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--- 184_notes:examples:week14_step_down_transformer [2018/08/09 19:20] – curdemma
+++ 184_notes:examples:week14_step_down_transformer [2021/07/22 13:56] (current) – [Solution] schram45
@@ Line 13: / Line 13: @@
 ===Approximations & Assumptions===
-  * We have access to the same materials as we did for the step-up transformer.
+  * We have access to the same materials as we did for the step-up transformer: This allows us to use some of the same relationships from the step-up transformer solution.
   * The step-down transformer we are building will have a similar design to the step-up transformer.
@@ Line 60: / Line 60: @@
  $V_S=V_P \frac{N_S}{N_P}$
-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$.