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Problem about source

vick5821

Jan 22, 2012
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hey, one very weird things. When my switch is off, my battery is 8V.But once I conneect it to the circuit and the switch is on, the voltage from the battery dropped to around 5V soemthing. Why ? Anything wrong there in my circuit ?
 

Harald Kapp

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The current your circuit draws from the battery is more than the battery can deliver at full voltage. Think of the batery as an ideal voltage source with a series resistor (inner resistance caused by the battery's chemistry, its state of charge etc.). The current will create a voltage drop at this internal resistor. You cannot see or measure directly the internal voltage drop, but you see the voltage at the battery's terminals go gown as a consequence.
 

vick5821

Jan 22, 2012
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The current your circuit draws from the battery is more than the battery can deliver at full voltage. Think of the batery as an ideal voltage source with a series resistor (inner resistance caused by the battery's chemistry, its state of charge etc.). The current will create a voltage drop at this internal resistor. You cannot see or measure directly the internal voltage drop, but you see the voltage at the battery's terminals go gown as a consequence.

Why causes this ? Any solving methods?
 

gorgon

Jun 6, 2011
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the rectangular battery :)

Yes, but is it a 9V battery?
If it is, it will be exhausted with 8V and no load, and not manage to keep the voltage up with a load applied. No mystery here.

TOK ;)
 

vick5821

Jan 22, 2012
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Yes, it is 9V battery. I measured it to be 9V but when switch on my circuit, the battery voltage becomes around 6V >< why ? circuit problem ?
 

davenn

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Yes, it is 9V battery. I measured it to be 9V but when switch on my circuit, the battery voltage becomes around 6V >< why ? circuit problem ?

Harald gave you the answer in post #2


Dave
 

Harald Kapp

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No. This is an intrinsic property of batteries.
Buy a new battery or use another battery with more capacity (e.g. 6*AA cell).
 
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