# Question relating to Kirchhoff's law

## 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 .

## 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.

#### Attachments

• Untitled.png
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phinds
Gold Member
2021 Award
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)

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.

#### Attachments

phinds
Gold Member
2021 Award
So how did you work the magic of making I1 not flow through the middle resistor?

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?

phinds