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Is there a way to calculate voltage drop with internal resistance and load values in the equation? Or is it somehow connected to ohms law? If so, how is it done? I have this test circuit below to illustrate what I mean.

schematic

simulate this circuit – Schematic created using CircuitLab

The following is what I mean when I say internal resistance:

When designing a circuit with a battery, we often assume that the battery is an ideal voltage source. This means that no matter how much or little load we attach to the battery, the voltage at the source's terminals will always stay the same. ... In reality, several factors can limit a battery's ability to act as an ideal voltage source. Battery size, chemical properties, age, and temperature all affect the amount of current a battery is able to source. As a result, we can create a better model of a battery with an ideal voltage source and a resistor in series. internal resistance
(from learn.sparkfun.com)


Stated in my previous post: electronics.stackexchange.com I understand that a batteries internal resistance + a resistor will reduce voltage by small degrees. I also know that with no internal resistance, voltage is unchangeable, except from the battery itself. If this phenomenon is somehow linked to ohms law, how is it linked? or is there a separate "law" that covers this?


My second question is very basic: Is the voltage the same thru an entire series circuit?

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    \$\begingroup\$ Not possible. 10V/5A = 2\$\Omega\$ = \$R_T\$, which is total resistance, not internal. You have to clarify this! If it was, there would be no voltage at load. \$\endgroup\$ Aug 22 '20 at 22:15
  • \$\begingroup\$ What do you mean? @StainlessSteelRat \$\endgroup\$
    – user256116
    Aug 22 '20 at 22:21
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    \$\begingroup\$ I mean Ohm's Law! If the current is 5A and the applied voltage is 10V, then the Total Resistance if 2\$\Omega\$. Shows us how you determined what you state. \$\endgroup\$ Aug 22 '20 at 22:26
  • \$\begingroup\$ What is the inertial energy? \$\endgroup\$ Aug 22 '20 at 22:47
  • \$\begingroup\$ The term we use for the resistor "hidden" inside a battery is "internal resistance", not "inertial resistance" \$\endgroup\$ Aug 23 '20 at 1:45
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There is only one current loop, so the magnitude of current will be 5A throughout the circuit. As there will be 5A through the internal 2ohm load, apply Ohm's Law (V=IR) to calculate the voltage dropped across the resistor.

This shows a voltage of 10V will be dropped across the internal resistance, leaving no voltage for the current sink. If this is a serious and correct question, the answer is that 0V are available at the current sink - i.e. your battery has completely browned out, leaving no voltage for the load.

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  • \$\begingroup\$ Of course, if the load voltage is 0, then it can't be drawing 5 amps unless it is a short circuit. \$\endgroup\$
    – Barry
    Aug 22 '20 at 23:38
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  • battery ESR for cars is measured in Cold Cranking Amps at 7.5V for 30 seconds. so 5V drop at Crank Amps =500A then ESR= 5V/500A

  • for a 1.5V battery they have thermal load color bars where the current changes color . with a DMM you do the same delta V/I = ESR.

  • For Supercaps they have a more complex timing as there is a double layer charge effect. (Memory effect) (Yet almost all e-caps and batteries have this, it’s bigger in Supercaps and Ni-Cads

  • the voltage is the same only at the same node but Batteries have an electrode-electrolyte resistance internally that causes the voltage drop for ESR or each series cell.

  • If a battery has 2 0hms ESR with a 5A load it’s is dropping 10V to a short cct . So it must be less than 10% of the minimum load R to maintain 90% of the no load or open-circuit voltage, Voc.

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