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I think that you have faced a conceptual problem because you have used the relationship $V=RI$, rather you should have been using the definition of resistance $R = \frac V I$. Then as $V$ goes up and $I$ goes down the value of $R$ increases. Also you must be careful about $I=0$ (current equals exactly zero) and $I\rightarrow0$ (current gets closer and closer ...


Bear in mind that infinite resistance is equal to an open circuit. In real life if you have a battery not connected to anything, there will always be a voltage across the terminals. Even a lead-acid car battery will produce sparks if you short the terminals!


At R→∞ voltage won't be 0 but little less than EMF off cell, since a series circuit with voltmeter and battery is complete. Yes, voltage across resistor reaches 0.Voltmeter reading is not EMF because it draw some current which pass through internal r and gets some potential across it. An Ideal voltmeter cannot tell you voltage across resistance in ...


-First, I'm trying to grasp the concept of why and how the voltage drop at the terminals of a battery depends on the resistance of the circuit. The EMF, $\mathcal{E}$, exists chemically and is always present. If there is a path for current to flow, then there will be a voltage drop across the internal resistance, $r$, and the external resistance, $R$. ...


I actually did this for a battery control driver for a tablet. The only way I found to do it was to take manufacturers charge/discharge curves as functions of temperature and number of cycles and encode them by hand in a lookup table. This does assume that the manufacturer has enough data, the data is accurate, and is willing to supply it.

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