Non Ideal Batteries in Electric Current

[h2obc33]

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.

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.

 

[ehild]

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.

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?

 

[h2obc33]

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)

 

[ehild]

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.

 

[h2obc33]

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.

 

[ehild]

Yws

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 :)