Math Problem Statement

A transformer has a primary winding of 600 turns and a secondary winding of 150 turns. When the load current on the secondary is 60 A at 0.8 power factor lagging, the primary current is 20 A at 0.707 power factor lagging. Determine the no-load current of the transformer and its phase with respect to the voltage. [

Solution

To solve the problem, we will calculate the no-load current (I0I_0) of the transformer and determine its phase angle with respect to the voltage.

Given:

  • Primary turns (NpN_p): 600600
  • Secondary turns (NsN_s): 150150
  • Secondary current (IsI_s): 6060 A
  • Power factor of secondary (pfs\text{pf}_s): 0.80.8 (lagging)
  • Primary current (IpI_p): 2020 A
  • Power factor of primary (pfp\text{pf}_p): 0.7070.707 (lagging)

We will use the following steps:


Step 1: Transformer Current Relationship

Using the turns ratio, the ideal relationship between the primary and secondary currents is given by: IsIp=NpNs\frac{I_s}{I_p} = \frac{N_p}{N_s} Substitute the given values: 60Ipideal=600150    Ipideal=60×150600=15 A\frac{60}{I_p^\text{ideal}} = \frac{600}{150} \implies I_p^\text{ideal} = \frac{60 \times 150}{600} = 15 \text{ A} Thus, the ideal primary current (IpidealI_p^\text{ideal}) is 15A15 \, \text{A}.


Step 2: Magnetizing Component of Primary Current (ImI_m)

The actual primary current (Ip=20AI_p = 20 \, \text{A}) consists of two components:

  1. The magnetizing (no-load) current (I0I_0).
  2. The ideal current (IpidealI_p^\text{ideal}) that corresponds to the load current.

Using vector addition in the phasor domain: I_p = \sqrt{I_0^2 + I_p^\text{ideal}^2} Rearrange for I0I_0: I_0 = \sqrt{I_p^2 - I_p^\text{ideal}^2} Substitute values: I0=202152=400225=175=13.23AI_0 = \sqrt{20^2 - 15^2} = \sqrt{400 - 225} = \sqrt{175} = 13.23 \, \text{A}


Step 3: Phase Angle of I0I_0

The no-load current (I0I_0) lags the primary voltage. The phase angle of I0I_0 can be found using the power factor (pfp=0.707\text{pf}_p = 0.707) of the primary: cos(ϕp)=0.707    ϕp=cos1(0.707)=45\cos(\phi_p) = 0.707 \implies \phi_p = \cos^{-1}(0.707) = 45^\circ

Thus, the no-load current (I0I_0) lags the voltage by 4545^\circ.


Final Answer:

  1. No-load current (I0I_0): 13.23A13.23 \, \text{A}
  2. Phase angle: 4545^\circ lagging

Do you want further clarification or a detailed explanation of any step?

Related Questions:

  1. How is the ideal primary current (IpidealI_p^\text{ideal}) derived in transformers?
  2. What does the power factor indicate about the phase relationship in AC systems?
  3. Why is the no-load current significant in transformer operations?
  4. How do transformer turn ratios affect primary and secondary currents?
  5. What is the importance of phasor addition in analyzing AC circuits?

Tip:

Always ensure the primary and secondary power equations match when calculating transformer parameters, as energy conservation principles are critical!

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Math Problem Analysis

Mathematical Concepts

Transformer Operation
Phasor Analysis
Power Factor
Current Calculation
Trigonometry

Formulas

\frac{I_s}{I_p} = \frac{N_p}{N_s}
I_p = \sqrt{I_0^2 + I_p^2}
I_0 = \sqrt{I_p^2 - I_p^\text{ideal}^2}
\cos(\phi_p) = 0.707

Theorems

Energy Conservation in Transformers
Phasor Addition

Suitable Grade Level

Grades 11-12