Math Problem Statement

Solve this sum please

The following data apply to a single-phase transformer: output: 100 kVA, secondary voltage: 400 V; Primary turns: 200; secondary turns: 40; Neglecting the losses, calculate: (i) the primary applied voltage (ii) the normal primary and secondary currents (iii) the secondary current, when the load is 25 kW at 0.8 power factor.

Solution

The problem provided gives information on a single-phase transformer. Here’s a structured approach to solving each part.

Given Data:

  1. Apparent Power, S = 100 kVA
  2. Secondary Voltage, V2V_2 = 400 V
  3. Primary Turns, N1N_1 = 200
  4. Secondary Turns, N2N_2 = 40
  5. Load Power = 25 kW with a power factor of 0.8
  6. Neglecting transformer losses.

We are asked to calculate:

  1. V1V_1: the primary applied voltage.
  2. I1I_1 and I2I_2: the normal primary and secondary currents.
  3. I2I_2: the secondary current for a 25 kW load at 0.8 power factor.

Step-by-Step Solution

Part (i): Calculating the Primary Applied Voltage V1V_1

We use the turns ratio formula for a transformer:

V1V2=N1N2\frac{V_1}{V_2} = \frac{N_1}{N_2}

Rearranging to solve for V1V_1:

V1=V2×N1N2V_1 = V_2 \times \frac{N_1}{N_2}

Substituting the values:

V1=400V×20040=400×5=2000VV_1 = 400 \, \text{V} \times \frac{200}{40} = 400 \times 5 = 2000 \, \text{V}

So, the primary applied voltage V1V_1 is 2000 V.


Part (ii): Calculating Normal Primary and Secondary Currents I1I_1 and I2I_2

Using the apparent power equation for both primary and secondary sides:

  1. Secondary Current I2I_2:

    S=V2×I2S = V_2 \times I_2

    Solving for I2I_2:

    I2=SV2=100,000VA400V=250AI_2 = \frac{S}{V_2} = \frac{100,000 \, \text{VA}}{400 \, \text{V}} = 250 \, \text{A}

  2. Primary Current I1I_1:

    Using the primary side with the apparent power equation:

    I1=SV1=100,000VA2000V=50AI_1 = \frac{S}{V_1} = \frac{100,000 \, \text{VA}}{2000 \, \text{V}} = 50 \, \text{A}

So, the normal primary current I1I_1 is 50 A, and the secondary current I2I_2 is 250 A.


Part (iii): Calculating Secondary Current I2I_2 for a 25 kW Load at 0.8 Power Factor

For this load condition, we first calculate the apparent power SS at 25 kW with a power factor of 0.8:

S=Real PowerPower Factor=25,000W0.8=31,250VAS = \frac{\text{Real Power}}{\text{Power Factor}} = \frac{25,000 \, \text{W}}{0.8} = 31,250 \, \text{VA}

Then, we calculate I2I_2 using this new apparent power:

I2=SV2=31,250VA400V=78.125AI_2 = \frac{S}{V_2} = \frac{31,250 \, \text{VA}}{400 \, \text{V}} = 78.125 \, \text{A}

So, the secondary current I2I_2 when the load is 25 kW at 0.8 power factor is 78.125 A.


Summary of Results:

  1. Primary Applied Voltage V1V_1: 2000 V
  2. Normal Primary Current I1I_1: 50 A
  3. Normal Secondary Current I2I_2: 250 A
  4. Secondary Current I2I_2 for 25 kW load at 0.8 power factor: 78.125 A

Would you like a detailed explanation of any part of the solution, or do you have further questions?

Here are some related questions to consider:

  1. How would the calculations change if the power factor was different?
  2. What happens to the transformer currents if there were losses in the transformer?
  3. How does changing the primary or secondary voltage affect the transformer’s operation?
  4. What is the significance of the turns ratio in transformer design?
  5. How does the transformer’s efficiency affect the current and voltage calculations?

Tip: Always verify power factor and load conditions, as they can significantly affect current calculations in transformers.

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Math Problem Analysis

Mathematical Concepts

Electrical Engineering
Transformer Calculations
Power Factor
Current and Voltage Calculations

Formulas

Turns Ratio Formula: V1/V2 = N1/N2
Apparent Power Formula: S = V * I
Power Factor Formula: S = Real Power / Power Factor

Theorems

Transformer Turns Ratio

Suitable Grade Level

Undergraduate Engineering