How does a dead battery look in a circuit?

I'm working on a Kirchhoff's problem where two car batteries are connected in parallel. One is dead and one is good. It's a basic three variable problem where I'm working out the three different currents (I1/I2/I3) in the circuit when the switch is closed.

My question is how do I treat the dead battery in the circuit? I would have thought that being dead you would make the voltage equal 0 Volts because it has no potential for current but my instructor is calculating using it at 12 Volts. Is that correct and if so why?

simulate this circuit – Schematic created using CircuitLab

• Did the circuit give the resistances in the problem? Nov 21, 2015 at 21:19
• In the real world this will depend on the battery chemistry, both for what "dead" means in terms of voltage and current capacity and for the internal resistance the cell would provide in your circuit above. Nov 21, 2015 at 21:19
• @David : no kidding! My parents had a 1930s electric clock, which a "dead" flashlight battery would happily run for another year or two...
– user16324
Nov 21, 2015 at 21:21
• R2 current in your circuit wont be high with the dead battery circuit values of 1 ohm and 12V.If the battery is in a intermediate state of charge its possible to get high currents . Nov 21, 2015 at 21:36
• what did it die of? if all the electolyte has evaporated the result will be different to if the plates have warped, and different again to just discharged. Nov 22, 2015 at 6:59

By the time a "car battery" gets down to 12 volts, it is considered dead (mostly dead, Princess Bride). Although I think I would agree with you that my first instinct was to think of 0 volts. When I thought about it scientifically, and looked it up :

Battery voltage can be higher than 12.6 volts immediately after charging.
Open Circuit Voltage of 12V battery after car is off for one hour | Relative charge
12.4V: 75%
12.2V: 50%
12.1V: 25%
Less than 12 volts: Dead

I should give credit to : http://www.w8ji.com/battery_and_charging_system.htm

View it this way: The internal resistance of the battery (when discharged) becomes high. You could model the "dead" battery as 12 volts with a high resistance in series with the battery.

• If you add the 100% rating (12.6V), the problem could be solvable from there. Definitely a sketchy problem because live and dead are rather nebulous. Nov 21, 2015 at 21:17
• Thank you @horta for adding a crucial data point. For other comments that have shown up, I used the humorous "mostly dead" to indicate that the definition of "dead" is not well defined. Nov 21, 2015 at 21:23
• "mostly dead" lol... good analogy. Instructor is a physics prof that doesn't take into account the variability of "dead" and "mostly dead". I like the approach of increasing the resistance to indicate "dead". Might be what she's done here as BAT2 is 1 Ohm while BAT1 is 0.01 Ohm. She just never discussed this in class so its confusing what her approach was. Nov 21, 2015 at 21:29