Transformer primary-to-secondary turn ratio problem

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SUMMARY

The discussion focuses on calculating the peak current in the primary coil of a step-down transformer with a primary-to-secondary turn ratio of 4:1 and a secondary peak current of 12 A. Using the transformer equation, the relationship between primary and secondary currents is established as (4/1) = (Ip/12 A). The conclusion drawn is that the peak current in the primary coil is 48 A, based on the conservation of power principle, where P1 = P2 and P = IV.

PREREQUISITES
  • Understanding of transformer equations, specifically Vp/Vs = Np/Ns
  • Knowledge of electrical power concepts, particularly P = IV
  • Familiarity with current and voltage relationships in electrical circuits
  • Basic grasp of ideal transformer characteristics
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  • Learn about the implications of non-ideal transformers on current and voltage
  • Explore advanced transformer equations and their applications in electrical engineering
  • Investigate power conservation in electrical systems and its practical implications
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Electrical engineering students, educators, and professionals involved in power systems and transformer design will benefit from this discussion.

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Homework Statement



A step down transformer has a primary to secondary ratio of 4:1. If the peak current in the secondary is 12 A, the peak current in the primary is what?


Homework Equations



Vp/Vs = Np/Ns


The Attempt at a Solution



(4/1) = (Ip/12 A)

what is the relation between I and V? (V= IR or V= IP does not seem to work).

thanks in advance!
 
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In an ideal transformer the power in the primary is the same as in the secondary, P1 = P2. You know that P = IV, so ...
 

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