I tried making a lead acid battery. After charging for about 10 minutes, I connect the voltmeter in parallel to the cell and see that there is around a 1.2V (which is quite lower than what I predicted). However, when connecting to a lightbulb, the voltage seems to be smaller. Is this supposed to happen? Shouldn't the voltage be the same with the cell providing an appropriate current to match the resistance of the bulb?

  • $\begingroup$ This is how is supposed to be. For instance refer to electronics.stackexchange.com/questions/190857/… $\endgroup$ – Alchimista Jan 18 '18 at 17:07
  • $\begingroup$ @Alchimista So does that mean if I put a lightbulb with a higher resistance, the voltage would be even lower because a higher current will be drawn out and therefore more voltage drop due to the internal resistance of the cell? $\endgroup$ – Dan Skyler Jan 18 '18 at 17:26
  • $\begingroup$ No . More resistance means less current is drawn. To the limit of unconnected electrodes when current is zero and V is at maximum (Voc). $\endgroup$ – Alchimista Jan 18 '18 at 18:02
  • $\begingroup$ "voltage seems to be smaller" => you mean it is measured smaller by the voltmeter, or the bulb's brightness appears to you less? $\endgroup$ – Gaurang Tandon Jan 19 '18 at 2:24
  • $\begingroup$ I mean that the voltage reading on the voltmeter decreases $\endgroup$ – Dan Skyler Jan 19 '18 at 2:30

Yes, this happens because there is an internal resistance within the battery. Look up how batteries are modeled within circuits.


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