well I solved all the questions ex[ect of the last one..

I need a help

A battery with an emf of 3.00 V has an internal resistance r. When connected to a resistor R, the terminal voltage is 2.90 V and the current is 0.18 A.
What is the value of the external resistor R?
1.61×101 Ohm

What is the internal resistance r of the battery?
5.56×10-1 Ohm

What is the energy dissipated in the battery's internal resistance in 2.5 minutes?
2.70 J

When a second identical battery is added in series and the external resistor is R = 29 Ohms what is the resulting current?
so I got a wrong answer, I tried to
Total voltage / Total resistance
but I got a wrong answer , I got 5.01 A
I don't know what was wrong with me

Susus, your calculations for the first three parts are correct. If the battery is supplying 2.9V to the external load, and that draws a current of 0.18A, that means the external resistor is 2.9/0.18 = 16.1 ohms.

And with an internal drop of 0.1V, that equates to an internal resistance of 0.1/0.18 = 0.56 ohms.

For the energy calculation, E = P*t = (I*V)*t = (0.18A * 0.1V) * 150s = 2.7J.

As for the fourth part, your approach is correct, so I'm not sure why you're getting an answer of ~5A. Total resistance = 16.1 + 29 + 2*0.556 = 46.2 ohms. Total voltage = 6V. I = 6/46.2 = 0.130 amps.

thanks, I need to crorrect , now
I used I= 2epsilon/(R+2r) = 6/(29+0.56*2 ) = 1.99E-1 A

Oh yeah. Good catch! Now that I reread the question, I see that the 16.1 ohm resistor was REPLACED with a 29 ohm resistor, not included in addition to it. So you're right. It should be 199 mA like you said. Sorry about that. :)