# Calculating current and voltage?

#### WBahn

Joined Mar 31, 2012
26,398
Looks like homework. Is it? If so, I can move it to Homework Help where it will get more attention.

What analysis techniques have you been introduced do so far? I presume you know Ohm's Law and KVL/KCL, have you learned node voltage and mesh current analysis? Superposition?

In general, it is best if you show YOUR best attempt to solve the problem. That let's us see how you are approaching things, what kind of techniques you seem to be trying to use, and where you are going right and wrong. Then we can make observations and offer hints and suggestions to help you move from where you are to where you need to get.

#### TimoG

Joined Nov 2, 2017
17
Looks like homework. Is it? If so, I can move it to Homework Help where it will get more attention.

What analysis techniques have you been introduced do so far? I presume you know Ohm's Law and KVL/KCL, have you learned node voltage and mesh current analysis? Superposition?

In general, it is best if you show YOUR best attempt to solve the problem. That let's us see how you are approaching things, what kind of techniques you seem to be trying to use, and where you are going right and wrong. Then we can make observations and offer hints and suggestions to help you move from where you are to where you need to get.
Yeah it is a homework. I know the basics of circuit analysis as you mentioned above, its just that I was unsure what equations to start with. If someone would be able to quickly write down the three or four starting equations I can take it from there. Thank you for the help.

#### WBahn

Joined Mar 31, 2012
26,398
If someone writes down the three or four starting equations for you, then they will have done ALL of the EE in the problem for you and the rest is just math. You will learn nothing from that.

The first thing you might do is annotate your circuit with other voltages and currents (say for each resistor), being sure to indicate polarity. Then try to apply KVL around every loop you can identify and KCL at every node you can identify.

Do that much and present your results. We can first make sure that all of the equations are correct.

Then we can talk about how many of those equations you actually need to solve for everything.

#### WBahn

Joined Mar 31, 2012
26,398
MOD NOTE: Moved to Homework Help.

#### shteii01

Joined Feb 19, 2010
4,644
Step 1.
Choose a reference node.

#### Tonyr1084

Joined Sep 24, 2015
6,067
Well, without giving you an answer I'd like to suggest you redraw the circuit. Just looking at the 1 volt and 2 volt sources, they are essentially in series with each other. You don't have to keep them where they are. You can combine them into a single source of 3 volts. You can do the same with the resistors in that left hand loop. The two resistors in series act like a larger single resistor. Do the same with the right loop. Once you know the current you can separate the circuit back into its original form, and since you know the currents calculating the voltages would be easy.

#### WBahn

Joined Mar 31, 2012
26,398
Just how are the 1 V and 2 V sources in series?

Series means that the same current flows in both components.

#### Tonyr1084

Joined Sep 24, 2015
6,067
This comment is being edited out by me, myself because it's misleading and, well, wrong. So the drawing has been removed and my comment has been altered.

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#### WBahn

Joined Mar 31, 2012
26,398
No, that does not work. Your claim that the same current flows through both of them relies on arbitrarily declaring that the current in V3 is zero, which is the only way that you can ignore it. News flash: The current in V3 is not zero.

You can twist and turn and wiggle and probably get to the right answer if you carefully keep track of everything that you have to completely ignore both when you do it and when you undo it, but your circuit at the bottom is not equivalent to the original circuit.

Even the left hand side of it is not equivalent. For instance, let's call the bottom center node 0 V. Let's call the top left node Vleft. If we solve for Vleft in the original circuit and Vleft in your circuit, what is the difference between them?

What we CAN say about the circuit is that V1 isolates the left side from the right side. I think that is what is prompting the approach you are trying to take. But consider if there were a resistor directly above V1. Would you still claim that V1 and V2 are in series? You'd HAVE to, since how can putting a component in series with two components that are in series not leave the original two components still in series. Yet would your approach still work?

Better that he learn to analyze a circuit using solid techniques grounded in the fundamental principles of circuit analysis.

#### Tonyr1084

Joined Sep 24, 2015
6,067
Oh. OK. Ignore my comment please.

#### RBR1317

Joined Nov 13, 2010
649
News flash: The current in V3 is not zero.
OK, but suppose the current in V3 is a different mesh current. Then the two mesh currents, I1 & I2, only overlap in the V1 source, making this a particularly easy problem to solve. There is no need to even write the mesh equations.

#### WBahn

Joined Mar 31, 2012
26,398
OK, but suppose the current in V3 is a different mesh current. Then the two mesh currents, I1 & I2, only overlap in the V1 source, making this a particularly easy problem to solve. There is no need to even write the mesh equations.
While this is very true, it relies on a fairly subtle understanding of how currents in ideal voltage sources interact. There are a couple (probably more) reasonable ways to explain it. Because an ideal voltage source maintains its terminal voltage independent of how much current is flowing in the source, different mesh currents within a voltage source don't interact. You can drive this point home by duplicating the source in parallel, moving them apart, arguing that there is no need for any current to flow in the bridging wire (though this argument itself has some subtle issues since, in actuality, the current flowing in that wire is completely undefined) and then cut the wire to illustrate the isolating effect of an ideal voltage source in that situation.

But if someone takes the simple explanation that I1 and I2 only overlap in V1 and therefore they can ignore one when finding the other and use that as the way to solve the problem, they are extremely likely to come away with the impression that you can just go around one mesh and ignore the other meshes when solving for that mesh current. I'm not guessing here, I've seen it done many times -- it is hard for lots of students to grasp that mesh currents interact in most components and that interaction must be taken into account (and recognizing that, in a particular instance, the interaction has no effect, is a perfectly legitimate way to take it into account).

Let's say that the middle source what a current source. Such folks would be tempted to say that I1 and I2 only overlap in the current source making the mesh currents easy to calculate. Some will do the same if there's a resistor there.

I have no problem with recognizing when meshes do not interact and exploiting that knowledge to make solving the problem easier. That is (or at least was) about first on my list of things to point out to the TS once they get the problem solved using the tools they already know.

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#### shteii01

Joined Feb 19, 2010
4,644
Peeps, since you are not going to solve this problem for OP, why are you wasting their time?

#### Tonyr1084

Joined Sep 24, 2015
6,067
Peeps, since you are not going to solve this problem for OP, why are you wasting their time?
Wasting time? Not if someone (like me) can learn something simply by following this thread. To be perfectly honest, I don't recall ever hearing about a "Mesh" equation. Now I know there's something I need to look up and learn. That way when someone else asks a similar question instead of opening my mouth and looking foolish I can speak from a base of knowledge. Sometimes the best way to learn something is to teach it to others. They say "Practice makes perfect." Well, if I figure this thing out then I can practice by helping others learn something as well.

I'd like to hear the thread starters comments on all this. If it's helping them or exasperating them. Sometimes I ask for help and only find more exasperation. With persistence comes clarity. And with clarity comes ability. I am, if nothing else, persistent. I'll stick to it until it's firmly glued down.

#### TimoG

Joined Nov 2, 2017
17
Wasting time? Not if someone (like me) can learn something simply by following this thread. To be perfectly honest, I don't recall ever hearing about a "Mesh" equation. Now I know there's something I need to look up and learn. That way when someone else asks a similar question instead of opening my mouth and looking foolish I can speak from a base of knowledge. Sometimes the best way to learn something is to teach it to others. They say "Practice makes perfect." Well, if I figure this thing out then I can practice by helping others learn something as well.

I'd like to hear the thread starters comments on all this. If it's helping them or exasperating them. Sometimes I ask for help and only find more exasperation. With persistence comes clarity. And with clarity comes ability. I am, if nothing else, persistent. I'll stick to it until it's firmly glued down.
Its all a bit confusing as there seems so many ways to solve the problem, but for all of them I run into problems. Using just KCL and KVL I came up with these equations:
IV1 = IR1 - IR2
IV1 = IR3 - IR4
> IR1 - IR2 = IR3 - IR4

R1(IR1) - V1 + R3(IR3) - V2 = 0
R2(RI2) - V3 + R4(IR4) + V1 = 0

However, there are not enough equations to solve for the variables so I guess I need another equation. I am unsure as to what this could be unless there is something I overlooked? I reckoned I would solve for the various currents and then use that to calculate V4.

#### shteii01

Joined Feb 19, 2010
4,644
Its all a bit confusing as there seems so many ways to solve the problem, but for all of them I run into problems. Using just KCL and KVL I came up with these equations:
IV1 = IR1 - IR2
IV1 = IR3 - IR4
> IR1 - IR2 = IR3 - IR4

R1(IR1) - V1 + R3(IR3) - V2 = 0
R2(RI2) - V3 + R4(IR4) + V1 = 0

However, there are not enough equations to solve for the variables so I guess I need another equation. I am unsure as to what this could be unless there is something I overlooked? I reckoned I would solve for the various currents and then use that to calculate V4.
Sorry mate, none of them make sense.

I would make the top left node my reference node (0 volts) for Node Voltage Method analysis.

Regarding Mesh Current Method. You have two meshes. That means you have two mesh currents. You, by some magic, have 4 mesh currents. At this point it is obvious that you don't know the proper way to setup mesh current equations.

#### The Electrician

Joined Oct 9, 2007
2,846
He has 4 branch currents. not 4 mesh currents.

#### RBR1317

Joined Nov 13, 2010
649
However, there are not enough equations to solve for the variables so I guess I need another equation.
Not true. I think you have too many names for the same variable. Try looking at where the same current flows through several components, then give that current a single name. Now try writing the KVL equations again. I'll bet you have enough equations to solve for the variables. My suggestion for the named currents is here: 