Help Needed: Solving KCL Equations with Picture Provided

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Homework Help Overview

The discussion revolves around applying Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to analyze a circuit, as illustrated in a provided picture. Participants are attempting to solve for currents and voltages in a circuit involving resistors of 5kOhm and 20kOhm.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between currents in the circuit, specifically noting that the current through the 20kOhm resistor is twice that of another current. They explore writing KVL equations despite concerns about the absence of resistors in some loops. There are attempts to simplify the circuit by equating voltages at certain nodes.

Discussion Status

Some participants have provided guidance on simplifying the circuit and writing equations for currents and voltage drops. There is an ongoing exploration of KVL equations, with various interpretations and approaches being discussed. No explicit consensus has been reached, but productive direction has been provided.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on deriving relationships and equations based on the circuit's configuration.

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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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