Transformers - Don't understand VpIp = VsIs

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The discussion centers on understanding the transformer equation VpIp = VsIs and its implications for connecting various loads. A user questions how a 220V-3V step-down transformer can power an LED without burning it, suggesting confusion about current levels in the secondary. It is clarified that while the secondary current is indeed higher, it only matches the load requirements, and the primary current is lower due to the turns ratio. Additionally, the primary current includes magnetizing current, which is out of phase with the secondary current. Understanding these fundamentals is essential for grasping transformer operation and load behavior.
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transformers -- Don't understand VpIp = VsIs

Homework Statement



If the equation below of any transformer is true, therefore it does not matter if i connect a lightbulb or anything else either to the primary or the secundary, it should be working the same, right? I have a 220v-3v step down transformer and i can connect an led without burning it. How is that possible? Do we lose power here? Shouldn't be the current much higher in the secondary?

Homework Equations



VpIp = VsIs

The Attempt at a Solution

 
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alexmath said:

Homework Statement



If the equation below of any transformer is true, therefore it does not matter if i connect a lightbulb or anything else either to the primary or the secundary, it should be working the same, right? I have a 220v-3v step down transformer and i can connect an led without burning it. How is that possible? Do we lose power here? Shouldn't be the current much higher in the secondary?

Homework Equations



VpIp = VsIs

The Attempt at a Solution


Why should the current be any higher than what the load requires? I think you need to study transformer fundamentals and perhaps electronics fundamentals such as ohm's law.
 
alexmath said:

Homework Statement



Shouldn't be the current much higher in the secondary?

The current IS much higher in the secondary. Your turns ratio is 220/3 = 73.3 = N1/N2.
So if your load current is say 100 mA then your primary current is 100/73.3 = 1.36 mA. However, the primary current also includes the magnetizing current which depends on the primary winding inductance and your frequency of operation, and which is 90 deg. (approx.) out of phase with the secondary current.
 

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