Solving for Ix: Find Current Given R & V

[xlu2]

Homework Statement

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

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!

 

[NascentOxygen]

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?

 

[Arkavo]

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

 

[xlu2]

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?

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)?

 

[xlu2]

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)?

 

[NascentOxygen]

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.

 

[xlu2]

Yes.

I got I = 139.0438 μA.

 

[haruspex]

I got I = 139.0438 μA.

Looks right.