emf of a battery

pb23me

1. The problem statement, all variables and given/known data
determine the terminal voltage in each battery.
Emf₁=18V
r₁=1 ohm
EMF₂=12V
r₂=2 ohm
R=6.6 ohms
also the batterys currents are opposing each other.
2. Relevant equations Vab=emf-Ir


3. The attempt at a solution i used kirchnoffs rule to find the current I , 18-I6.6-12-I2-I=0 and got I= .625 then i thought i would just use Vab=emf-Ir to find the terminal voltage of each battery however im not getting the correct answer... could someone please explain this to me i have a test tomorrow morning and i really need to figure this out! the answers i got were 17.375 for the 18V batt and 10.75 for the 12V batt

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gneill

Watch out for the current direction and the thus the polarity of the voltage drop it creates on the internal resistors. If current is being forced into a battery's positive terminal it's possible for it to have a higher terminal voltage than its internal EMF.

pb23me

so on the larger voltage batt would i use equation Vab=Emf-Ir = 18-(.625*1) and on the batt with voltage being forced across it use Vab=Emf+... idk

pb23me

what are the equations you would use in this situation?

gneill

Yes. It can be helpful to sketch in your current and the polarities of the voltage drops across the resistors. You did fine writing the KVL loop equation.

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