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

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SUMMARY

The discussion addresses the phenomenon of voltage drop in a simple electronic circuit when a 5V DC power supply is connected to a load, resulting in a measured voltage of 3V. This voltage drop occurs due to the formation of a voltage divider between the internal resistance of the power supply and the load resistance. It is established that the load's resistance significantly influences the voltage drop, particularly when the load has low resistance, leading to a substantial voltage difference across the circuit.

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  • Knowledge of voltage dividers and their role in circuit behavior.
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momomo_mo
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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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