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Calculating turns in a primary coil of a transformer

  • Thread starter Bobazoide
  • Start date
  • #1
6
0

Homework Statement



A 100 W transformer (Ns = 1500) has an input voltage of 9.0V and an output current of 0.65 A

Homework Equations


P = IV
Np/Ns = Is/Ip = Vp/Vs


The Attempt at a Solution


The problem is that I'm not sure what a 100 watt transformer means. Assuming it means that it outputs, not inputs, 100 watts of power, then:

P = IV
100 = 0.65 V
V = 154 V

Np/Ns = Vp/Vs

Np/1500 = 9/154
Np = 87.66 turns.

I feel that mathematically I am doing everything correctly, but I thought that because voltage was being increased then the amount of coils in the primary coil should be more than in the secondary.

Thanks.
 

Answers and Replies

  • #2
gneill
Mentor
20,800
2,778

Homework Statement



A 100 W transformer (Ns = 1500) has an input voltage of 9.0V and an output current of 0.65 A

Homework Equations


P = IV
Np/Ns = Is/Ip = Vp/Vs


The Attempt at a Solution


The problem is that I'm not sure what a 100 watt transformer means. Assuming it means that it outputs, not inputs, 100 watts of power, then:

P = IV
100 = 0.65 V
V = 154 V

Np/Ns = Vp/Vs

Np/1500 = 9/154
Np = 87.66 turns.

I feel that mathematically I am doing everything correctly, but I thought that because voltage was being increased then the amount of coils in the primary coil should be more than in the secondary.

Thanks.
Looks like you've done just fine. A step-up transformer has more turns on the secondary than on the primary. It's the ratio of turns that sets the "magnification" factor for the voltage.
 
  • #3
6
0
Oh okay, I wasn't clear on that. Thank you very much!
 

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