Electrical nodal analysis, find current

[jshoe]

Homework Statement

ive tried many things and can't seem to grasp the concept of KCL.i know the node I am looking for is between r1 & r3. i assumed i1+i2=i3. also looking for directions of currents.

V1=30v
V2=40v
R1=82
R2=68
R3=47
Vx = ?

Homework Equations

i1 + i2 + i3 = 0 ?

The Attempt at a Solution

(30-Vx)/82 + (Vx-40)/68 = Vx/147

anyone help me out with KCL and Nodal analysis? I've been trying for awhile and can't seem to wrap my head around it.

 

[gneill]

Homework Statement

ive tried many things and can't seem to grasp the concept of KCL.i know the node I am looking for is between r1 & r3. i assumed i1+i2=i3. also looking for directions of currents.

V1=30v
V2=40v
R1=82
R2=68
R3=47
Vx = ?

Homework Equations

i1 + i2 + i3 = 0 ?

The Attempt at a Solution

(30-Vx)/82 + (Vx-40)/68 = Vx/147

anyone help me out with KCL and Nodal analysis? I've been trying for awhile and can't seem to wrap my head around it.

The trick is to make sure that the current directions that you choose for the sum will satisfy KCL at the given node. If you write: i1 + i2 + i3 = 0 then there is a tacit assumption that either all the currents are assumed to be flowing into the node, or all are assumed to be flowing out of the node.

Now take a look at your next equation. The first term (30-Vx)/82 is making the assumption that it the current is directed INTO the node. The next term (Vx-40)/68 makes the opposite assumption for current direction (it's directed OUT of the node). Placing the last term on the right hand side effectively changes its sign, so Vx/147 implies a current flowing INTO the node. So one out of three currents is mismatched in direction if you want to satisfy i1 + i2 + i3 = 0.

A simple way to avoid such current direction problems is to apply a fixed procedure to writing the equation for the node:

1) Assume that all currents flow OUT of the node
2) Write all the terms on one side of the equation, setting the other to zero.

Step 1 is accomplished by always taking the node potential first in the terms, and subtracting the fixed offset from the reference:

(Vx - 30)/82 + (Vx - 40)/68 + Vx/147 = 0

 

[jshoe]

thanx for the reply. i asked my professor same question and it is becoming a little clearer. however I am still a little confused on my results, current flow and polarities. my answers are:

i1= -20.6 mA
i2=-172 mA
i3= 193 mA

these answer match up in the back of the book however i1= 20.6 ?

so does i3 + i2 = -i1 ? is this how i would analyze my answer?

how do i determine my results for current flow direction?
how do i know which currents to switch from - to + ?
do + currents flow into node and - flow out of node ?

 

[gneill]

thanx for the reply. i asked my professor same question and it is becoming a little clearer. however I am still a little confused on my results, current flow and polarities. my answers are:

i1= -20.6 mA
i2=-172 mA
i3= 193 mA

these answer match up in the back of the book however i1= 20.6 ?

so does i3 + i2 = -i1 ? is this how i would analyze my answer?

how do i determine my results for current flow direction?
how do i know which currents to switch from - to + ?
do + currents flow into node and - flow out of node ?

Well, it looks like you took my advice and solved for the currents assuming that they are all leaving node A. That means the assumed current directions are shown by arrows pointing away from node A along each branch. Now, if the problem has a particular direction indicated for any given current by which you are meant to judge the current, and the indicated direction happens to match your calculation assumptions then no problem, the calculated current signs match. If on the other hand the problem wants you to report the current as measured in the opposite direction, then you change the sign of your calculated result to match.

So for example, suppose the problem had certain currents and directions specified on the circuit diagram, but you solved mathematically with the assumption that all currents flow out of the node:

Note that the calculated current i1 has the opposite direction from the current label in the problem diagram, while the other currents match in direction. That means you change the sign on your calculated value for i1 to match the diagram and leave i2 and i3 alone because they already match.

 

[rezeew]

Hello,

I understand that you are struggling with understanding KCL and nodal analysis. Here are some steps that may help you:

1. First, let's define the nodes in the circuit. In this case, there are three nodes: V1, V2, and Vx.

2. Next, let's apply KCL at the node Vx. KCL states that the sum of all currents entering and leaving a node must equal zero. In this case, we have three currents entering the node Vx: i1, i2, and i3.

3. We can use Ohm's Law to express each current in terms of the voltage difference and resistance. For example, i1 = (30-Vx)/82, i2 = (Vx-40)/68, and i3 = Vx/47.

4. Now, we can plug these expressions into our KCL equation: (30-Vx)/82 + (Vx-40)/68 + Vx/47 = 0.

5. Simplify the equation to solve for Vx. You should get Vx = 28.53 V.

6. To find the directions of the currents, we can use the same expressions from step 3. We know that current flows from higher voltage to lower voltage. So, for example, i1 will flow from V1 (higher voltage) to Vx (lower voltage).

I hope this helps you understand KCL and nodal analysis better. Remember to always define your nodes and apply KCL at each node to solve for unknown currents. Let me know if you have any other questions. Good luck!