Did I count this current in a circuit correctly?

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Homework Help Overview

The discussion revolves around a circuit analysis problem involving three resistors (R1, R2, R3) and the calculation of currents (I1, I2, I3) without the use of a calculator. The original poster expresses uncertainty about their calculation of I1, which they found to be -1/7, and seeks to determine I3.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations leading to I1 and question the transcription of terms in the equations. There is also mention of the difficulty in reading the circuit diagram and a request for clearer images. Some participants explore the implications of the signs of the currents and suggest that I1 and I2 have different signs.

Discussion Status

The discussion is ongoing, with participants verifying calculations and suggesting alternative approaches, such as Thevenin and Norton equivalent transformations. There is no explicit consensus, but several participants are engaging with the problem and providing supportive feedback.

Contextual Notes

Participants note the constraints of not using calculators and the potential for transcription errors in the equations. The original poster is working under test conditions, which may influence their approach and reasoning.

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
[/B]
 

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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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