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Zorluk: OrtaInductors and Transformers

A step-down transformer installed in an industrial power supply has 1200 turns1{}200\text{ turns} on its primary winding and 300 turns300\text{ turns} on its secondary winding. If the secondary coil delivers a current of 8 A8\text{ A} to an operating load, what is the primary current (IpI_p) drawn from the alternating current (AC\text{AC}) source?

  1. A
    32 A32\text{ A}
  2. 2 A2\text{ A}Cevap
  3. C
    0.5 A0.5\text{ A}
  4. D
    200 mA200\text{ mA}

Cevap

The primary current drawn from the source is 2 A2\text{ A}.
In an ideal transformer, input power equals output power (Vp×Ip=Vs×IsV_p \times I_p = V_s \times I_s). Therefore, current is inversely proportional to the turns ratio: IpIs=NsNp\frac{I_p}{I_s} = \frac{N_s}{N_p}. With a 4:14:1 step-down turns ratio (1200 turns1{}200\text{ turns} to 300 turns300\text{ turns}), the secondary current of 8 A8\text{ A} corresponds to a primary current of 8 A÷4=2 A8\text{ A} \div 4 = 2\text{ A}.

Adım Adım Çözüm

1
Calculate the turns ratio of the transformer (Np/NsN_p / N_s).
Turns Ratio=NpNs=1200300=4\text{Turns Ratio} = \frac{N_p}{N_s} = \frac{1{}200}{300} = 4
The turns ratio defines the proportional relationship between primary and secondary electrical parameters.
2
Apply the inverse current relationship for an ideal transformer (Ip/Is=Ns/NpI_p / I_s = N_s / N_p).
Ip=Is×(NsNp)=8 A×(3001200)=8 A×14=2 AI_p = I_s \times \left(\frac{N_s}{N_p}\right) = 8\text{ A} \times \left(\frac{300}{1{}200}\right) = 8\text{ A} \times \frac{1}{4} = 2\text{ A}
Because transformers conserve power (Pin=PoutP_{\text{in}} = P_{\text{out}}), stepping down voltage by a factor of 4 causes current to step up by a factor of 4, meaning the primary current must be one-fourth of the secondary current.

Anahtar Kavram

Transformer Current and Turns Relationship
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