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Hi all. Plz could someone explain what is the voltage drop? If i use a 9V battery and 3ohm resistor in series connection and nothing more,
what will be the voltage drop at the resistor? Sry for bad English.
Thx for help.
If you have just a 9v battery with that 3Ω resistor across it's terminals then the voltage across the resistor = full voltage of the battery.
Just don't expect that battery to last long as 3Ω resistor would drain a fresh 9V in less than 15 minutes.
Also, a battery's voltage reduces as you draw more current from it.
If you have just a 9v battery with that 3Ω resistor across it's terminals then the voltage across the resistor = full voltage of the battery.
Just don't expect that battery to last long as 3Ω resistor would drain a fresh 9V in less than 15 minutes.
Also, a battery's voltage reduces as you draw more current from it.
As I recently measured 9V alkaline battery has 3 Ohm internal resistance so
with 3 Ohm load the voltage on the resistor would be 4.5 V .
If you use 100 Ohm resistor or higher then it would be 9V.
If you actually connected the resistor to the battery some of the 9V would drop across the 3 ohm resistor and the rest would drop across whatever the battery internal resistance is. If you had a 9V rectangular transistor, a large percentage of the voltage would drop across the battery internal resistance, since you would be trying to draw a 3A load from a small battery, which has significant internal resistance.
ANY power supply source has so called internal resistance (an supposed resitance which indicates the inability of a real voltage source to act as an ideal voltage source to give ANY current to the load with a constant voltage across its leads). the said resistance is supposed to be connected in series with an ideal battery or the volatge sourece. If you connect ANY load across your battery or mains it would be in series with internal resistance and the ideal voltage source and composes so called resistive VOLTAGE DIVIDER. You HAVE TO know the conseot of voltage dividing to understand the responses to your question and to what is called Voltage Drop. Please take a look at the bellow links:
If you have a 9V battery with an internal resitance of 3 ohms then you will see 9V across its terminal when there is no load connected to the battery. The reason is clear, you do not have ANY current draw from the battery and so there would not be any voltage drop across the iternal resistance [Voltage drop in the internal resistance of a battery = (internal resistance value) X (Current which passes through this resistance)]. when no current (current=0) is passing through the internal resistance then the voltage drop across this resistance would be 0V and so the whole 9V will be viewed across its terminals.
Makes sense?
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