Non Ideal Batteries in Electric Current

AI Thread Summary
The discussion revolves around calculating the current (I1) through resistor R1 in a circuit with five resistors and a real battery. The measured voltage across the battery terminals is 11.62 V, and the user initially calculated I1 using an incorrect equivalent resistance, leading to confusion. After feedback, the user realized they had mistakenly referenced resistor values from a previous problem, which affected their calculations. Once the correct resistor values were applied, the user was able to arrive at the right answer for I1. The conversation highlights the importance of accurate resistor values in circuit calculations.
h2obc33
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Homework Statement


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.
h10_realbatteryA.png


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

Homework Equations


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

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.
 
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h2obc33 said:

Homework Statement


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.
h10_realbatteryA.png


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

Homework Equations


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

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.
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?
 
ehild said:
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?
For my equivalent resistance I got 133.5606Ω and I checked I divided 11.62/133.5606 = 0.087A (87mA)
 
h2obc33 said:
For my equivalent resistance I got 133.5606Ω and I checked I divided 11.62/133.5606 = 0.087A (87mA)
Your equivalent resistance is not correct, I am afraid.
 
ehild said:
Your equivalent resistance is not correct, I am afraid.
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.
 
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h2obc33 said:
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.
Yes, it was funny, that your result was as if you divided 12 V with the correct resistance :)
 
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