# Find current

1. May 22, 2013

### xlu2

1. The problem statement, all variables and given/known data
What is ix (in μA)? Given all R and V.

2. Relevant equations
V=IR

3. 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?

2. May 22, 2013

### Staff: Mentor

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?

3. May 22, 2013

### 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 kirchoff's law

4. May 22, 2013

### xlu2

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)?

5. May 22, 2013

### xlu2

I did. I just need to practice more. So I thru 15 = I thru 10 + I thru combined (10, 4, 47)?

6. May 22, 2013

### Staff: Mentor

Yes.

7. May 22, 2013

### xlu2

I got I = 139.0438 μA.

8. May 22, 2013

Looks right.