184_notes:examples:week14_step_down_transformer

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184_notes:examples:week14_step_down_transformer [2021/07/13 13:38] schram45184_notes:examples:week14_step_down_transformer [2021/07/22 13:56] (current) – [Solution] schram45
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 $$V_S=V_P \frac{N_S}{N_P}$$ $$V_S=V_P \frac{N_S}{N_P}$$
  
-Remember, we need to step down from the $240 \text{ kV}$ power line to a $120 \text{ V}$ line. This is a factor of 2000. One way to achieve this would be to set $N_P = 10$, and $N_S = 20000$. Due to the huge step down, it may be even easier to design a series of step-down transformers, so that we don't have to have such a large number of turns for the secondary solenoid. Maybe apply a factor of $N_S/N_P=40$ for one transformer, and then $N_S/N_P=50$ for a second transformer. You should be able to convince yourself that this would be physically equivalent to just one step-down transformer with $N_S/N_P=2000$.+Remember, we need to step down from the $240 \text{ kV}$ power line to a $120 \text{ V}$ line. This is a factor of 2000. One way to achieve this would be to set $N_P = 20000$, and $N_S = 10$. Due to the huge step down, it may be even easier to design a series of step-down transformers, so that we don't have to have such a large number of turns for the secondary solenoid. Maybe apply a factor of $N_S/N_P=1/40$ for one transformer, and then $N_S/N_P=1/50$ for a second transformer. You should be able to convince yourself that this would be physically equivalent to just one step-down transformer with $N_S/N_P=1/2000$.
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