A power transmission line feeds input power at 2300 V a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is `90%`, the output current would be :

Options

A. 20 A
B. 50 A
C. 25 A
D. 45 A

Answer

To solve the problem, we need to find the output current of a step-down transformer given the input parameters. Let's break down the solution step by step.

### Step 1: Understand the given data
- Input voltage (\(V_i\)) = 2300 V
- Output voltage (\(V_o\)) = 230 V
- Primary current (\(I_i\)) = 5 A
- Efficiency (\(\eta\)) = 90% = 0.9

### Step 2: Write the formula for efficiency
The efficiency of the transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} \]
Where:
- \(P_{out} = V_o \times I_o\) (Output Power)
- \(P_{in} = V_i \times I_i\) (Input Power)

### Step 3: Substitute the power formulas into the efficiency equation
Substituting the power expressions into the efficiency formula gives:
\[ \eta = \frac{V_o \times I_o}{V_i \times I_i} \]
Now, substituting the known values:
\[ 0.9 = \frac{230 \times I_o}{2300 \times 5} \]

### Step 4: Simplify the equation
First, calculate the denominator:
\[ 2300 \times 5 = 11500 \]
Now, substitute this back into the equation:
\[ 0.9 = \frac{230 \times I_o}{11500} \]

### Step 5: Solve for \(I_o\)
Rearranging the equation to solve for \(I_o\):
\[ 0.9 \times 11500 = 230 \times I_o \]
Calculating \(0.9 \times 11500\):
\[ 0.9 \times 11500 = 10350 \]
Now, we have:
\[ 10350 = 230 \times I_o \]
Dividing both sides by 230 to find \(I_o\):
\[ I_o = \frac{10350}{230} \]

### Step 6: Calculate \(I_o\)
Calculating \(I_o\):
\[ I_o = 45 \text{ A} \]

### Conclusion
The output current \(I_o\) is 45 A.