Voltage measured across battery terminals

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SUMMARY

The discussion centers on the voltage drop observed across a DC voltage source when a load, specifically an LED circuit, is connected. Initially, the voltage measured at the terminals was 5.17V without a load, but it dropped to 4.97V when the LED was activated. This voltage drop is attributed to the internal resistance of the power source and the switch, which has a resistance of 0.75 ohms. The phenomenon illustrates the concept of voltage division and the limitations of current supply in electrical circuits.

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ramonegumpert
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Hello Experts! :!)

I have a LED light with 3 leds, a switch and a resistor i added to adjust for higher supply voltage.

I removed the batteries and connect it to a DC voltage source from power adaptor.

Without anything connected as load to this DC source, the voltage read 5.17V

with the LED connect to the DC source and LED switch turned on, the voltage measured across the same terminals is lower at 4.97v.

I found that the switch itself has a resistance of 0.75 ohm.

I could not understand why the voltage supplied is 5.17v measured at the battery terminals but when the LEDs are switched on, it drops a bit. Should not the voltage source measured at the terminals remain constant?

Sorry if this is a noob question. I am beginning to explore this field.

Hope to receive your advice.

sincerely
Ramone :smile:
 
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That's a good question.

Basically, the reason the voltage drops is because of the internal resistance of the battery.

Batteries can't supply infinite amount of current. If you attempt to draw more current, then more voltage will drop. And if you were to connect a battery to a load, then you will get a voltage divider. Here is a model of its internal behavior:


http://people.sinclair.edu/nickreeder/EET150/PageArt/voltageSourceResistance.gif
 
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Dear what,
thanks for your clear explanation. I really appreciate it.

Wishing you a great weekend ahead.

sincerely
Ramone
 

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