6. 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 source voltage and internal resistance of the unknown battery. Assume that in each case the direction of the current is negative to positive in the 5.60-V battery.
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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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