Question relating to Kirchhoff's law

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SUMMARY

The discussion focuses on applying Kirchhoff's laws to determine the currents I1 and I2 in a circuit with two resistors and two batteries, each having an internal resistance of 1.2 Ω. The user initially derived the equations I3 = I1 + I2 and -18*I2 - 22*I1 + 9V = 0 for the top loop, and 6V + 18*I2 = 0 for the bottom loop. After isolating I2 and substituting it back into the top loop equation, the user arrived at I1 = 0.6818, which was marked incorrect by the masteringphysics tool. The discussion emphasizes the importance of correctly identifying current directions and magnitudes, noting that magnitudes cannot be negative.

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Homework Statement


Determine the magnitudes of the currents through R1 and R2 in (Figure 1) , assuming that each battery has an internal resistance r = 1.2 Ω .
Express your answers using two significant figures separated by commas.

Homework Equations


Junction rule: I3 = I1 + I2 (any current going into one junction must come out with the same amount)
loop rule: The sum of all the potential differences around a complete loop is equal to zero.

The Attempt at a Solution


Please refer to the image I attached.
First I set up the directions for current flow. I got the equations:

I3 = I1 + I2.
For the top loop, I got -18*I2 - 22*I1 +9V = 0.

For the bottom loop, I got 6V + 18*I2 = 0. I isolated I2 to get -6/18 = -0.33333.

Then I plugged in I2(-0.33333) into the top loop's equation to isolate I1. I got I1 = 0.6818.

However, the masteringphysics keeps telling me it's wrong. Can anyone help me with this?

*I've randomly tried changing up the signs of the two numbers, which didn't work so I don't know where I went wrong.
 

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So where are I1, I2, and I3, and in what direction? I'd suggest dropping I3 as you only need 2 (and you don't use it anyway)
 
phinds said:
So where are I1, I2, and I3, and in what direction? I'd suggest dropping I3 as you only need 2 (and you don't use it anyway)
Yes haha I dropped the first equation basically since I3 wasnt needed. I1 i drew to point towards the left from the top loop. As for I2, I drew it pointing towards the right on the middle segment. I3 I drew it point towards the right from the battery on the bottom loop.
 

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So how did you work the magic of making I1 not flow through the middle resistor?
 
phinds said:
So how did you work the magic of making I1 not flow through the middle resistor?
Sorry, not too sure what you mean. I included 18*I2 for the first loop. Is that not the current going through resistor in the middle?
 
johnknee said:
Sorry, not too sure what you mean. I included 18*I2 for the first loop. Is that not the current going through resistor in the middle?
Draw a full loop current in each loop. Where do you think I1 goes when it hits the node? Does it just evaporate?
 
The problem asks for the current magnitudes. Magnitudes are never negative.
 

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