Did I count this current in a circuit correctly?

AI Thread Summary
The discussion focuses on verifying the calculation of current I1 in a circuit, which was found to be -1/7. The user is uncertain about their approach and seeks confirmation, particularly regarding the transcription of equations. A participant suggests that a term in the equation may have been incorrectly noted as 20I1 instead of 10I1, indicating a possible typo. The conversation also touches on the signs of the currents, confirming that I1 is negative while I3 is positive, and mentions the potential usefulness of Thevenin and Norton equivalent circuit transformations for simplifying such problems. Overall, the user is encouraged to continue their calculations with the provided insights.
Duc Anh Le
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This i a test question and we are not allowed to use a calculator, so I was wondering whether I did everything correctly, since the I1 = - 1/7. I am supposed to find out I3, but stopped at I1, since the result is a bit tricky.

1. Homework Statement

R1 = 10 ohm, R2 = 20 ohm, R3 = 40 ohm
I3 = ?

Homework Equations


I3 = I2 + I1
V1 = I1 * R1 + I3 * R3
V2 = R2 * I2 + I3 * R3

The Attempt at a Solution


10 = 20I1 + 40 I2 + 40I1
20 = 20I2 + 40I1 + 40I2
-10 = 70I1
I1 = -1/7

5ly1os8imkw11.jpg
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The circuit in the image is very difficult to read the component labels. Please post a closeup or draw and photograph a larger version.
Duc Anh Le said:
3. The Attempt at a Solution
10 = 20I1 + 40 I2 + 40I1
20 = 20I2 + 40I1 + 40I2
-10 = 70I1
I1 = -1/7
From what I can tell, the first term on the RHS should be 10I1 rather than 20I1. This is likely a transcription typo since your value for the current I1 looks good to me.
 
gneill said:
The circuit in the image is very difficult to read the component labels. Please post a closeup or draw and photograph a larger version.

From what I can tell, the first term on the RHS should be 10I1 rather than 20I1. This is likely a transcription typo since your value for the current I1 looks good to me.
6uVlA3E.png

Here's a better picture
 

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It is indeed much better! Thanks.
 
Duc Anh Le said:
This i a test question and we are not allowed to use a calculator, so I was wondering whether I did everything correctly, since the I1 = - 1/7. I am supposed to find out I3, but stopped at I1, since the result is a bit tricky.
...

I will show that ##I_1## is negative:

$$50 I_1 + 40 I_2 = 10;\ 40 I_1 + 60 I_2 = 20 \\
5 I_1 + 4 I_2 = 1; 2 I_1 + 3 I_2 = 1\\
3 I_1 + I_2 = 0$$
Note that ##I_1## and ##I_2## have different signs and that ##I_3 = -2 I_1##, so ##I_3## and ##I_1## have different signs. And since ##I_2## is clearly positive, we know all the signs and ##I_1## is indeed negative.
 
You are on the right track, good work so far!
Do you know about Thevinin & Norton equivalent circuit transformations yet? I think they make problems like this much easier to do.
 
DaveE said:
You are on the right track, good work so far!
Do you know about Thevinin & Norton equivalent circuit transformations yet? I think they make problems like this much easier to do.
Alright, thanks for the verification. So far we've only been taught ohm's and kirchhoff's laws.
 
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