Help Needed: Solving KCL Equations with Picture Provided

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The discussion revolves around solving Kirchhoff's Current Law (KCL) equations in an electrical circuit. The user initially identifies that the current ix equals i1 and seeks assistance in progressing further. Participants confirm the current through the 20kOhm resistor is 2i1 and suggest writing Kirchhoff's Voltage Law (KVL) equations. Simplifications are made, leading to the conclusion that V1 equals V3 and V2 equals V4, which helps in formulating the necessary equations. Ultimately, the user derives an equation involving i1, indicating progress in solving the problem.
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Homework Statement



QQ??20130527135512.jpg


Homework Equations



KCL

The Attempt at a Solution



See picture. I got ix=i1. And I am stuck. Would someone please help me?

Many thanks in advance.
 
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I got ix=i1

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

How about writing some KVL equations.
 
CWatters said:
Thats correct. The current through the 20kOhm must be 2i1

How about writing some KVL equations.

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
 
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.
 
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CWatters said:
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.

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?
 
CWatters said:
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.

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

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

which can be solved for i1
 
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