In a transformer, number of turns in the primary coil are 140 and that in the secondry coil are 280. If current i primary coil is 4A, then that in the secondary coil is

Options

A. 4A
B. 2A
C. 6A
D. 10A

Explanation

To solve the problem, we will use the relationship between the currents in the primary and secondary coils of a transformer, which is given by the formula:

\[ \frac{I_S}{I_P} = \frac{N_P}{N_S} \]

Where:
- \(I_S\) = Current in the secondary coil
- \(I_P\) = Current in the primary coil
- \(N_P\) = Number of turns in the primary coil
- \(N_S\) = Number of turns in the secondary coil

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns in the primary coil, \(N_P = 140\)
- Number of turns in the secondary coil, \(N_S = 280\)
- Current in the primary coil, \(I_P = 4 \, A\)

2. **Set up the equation using the transformer formula:**
\[ \frac{I_S}{I_P} = \frac{N_P}{N_S} \]

3. **Substitute the known values into the equation:**
\[ \frac{I_S}{4} = \frac{140}{280} \]

4. **Simplify the fraction on the right side:**
\[ \frac{140}{280} = \frac{1}{2} \]

5. **Now the equation looks like this:**
\[ \frac{I_S}{4} = \frac{1}{2} \]

6. **Cross-multiply to solve for \(I_S\):**
\[ I_S = 4 \times \frac{1}{2} \]

7. **Calculate \(I_S\):**
\[ I_S = 2 \, A \]

### Final Answer:
The current in the secondary coil \(I_S\) is **2 A**.