Much like my last thread, I have another homework problem in which I must solve for a current value using nodal analysis. I've attempted the problem using some techniques to visualize the circuit from the last forum post and felt confident in my answer, however it is again incorrect and I'm not sure where I went wrong. Some guidance to point me in the right direction would be much appreciated. Attached are the problem and my attempt at a solution, I have a feeling posting to this forum may become a regular occurrence for me. Thank You.

 

When you use an arrow to show current direction you are also defining the order the voltages must be used in calculating the current. the current through the resistors are defined as (V2 - V1)/Resistance, where V2 is the tail of the arrow and V1 is the head of the arrow.

Look at your definition of the current flowing through the 8kOhm resistor.

Also, since there are only two unknown voltages, only two equations are required.

 

The given circuit has 3 nodes, one of which is selected as the ground node. So there will only be two node equations. One does not write a node equation for the ground because we already know the voltage on the ground node (it's zero).

Here are my rules for nodal analysis:

* Sum all the currents at a node and set it equal to zero.

* Each term of the node equation begins with the voltage of that node (this makes it simpler to avoid errors).

* Each term has the form: (Vnode - Vadjacent)/(Resistance to adjacent node).

* Specified currents (i.e. current source) are positive if leaving the node.

Never vary from these rules, and you will find it hard to make a mistake.

 

When you use an arrow to show current direction you are also defining the order the voltages must be used in calculating the current. the current through the resistors are defined as (V2 - V1)/Resistance, where V2 is the tail of the arrow and V1 is the head of the arrow.

Look at your definition of the current flowing through the 8kOhm resistor.

Also, since there are only two unknown voltages, only two equations are required.

So would the equation describing the current regarding the 8kOhm resistor actually be (0-V1)/8 which equals -V1/8?

 

Remember that nodal analysis is nothing more than the systematic application of Kirchhoff's Current Law, which can be stated either as the sum of the currents entering a node is zero, or equivalently that the sum of the currents leaving a node is equal to the sum of the currents entering the node. With that in mind, write the currents leaving the node through resistors on the left side and currents entering the node through current sources on the other. This puts most of the quantities on the correct side of the equation from the beginning, lessening the likelihood for silly math errors.

So for the first node you need

(the current LEAVING the node through the 8 kΩ resistor) + (the current LEAVING the node through the 4 kΩ resistor) = (the current ENTERING the node from the left 6 mA current source) + (the current ENTERING the node from the right 6 mA current source)

[(V1 - 0V) / 8 kΩ] + [(V1 - V3) / 4 kΩ] = (+6 mA) + (-6 mA)

Until you get more proficient and comfortable do the simplifying steps in your head, write your setup equations this way. ALL of the electrical engineering principles represented by the circuit is embodied by the two node equations. Everything after that is just math. Mistakes in setting up the equation can't be caught for fixed in the math, no matter how carefully you do the math from that point on. So set up the equations in a form that lets you look at them and verify that they are correct, term by term, before proceeding to solve them.

 

So would the equation describing the current regarding the 8kOhm resistor actually be (0-V1)/8 which equals -V1/8?

Yes, with the current defined in the direction of your arrow the minus sign is needed as you have shown above.

 

So would the equation describing the current regarding the 8kOhm resistor actually be (0-V1)/8 which equals -V1/8?

In a loose sense, yes.

When you define a current direction for each branch and then write the node equations based on those branch currents, you really aren't doing nodal analysis, you are really doing the more fundamental branch current analysis. True node voltage analysis doesn't specify individual branch currents, but instead embeds them implicitly into the nodal equations. The result is that if you have two nodes A and B, then the implied branch current flows from A to B in one of the node equations and from B to A in the other -- which is why we don't annotate the branch current on the diagram as it would be confusing since it applies to one of the node equations and not the other.

Aside from this, -V1/8 is a voltage, namely the additive inverse of one-eighth of whatever V1 is. It is NOT a current. The current is -V1/8kΩ.

You need to start tracking your units properly and throughout your work from beginning to end. Tracking units is perhaps the single most powerful error detection technique available to the engineer since most mistakes that we make, either conceptual or just math errors, will mess up the units allowing us to immediately identify that an error has been made and quickly determine what it was, where it is, and how to fix it before moving on and wasting a bunch of time and effort on work that, from that point on, is pretty much guaranteed to produce a garbage answer.

 

In a loose sense, yes.

When you define a current direction for each branch and then write the node equations based on those branch currents, you really aren't doing nodal analysis, you are really doing the more fundamental branch current analysis. True node voltage analysis doesn't specify individual branch currents, but instead embeds them implicitly into the nodal equations. The result is that if you have two nodes A and B, then the implied branch current flows from A to B in one of the node equations and from B to A in the other -- which is why we don't annotate the branch current on the diagram as it would be confusing since it applies to one of the node equations and not the other.

Aside from this, -V1/8 is a voltage, namely the additive inverse of one-eighth of whatever V1 is. It is NOT a current. The current is -V1/8kΩ.

You need to start tracking your units properly and throughout your work from beginning to end. Tracking units is perhaps the single most powerful error detection technique available to the engineer since most mistakes that we make, either conceptual or just math errors, will mess up the units allowing us to immediately identify that an error has been made and quickly determine what it was, where it is, and how to fix it before moving on and wasting a bunch of time and effort on work that, from that point on, is pretty much guaranteed to produce a garbage answer.

What you're saying makes perfect sense to me and is very good for making sure there are no errors in the set-up which would lead to a useless answer. However my answer(s) are still incorrect for an unknown reason which is part of the reason I struggle with circuit analysis, because I currently lack the intuition to see where I went wrong and learn from it. My other attempt is attached which was incorrect.

 

What you're saying makes perfect sense to me and is very good for making sure there are no errors in the set-up which would lead to a useless answer. However my answer(s) are still incorrect for an unknown reason which is part of the reason I struggle with circuit analysis, because I currently lack the intuition to see where I went wrong and learn from it. My other attempt is attached which was incorrect.

The right side of the equation you've labeled N3: should be -2/1000, not just -2

 

However my answer(s) are still incorrect for an unknown reason

I put your two node equations (N1, N3) into Maple, but the Maple solver returns a different solution for V3 than what you have. Have you checked your algebra?

 

The right side of the equation you've labeled N3: should be -2/1000, not just -2

But why is that?

 

I put your two node equations (N1, N3) into Maple, but the Maple solver returns a different solution for V3 than what you have. Have you checked your algebra?

I see now, I forgot to multiply the -2mA by 4kOhm in my second equation.

 

I see now, I forgot to multiply the -2mA by 4kOhm in my second equation.

If you had labeled the right side of N3: as -2 mA, rather than just -2 you might have noticed your error.

 

 

What you're saying makes perfect sense to me and is very good for making sure there are no errors in the set-up which would lead to a useless answer. However my answer(s) are still incorrect for an unknown reason which is part of the reason I struggle with circuit analysis, because I currently lack the intuition to see where I went wrong and learn from it. My other attempt is attached which was incorrect.

If what I'm saying makes perfect sense, then why did you ignore it and proceed to make a mistake in your set up equations precisely because you ignored what you claim made perfect sense to you?????

Let me reiterate: You need to start tracking your units properly and throughout your work from beginning to end. Tracking units is perhaps the single most powerful error detection technique available to the engineer since most mistakes that we make, either conceptual or just math errors, will mess up the units allowing us to immediately identify that an error has been made and quickly determine what it was, where it is, and how to fix it before moving on and wasting a bunch of time and effort on work that, from that point on, is pretty much guaranteed to produce a garbage answer.

Your set up equation for N3 should have been

\(\frac{V_3 \; - \; V_1}{4k\Omega} \; + \; \frac{V_3}{2k\Omega} \; + \; \frac{V_3}{2k\Omega} \; = \; 4 mA \; - \; 6 mA \; = \; -2 mA\)

When you multiplied both sides by 4 kΩ two lines later you were so focused on the left hand side that you simply forgot to do the right hand side -- a pretty easy mistake to make. But let's see what would have happened had you bothered to track your units.

\(V_3 \; - \; V_1 \; + \; 2V_3 \; + \; 2V_3 \; = \; -2 mA\)

And now youi have a bunch of voltages on the left equal to a current on the right.

RED FLAG! Something is wrong. Fix it right now. Do not go further until you fix it because all further work is a pure waste of time and effort.

So walk back up the work and check the units very carefully. In the set up equation we have three terms with voltage divided by current on the left, so the sum of currents. On the right we have the sum of two currents. So the units are good. The problem HAS to be in what we did to get from the first equation to the second. Oh... we forgot to multiply the right hand side. It should have been

\(V_3 \; - \; V_1 \; + \; 2V_3 \; + \; 2V_3 \; = \; (-2 mA)(4 k\Omega) \; = \; - 8 V\)

Let me reiterate: You need to start tracking your units properly and throughout your work from beginning to end. Tracking units is perhaps the single most powerful error detection technique available to the engineer since most mistakes that we make, either conceptual or just math errors, will mess up the units allowing us to immediately identify that an error has been made and quickly determine what it was, where it is, and how to fix it before moving on and wasting a bunch of time and effort on work that, from that point on, is pretty much guaranteed to produce a garbage answer.

You need to STOP just ignoring units and then paying lip service to them at the end by tacking on the units that you hope, pray, and would like the answer to have.