Connecting a known and unknown battery

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SUMMARY

The discussion addresses the calculation of the electromotive force (emf) and internal resistance of an unknown battery connected to a known 5.60 V battery. Using the loop rule, the user derived two equations based on the current readings of 10.0 mA and 25.0 mA, leading to the conclusion that the emf of the unknown battery is 2.4 V and its internal resistance is 320 ohms. The connection method of the batteries, positive to positive, does not affect their series configuration.

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  • Understanding of Ohm's Law and circuit analysis
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[SOLVED] Connecting a known and unknown battery

Homework Statement


An unmarked battery has an unknown internal resistance. If the battery is connected to a fresh 5.60 V battery (negligible internal resistance) positive to positive and negative to negative, the current through the circuit is 10.0 mA. If the polarity of the unknown battery is reversed, the current increases to 25.0 mA. Determine the emf and internal resistance of the unknown battery. Assume that in each case the direction of the current is negative to positive within the 5.60 V battery.


Homework Equations




The Attempt at a Solution


I tried making two equations using the loop rule for each setup.

5.6 - .01*r - e = 0
5.6 - .025*r + e = 0

e = 2.4 V
r = 320 ohms

Would that be right? The resistance looks a little too high to me.

When it says the batteries are connected positive to positive, does that mean the batteries are connected in series in the circuit?
 
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mike115 said:
e = 2.4 V
r = 320 ohms

Would that be right? The resistance looks a little too high to me.

When it says the batteries are connected positive to positive, does that mean the batteries are connected in series in the circuit?

The answer is right. Positive to positive just means that the positive sides are connected to each other.

This has nothing to do with being in series, because the batteries will be in series no matter how you connect them.
 
All right, thanks kamerling!
 

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