Solving the Mystery of a Simple Electronic Circuit's 2V Drop

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A DC power supply set to 5V dropped to 3V when connected to a circuit, raising questions about the cause of the voltage drop. The drop is attributed to a voltage divider effect created by the internal resistance of the power supply and the load resistance. When the load has low resistance, the voltage drop becomes more pronounced. The power supply continues to output 5V, but the effective voltage across the load is reduced due to this interaction. Understanding these principles is crucial for analyzing circuit behavior.
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hi
That's the question.
I have done an experiment.
The voltage of a free DC power supply was set to 5V. When it was connected to a circuit, its voltage dropped to 3V. What happened?

Isn't the 2V is lost in the internal resistance of the power supply.
However,the power supply will continuously supply 5V.So why do the voltage will drop to 3V.And, in a simple circuit,the voltage will be 0V when it comes to the end after passing through those resistance,so why the voltage will drop to 3V.

Thanks everyone!
 
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The voltage drops at the point where you connect the load because a voltage divider is formed of the internal resistance of the supply and the resistance of the load. The voltage drop will be significant if the load has a very low resistance.
 

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