An ideal transformer is used to step up an alternating emf of `220V` to `4.4kV` to transmit `6.6kW` of power The current rating of the secondary is

Options

A. `30A`
B. `3A`
C. `1.5A`
D. `1A`

Explanation

To find the current rating of the secondary in an ideal transformer, we can use the relationship between power, voltage, and current. The power in an electrical circuit can be expressed as:

\[ P = V \times I \]

where \( P \) is the power, \( V \) is the voltage, and \( I \) is the current.

### Step-by-Step Solution:

1. **Identify the given values:**
- Primary voltage (\( V_p \)) = 220 V
- Secondary voltage (\( V_s \)) = 4.4 kV = 4400 V
- Power (\( P \)) = 6.6 kW = 6600 W

2. **Use the power formula for the secondary side:**
Since we are interested in the current on the secondary side, we can rearrange the power formula to solve for current:

\[ I_s = \frac{P}{V_s} \]

where \( I_s \) is the secondary current.

3. **Substitute the known values into the formula:**
\[ I_s = \frac{6600 \, \text{W}}{4400 \, \text{V}} \]

4. **Calculate the secondary current:**
\[ I_s = \frac{6600}{4400} = 1.5 \, \text{A} \]

### Final Answer:
The current rating of the secondary is **1.5 A**.