- #1
That's correct.Marcin H said: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
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 said:Looks good to me.
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.)Marcin H said:I'm haven't seen this kind of circuit yet, so I'm not sure how to apply kirchhoffs current law.
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.Marcin H said: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.
It looks like you specified the units of the result as mA. That would be milliamps. Was that your intention?Marcin H said:Did I do number 14 correctly? I have a feeling it's not that simple.
Good eye!gneill said:It looks like you specified the units of the result as mA. That would be milliamps. Was that your intention?
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.Doc Al said:for any point, the currents going into the point must equal those going out of the point.
Fixed. Thanks!gneill said:It looks like you specified the units of the result as mA. That would be milliamps. Was that your intention?
Sorry, I'm new here. How did you do that exactly?SammyS said: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.
The currents have directions, specified by the arrows.Marcin H said:How do you know what is going in and what is coming out exactly.
OK.Marcin H said:Sorry, I'm new here. How did you do that exactly?
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).Doc Al said:The currents have directions, specified by the arrows.
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 said: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 .
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.Marcin H said: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.
You have four options. Check each option. You are asked to find which option(s) is(are) correct.Marcin H said: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).
There are three 'junctions', each marked by a dot. So you can have three equations.Marcin H said: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.
Good! But there's one more.Marcin H said: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).
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=0Doc Al said: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.
Do you agree that the diagram shows three junctions? Each one deserves its own equation.Marcin H said:One more? I can't find another.
Good.Marcin H said:i1+i2=i3+i4,
Almost, but not quite.Marcin H said:i5=i1+i2,
Good.Marcin H said:i3+i4+i5=0
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 said:Do you agree that the diagram shows three junctions? Each one deserves its own equation.
Exactly!Marcin H said: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?
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 said:Exactly!
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.Doc Al said:Go ahead and ask it here, if it's closely related.
Sounds good to me.Marcin H said: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.
Yes, standard units are amperes for current, so 100 mA = .1A. Good!Marcin H said: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?
What you found is the total resistance of the circuit.Marcin H said:I think I did 18A correctly,
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 said:What you found is the total resistance of the circuit.
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!Doc Al said: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 said:Well, you found the total resistance. The only unknown is R, so solve for it.