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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.
the rectangular battery
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 ?
I mean any method to overcome this problem ?