_A \cdot R = I_s \cdot R_s \]

3. **Defining the Currents**:
- Let \( I \) be the total current flowing through the circuit.
- The current through the ammeter \( I_A \) is 10 A (the desired reading).
- The current through the shunt \( I_s \) can be expressed as:
\[ I_s = I - I_A = 30 A - 10 A = 20 A \]

4. **Substituting Values**:
- Now substituting \( I_A \) and \( I_s \) into the equation:
\[ 10 A \cdot R = 20 A \cdot R_s \]

5. **Solving for Shunt Resistance**:
- Rearranging the equation to find \( R_s \):
\[ R_s = \frac{10 A \cdot R}{20 A} = \frac{R}{2} \]

6. **Conclusion**:
- The required shunt resistance \( R_s \) to decrease the deflection from 30 A to 10 A is:
\[ R_s = \frac{R}{2} \]