A power transformer is used to step up an alternating e.m.f. of `220 V` to `11kv` to trannsmit `4.4 kW` of power. If the primary coil has `1000` turns, what is the current rating of the secondary?(Assume `100%` efficiency for the transformer)

Options

A. 4A
B. 0.4A
C. 0.04A
D. 0.2A

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Primary voltage (V_p) = 220 V
- Secondary voltage (V_s) = 11 kV = 11,000 V
- Power transmitted (P) = 4.4 kW = 4,400 W
- Number of turns in the primary coil (N_p) = 1000 turns

### Step 2: Use the power formula for the secondary side
Since we assume 100% efficiency for the transformer, the power in the primary (P_p) is equal to the power in the secondary (P_s):
\[ P_p = P_s = 4,400 \text{ W} \]

### Step 3: Calculate the current in the secondary coil
The power in the secondary coil can be expressed in terms of voltage and current:
\[ P_s = V_s \times I_s \]
Where:
- \( I_s \) = current in the secondary coil

Rearranging the formula to find \( I_s \):
\[ I_s = \frac{P_s}{V_s} \]

### Step 4: Substitute the values
Now, substitute the known values into the equation:
\[ I_s = \frac{4,400 \text{ W}}{11,000 \text{ V}} \]

### Step 5: Perform the calculation
Calculating \( I_s \):
\[ I_s = \frac{4,400}{11,000} = 0.4 \text{ A} \]

### Step 6: Conclusion
The current rating of the secondary coil is:
\[ I_s = 0.4 \text{ A} \]

### Final Answer
The current rating of the secondary is **0.4 A**.

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