Question

A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is_____A.

Answer 1

Correct Answer: 45

To find the output current of the transformer, we can use the principles of conservation of energy and the efficiency of the transformer.

Given Values: Input power \( P_{\text{in}} = V_{\text{in}} \times I_{\text{in}} \).

\( V_{\text{in}} = 2300 \, V \) (input voltage).

\( I_{\text{in}} = 5 \, A \) (input current).

Efficiency \( \eta = 90\% = 0.9 \).

Output voltage \( V_{\text{out}} = 230 \, V \).

Calculating Input Power:

\( P_{\text{in}} = V_{\text{in}} \times I_{\text{in}} = 2300 \, V \times 5 \, A = 11500 \, W \).

Calculating Output Power: Since the transformer is 90% efficient:

\( P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 11500 \, W = 10350 \, W \).

Calculating Output Current: Using the output power and output voltage, the output current \( I_{\text{out}} \) can be calculated as:

\( P_{\text{out}} = V_{\text{out}} \times I_{\text{out}} \Rightarrow I_{\text{out}} = \frac{P_{\text{out}}}{V_{\text{out}}} = \frac{10350 \, W}{230 \, V} \).

Final Calculation:

\( I_{\text{out}} = 45 \, A \).

Answer 2

The problem asks for the output current of a step-down transformer. We are given the input voltage and current, the output voltage, and the efficiency of the transformer.

Concept Used:

The solution involves the concepts of electrical power and transformer efficiency.

1. Input Power (\(P_{in}\)): The power supplied to the primary winding of the transformer is the product of the primary voltage (\(V_p\)) and the primary current (\(I_p\)).

\[ P_{in} = V_p I_p \]

2. Output Power (\(P_{out}\)): The power delivered by the secondary winding is the product of the secondary voltage (\(V_s\)) and the secondary current (\(I_s\)).

\[ P_{out} = V_s I_s \]

3. Transformer Efficiency (\(\eta\)): The efficiency of a transformer is the ratio of the output power to the input power, usually expressed as a percentage.

\[ \eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{V_p I_p} \]

Step-by-Step Solution:

Step 1: List all the given parameters and convert them to standard units.

Input (primary) voltage, \( V_p = 2.3 \, \text{kV} = 2300 \, \text{V} \)
Input (primary) current, \( I_p = 5 \, \text{A} \)
Output (secondary) voltage, \( V_s = 230 \, \text{V} \)
Efficiency, \( \eta = 90\% = 0.90 \)

(Note: The number of primary turns, 3000, is not required for this specific calculation.)

Step 2: Calculate the input power (\(P_{in}\)) supplied to the transformer.

\[ P_{in} = V_p \times I_p \] \[ P_{in} = 2300 \, \text{V} \times 5 \, \text{A} = 11500 \, \text{W} \]

Step 3: Use the efficiency to calculate the output power (\(P_{out}\)) delivered by the transformer.

The efficiency formula is \( \eta = \frac{P_{out}}{P_{in}} \). Rearranging for \(P_{out}\):

\[ P_{out} = \eta \times P_{in} \] \[ P_{out} = 0.90 \times 11500 \, \text{W} = 10350 \, \text{W} \]

Step 4: Calculate the output current (\(I_s\)) using the output power and output voltage.

The formula for output power is \( P_{out} = V_s \times I_s \). Rearranging for \(I_s\):

\[ I_s = \frac{P_{out}}{V_s} \]

Final Computation & Result:

Substitute the values of \(P_{out}\) and \(V_s\) to find the output current.

\[ I_s = \frac{10350 \, \text{W}}{230 \, \text{V}} \] \[ I_s = \frac{1035}{23} \, \text{A} \]

Performing the division:

\[ I_s = 45 \, \text{A} \]

The output current of the transformer is 45 A.