Solving for Ix: Find Current Given R & V

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Homework Help Overview

The discussion revolves around finding the current Ix in a circuit given resistances and voltage. The original poster attempts to calculate Ix based on a reduced equivalent resistance but questions the validity of their approach.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the reduction of the circuit to a single equivalent resistance and the implications of current splitting in parallel branches. Questions arise about the original circuit configuration and the correct application of Kirchhoff's laws.

Discussion Status

Some participants provide guidance on the need to consider current division in parallel paths, while others express uncertainty about the original poster's calculations. There is an ongoing exploration of the circuit's structure and the relationships between the currents in different branches.

Contextual Notes

Participants note the importance of understanding Kirchhoff's laws for solving the problem effectively. The original poster acknowledges a need for more practice with these concepts.

xlu2
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Homework Statement


What is ix (in μA)? Given all R and V.

1.jpg


Homework Equations


V=IR


The Attempt at a Solution


I reduced the circuit into just one Req = 20.778 kohms and find the current through the reduced circuit to be 240.6375 uA (5V/20778 ohms). Since current flowing in series is the same, Ix through the 10 kohms is also 240.6375 uA.

Would anyone please tell me where I am doing wrong here?

Many thanks in advance!
 
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The element carrying Ix has a path in parallel to it: the current through the 15k resistor is shared between two parallel paths.

What did your circuit look like just before you collapsed it all into one resistance? :smile:
 
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what u r finding is the total current. the current however splits into 3 branches at the junctions. i wouldn't recommend this problem until u have covered kirchhoffs law
 
NascentOxygen said:
The element carrying Ix has a path in parallel to it: the current through the 15k resistor is shared between two parallel paths.

What did your circuit look like just before you collapsed it all into one resistance? :smile:


It was 15 kohms + the reduced resistance (including 10, 10, 4, and 47 kohms resistors). Do you mean that the current I found has to be divided between the 10 and the rest of the reduced resistors (10, 4, 47)?
 
Arkavo said:
what u r finding is the total current. the current however splits into 3 branches at the junctions. i wouldn't recommend this problem until u have covered kirchhoffs law

I did. I just need to practice more. So I thru 15 = I thru 10 + I thru combined (10, 4, 47)?
 
xlu2 said:
It was 15 kohms + the reduced resistance (including 10, 10, 4, and 47 kohms resistors). Do you mean that the current I found has to be divided between the 10 and the rest of the reduced resistors (10, 4, 47)?

Yes.
 
NascentOxygen said:
Yes.

I got I = 139.0438 μA.
 
xlu2 said:
I got I = 139.0438 μA.

Looks right.
 
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