# Homework Help: Non Ideal Batteries in Electric Current

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1. Sep 27, 2015

### h2obc33

1. The problem statement, all variables and given/known data
A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf V = 12 V in series with an internal resistance r as shown above left. The values for the resistors are: R1 = R3 = 57 Ω, R4 = R5 = 75 Ω and R2 = 133 Ω. The measured voltage across the terminals of the batery is Vbattery = 11.62 V.

1) What is I1, the current that flows through the resistor R1?

2. Relevant equations
Ohm's Law V=IR
Kirchoff's Laws: At Junction Iin = Iout and loop ΣVn=0

3. The attempt at a solution
To solve for I1 I did Vb/Requivalent. That would be R345 = R3+R4+R5 then I1 = Vb/(R1+(1/R345+1/R2)^-1) = 87mA
I got the message "It looks like you've calculated the current through R1 by dividing the emf of the battery (12 V) by the equivalent resistance of the external resistances in the circuit. This is not quite right. Look at the circuit more carefully to correct your mistake."
I made sure to used 11.62V instead of 12V so I still don't understand what I did wrong.

2. Sep 27, 2015

### ehild

Your method is correct, Your equivalent resistance might be is wrong, or you really divided 12 V with it. What did you get as equivalent resistance?

3. Sep 27, 2015

### h2obc33

For my equivalent resistance I got 133.5606Ω and I checked I divided 11.62/133.5606 = 0.087A (87mA)

4. Sep 27, 2015

### ehild

Your equivalent resistance is not correct, I am afraid.

5. Sep 27, 2015

### h2obc33

Thank you so much, I have it right now. I realized my error, I mistook the R values for those of a previous problem. I even checked and if I had used 12 with the correct R equivalence it would have been nearly the same as my original answer.

6. Sep 27, 2015

### ehild

Yws
Yes, it was funny, that your result was as if you divided 12 V with the correct resistance :)