![Mangesh4438](https://www.crazyengineers.com/img/avatar.jpg)
Member • Sep 23, 2011
Member • Sep 23, 2011
Member • Sep 23, 2011
Member • Sep 23, 2011
check out the limitations of ohm's law....Mangesh4438Ohm says i=v/r but at some condition current increases then voltage decreases how is that?
Member • Sep 23, 2011
Member • Sep 23, 2011
Source resistance. When load resistance decreases, current increases, and voltage drop at source side appears. For example:V battery= 14,4VR lamp = 4 OhmR starter = 0.144 OhmR internal of a battery= 0.024 OhmI1=Vbat/(Rlamp+Rbat)=3,57A Vlamp=Vbat-I1*Rbat=14.28VI2= Vbat/(Rstarter+Rbat)=85,7AVstarter= Vbat-I2*Rbat=12,34VMangesh4438Ohm says i=v/r but at some condition current increases then voltage decreases how is that?
Member • Sep 24, 2011
Member • Sep 25, 2011
If you are talking about real circuits the culprit may be the impedance of the source. The out put voltage of some power supplies could be higher when the load has a high resistance than when it has a low resistance. Ohm's law is still obeyed if one considers the total resistance, that is, source resistance plus load resistance.Mangesh4438Ohm says i=v/r but at some condition current increases then voltage decreases how is that?
Member • Sep 29, 2011