# Homework Help: Kcl problem

1. May 27, 2013

### xlu2

1. The problem statement, all variables and given/known data

2. Relevant equations

KCL

3. The attempt at a solution

2. May 27, 2013

### CWatters

Thats correct. The current through the 20kOhm must be 2i1

How about writing some KVL equations.

3. May 27, 2013

### xlu2

How can I write KVL without any resistors in some loops?

For the leftmost loop: (v2-v1)/5000=0
For the rightmost loop: (v4-v3)/20000=0

4. May 27, 2013

### CWatters

First off you can simplify things..

V1 = V3
V2 = V4

Then write equations for the current through the 5k and 20k.

Edit: oh heck I've given you the current through the 20K already. Just write an eqn for the current in the 5k.

Then using those write equations for the voltage drop across the 5k and 20K.

Then write your KVL equation. Try the outer loop.

Last edited: May 27, 2013
5. May 27, 2013

### xlu2

Since the current through 20K is 2i, and V1=V3 and V2=V4, the current through 20K can be expressed as (V4-V3)/20K=(V2-V1)/20K=2i

Substitute that back to my node V1: 8i-i=0.004. So i=0.000571429A?

6. May 27, 2013

### xlu2

Got it. Thanks!

7. May 27, 2013

### CWatters

For what it's worth the equation I got was..

(4mA + i1)*5000 - 2i1*20,000 = 0

which can be solved for i1