Checking Your Work: Did I Solve #14 Correctly?

[Marcin H]

Homework Statement

Did I do number 14 correctly? I have a feeling it's not that simple.

Homework Equations

V=IR
i1=i2+i3+...In

The Attempt at a Solution

on picture

 

[Doc Al]

Looks good to me.

 

[SammyS]

Homework Statement

Did I do number 14 correctly? I have a feeling it's not that simple.

Homework Equations

V=IR
i1=i2+i3+...In

The Attempt at a Solution

on picture

That's correct.

 

[Marcin H]

Looks good to me.

Ok good. I am stuck on #15 now. I'm haven't seen this kind of circuit yet, so I'm not sure how to apply kirchhoffs current law. I think A, B, and D are correct, but I'm not sure. I think all the squares in the circuit are power sources, but I'm not sure where to start in this circuit to make my loop and write out my equation.

 

[Doc Al]

I'm haven't seen this kind of circuit yet, so I'm not sure how to apply kirchhoffs current law.

Kirchoff's current law just says, for any point, the currents going into the point must equal those going out of the point. (You don't need to know anything about that kind of circuit to apply the law to each point.)

 

[SammyS]

Ok good. I am stuck on #15 now. I'm haven't seen this kind of circuit yet, so I'm not sure how to apply kirchhoffs current law. I think A, B, and D are correct, but I'm not sure. I think all the squares in the circuit are power sources, but I'm not sure where to start in this circuit to make my loop and write out my equation.

Do you know how to do screen shots, or use the snipping tool ? It's much more likely that you get help if we don't have to open new tabs/windows on our browser.

All that is required for this problem is to apply Kirchhoff's Current Law.

You do not need to know what the squares represent at all.

 

[Doc Al]

Hint: Write an equation for each point where wires connect. (The three dots in the diagram.) Then compare those equations with the choices given.

 

[gneill]

Did I do number 14 correctly? I have a feeling it's not that simple.

It looks like you specified the units of the result as mA. That would be milliamps. Was that your intention?

 

[Doc Al]

It looks like you specified the units of the result as mA. That would be milliamps. Was that your intention?

Good eye!

 

[Marcin H]

for any point, the currents going into the point must equal those going out of the point.

I'm having a hard time with this. How do you know what is going in and what is coming out exactly. All the currents can come out of one point. I was trying to write an equation for the point above box c, but how can I tell what is going in and out. Also, I thought current flows to the spot with least resistance.
Edit: Also, I think I would understand this there were 2 points/junctions, but how do I do this with three?

 

[Marcin H]

It looks like you specified the units of the result as mA. That would be milliamps. Was that your intention?

Fixed. Thanks!

 

[Marcin H]

Do you know how to do screen shots, or use the snipping tool ? It's much more likely that you get help if we don't have to open new tabs/windows on our browser.
View attachment 94717
View attachment 94718

All that is required for this problem is to apply Kirchhoff's Current Law.

You do not need to know what the squares represent at all.

Sorry, I'm new here. How did you do that exactly?

 

[Doc Al]

How do you know what is going in and what is coming out exactly.

The currents have directions, specified by the arrows.

 

[SammyS]

Sorry, I'm new here. How did you do that exactly?

OK.

One of the first things to learn here is to make it clear as to what post (you did that fine) or what part of a post you are responding to.

If you were asking me, about that thing I did, and that thing was posting those images, see the following. Otherwise, Doc Al and others, will continue working with you.

In (Microsoft) Windows, there is a "Snipping Tool" application. Find it in the Applications folder in the list of programs.

It will take a 'snapshot' of the portion of your monitor's screen, that portion being determined by what you highlight. You may need to use the help feature provided by the Snipping Tool or google for more details.

Once you take a snapshot of whatever is of interest, you simply paste it into the reply window of PF .

 

[Marcin H]

The currents have directions, specified by the arrows.

I know, but I still don't see how I can find my equations. This is a very weird circuit. It looks like you can have many possibilities. EX. i1=i3+i4-i2 or i2=i3+i4-i1 (these 2 are the same thing actually). i5=-i3-i4 or i3 = i5+i4 (not sure about this one).

 

[Marcin H]

OK.

One of the first things to learn here is to make it clear as to what post (you did that fine) or what part of a post you are responding to.

If you were asking me, about that thing I did, and that thing was posting those images, see the following. Otherwise, Doc Al and others, will continue working with you.

In (Microsoft) Windows, there is a "Snipping Tool" application. Find it in the Applications folder in the list of programs.

It will take a 'snapshot' of the portion of your monitor's screen, that portion being determined by what you highlight. You may need to use the help feature provided by the Snipping Tool or google for more details.

Once you take a snapshot of whatever is of interest, you simply paste it into the reply window of PF .

So I just copy paste a picture from a word document and it will display in a post like in your first post? Can I drag and drop pics from my desktop? Or is there a way to post them the way you did. i usually take a screenshot on my mac and post that.

 

[SammyS]

So I just copy paste a picture from a word document and it will display in a post like in your first post? Can I drag and drop pics from my desktop? Or is there a way to post them the way you did. i usually take a screenshot on my mac and post that.

The Snipping Tool allows you to take a snap shot of a relatively small portion of the screen, rather than snapping the whole screen. I'm not very familiar with the details of what's available on the Mac.
I used the pdf file viewed in adobe reader and snapped a portion of that.

Yes, you can do the same with a word document.

 

[cnh1995]

I know, but I still don't see how I can find my equations. This is a very weird circuit. It looks like you can have many possibilities. EX. i1=i3+i4-i2 or i2=i3+i4-i1 (these 2 are the same thing actually). i5=-i3-i4 or i3 = i5+i4 (not sure about this one).

You have four options. Check each option. You are asked to find which option(s) is(are) correct.

 

[Doc Al]

I know, but I still don't see how I can find my equations. This is a very weird circuit. It looks like you can have many possibilities.

There are three 'junctions', each marked by a dot. So you can have three equations.

EX. i1=i3+i4-i2 or i2=i3+i4-i1 (these 2 are the same thing actually). i5=-i3-i4 or i3 = i5+i4 (not sure about this one).

Good! But there's one more.

Now just check to see which of the given equations are consistent with those.

 

[Marcin H]

There are three 'junctions', each marked by a dot. So you can have three equations.Good! But there's one more.

Now just check to see which of the given equations are consistent with those.

One more? I can't find another. Re-arraging them i get A,B, and D fro my answers. i1+i2=i3+i4, i5=i1+i2, and i3+i4+i5=0

 

[Doc Al]

One more? I can't find another.

Do you agree that the diagram shows three junctions? Each one deserves its own equation.

 

[Doc Al]

i1+i2=i3+i4,

Good.

i5=i1+i2,

Almost, but not quite.

i3+i4+i5=0

Good.

 

[Marcin H]

Do you agree that the diagram shows three junctions? Each one deserves its own equation.

Hmm. So am I missing an equation for the junction above box C? it looks like i1, i2, and i5 would be coming out of that junction. So Does that mean i1+i2+i5=0?

 

[Doc Al]

Hmm. So am I missing an equation for the junction above box C? it looks like i1, i2, and i5 would be coming out of that junction. So Does that mean i1+i2+i5=0?

Exactly!

 

[Marcin H]

Exactly!

Ok I see that now. Thanks! I have another question about Kirchoffs Voltage Law related to a different question. Should I make a new thread or may I ask here?

 

[Doc Al]

Go ahead and ask it here, if it's closely related.

 

[Marcin H]

Go ahead and ask it here, if it's closely related.

Ok. Did I do this problem correctly? I looked at the equations in the answers and then made loops that included those voltages and checked if they make sense. I don't know if this is the best way of looking at it. I think the correct answers are A, D, E.

Also, quick side question about units. When using equations like P=IV or V=IR we use AMPS for our unit for current right? So if we are given 100mA we have to use .1A in the equation right?

 

[Doc Al]

Ok. Did I do this problem correctly? I looked at the equations in the answers and then made loops that included those voltages and checked if they make sense. I don't know if this is the best way of looking at it. I think the correct answers are A, D, E.

Sounds good to me.

Also, quick side question about units. When using equations like P=IV or V=IR we use AMPS for our unit for current right? So if we are given 100mA we have to use .1A in the equation right?

Yes, standard units are amperes for current, so 100 mA = .1A. Good!

 

[Marcin H]

Awesome! Thanks!

 

[Marcin H]

Couple more questions. I think I did 18A correctly, but it looks like I did something wrong in 18B and 18C. Shouldn't my three answers from (C) add up to 1W, my asnwer in (B)? Not sure where I went wrong here.

 

[Doc Al]

I think I did 18A correctly,

What you found is the total resistance of the circuit.

 

[Marcin H]

What you found is the total resistance of the circuit.

Oh, true... Hmm. So how else can I find the resistance? Should I find the voltage drop across the 20 ohm resistor and then subtract that from the source voltage and use that V in the equation? Or is their a different way.

 

[Doc Al]

Well, you found the total resistance. The only unknown is R, so solve for it.

 

[Marcin H]

Well, you found the total resistance. The only unknown is R, so solve for it.

Oh... lol. I keep over thinking everything. 100-20-60=20ohms. I am so used to harder circuits from E&M that now I'm making dumb mistakes over easier problems. It all works out now. Thanks!

 

[Marcin H]

Well, you found the total resistance. The only unknown is R, so solve for it.

Hello again! Sorry have one last question! I thought I had this right, but then I started thinking about it and I'm not sure anymore. My thought was since the voltage was changing and the current was constant, that the source was a voltage source. I kinda thought of it as you changing the voltage on a power supply or something keeping the current constant. Is that right? I feel like it's not.

 

[Doc Al]

Hello again! Sorry have one last question! I thought I had this right, but then I started thinking about it and I'm not sure anymore. My thought was since the voltage was changing and the current was constant, that the source was a voltage source. I kinda thought of it as you changing the voltage on a power supply or something keeping the current constant. Is that right? I feel like it's not.

You had it right the first time. Read this: Voltage and Current Sources

 

[Marcin H]

You had it right the first time. Read this: Voltage and Current Sources

I am very confused right now. So are you saying that the graph shows a constant voltage source? My professor said that if you have a constant current, then it is a current source, but i don't think that's right. The slope for the graph on that problem would be zero which means zero resistance. The hyperphysics link agrees with that too. Here are my professors notes:

Edit: Also, I asked this here yesterday and gneill seems to say otherwise.
https://www.physicsforums.com/threads/voltage-or-current-source.854007/#post-5356009

 

[Doc Al]

So are you saying that the graph shows a constant voltage source?

No. Current is on the vertical axis and is constant.

My professor said that if you have a constant current, then it is a current source, but i don't think that's right.

Your professor is correct.

The slope for the graph on that problem would be zero which means zero resistance.

Note that is is a Current vs Voltage graph, so be careful how you interpret the slope.

The hyperphysics link agrees with that too.

Not sure what you mean here. Hyperphysics agrees with your professor (and with me).

 

[Marcin H]

Note that is is a Current vs Voltage graph, so be careful how you interpret the slope.

How can you interpret the slope then? Slope is rise over run, so i/v. Oh, which is not resistance. REsistance is v/i. So what does the slope tell us exactly?

 

[Doc Al]

So what does the slope tell us exactly?

You can think of it as 1/R going to zero.

Or just imagine how you'd draw the v/i graph.

 

[Marcin H]

Ohhhhhhh. Ok, that makes sense now. I had it flipped. Now makes me question why show the graph like that -_- Oh well. Thanks for clarifying!