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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

I am trying to solve this problem but the units dont cancel out, maybe someone can shed some light on where I am going wrong.

These are equations from a circuits problem, applying KCL(nodal analysis) on a circuit. Here are the equations:

Now when I try to solve for Ix, after substituting for V1 and V2 into the second equation, I get stuck here:

I cannot cancel out the Ohm from the denominator of the first term. If I ignore it I get the right answer but I was just wondering if there was something I am missing here.

Thanks,

C25

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi careless25

Without seeing your circuit I cannot be 100% sure, but it looks like k ohms and m A are just the units for the problem.

By Ohms law

So without the units:

so try it like this:

k ohms = 1000 ohms. m A = milli amp

Hope that helps.

Bob

*Last edited by Bob (2011-11-26 20:35:45)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

Thanks for the explanation, I eventually figured out how to cancel the units.

I need help with circuit problems, will you be able to help or do you know any good forums where I can ask my questions?

Thanks,

C25

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi C25

Not sure if I'll always be able to help. It'll depend on the problem. Post here and I'll say if I can help.

There are physics forums but I haven't used them myself, so I won't comment on whether they'd give you what you want.

Bob

*Last edited by Bob (2011-11-27 20:03:45)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

OK heres a question I am having trouble with:

Circuit: https://docs.google.com/open?id=0By2PVt4Tdl6VZjNmMGRmODgtOWVmYy00MTUxLWExNjItZmQ4MTMxMzYyZDAy

I have to solve for

using Thevenin's Theorem.I dont know how to break apart the circuit, I tried just removing the 1kOhm resistor and trying to find V(Open Circuit) and R(thevenin) but I couldnt solve for V(OC).

Thanks in advance,

C25

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi careless25,

Ok, I've had a look at that circuit. Mathematically, I don't have a problem with getting some equations but I'm very confused about the 8 mA and 2 mA symbols. I took those to be ammeters measuing the current passing through, but they are connected in PARALLEL in the circuit so they will split the current and I'll need to take account of their internal resistance.

Is that what you'd expect?

I asked my wife who is (was) a science teacher. She said she'd expect ammeters to be connected in SERIES in order to show the current flowing in that part of the circuit. So, am I to take those as diagrammatic only with no real device at that position.

I've re-drawn the circuit with the two alteratives shown.

I think the second must be the correct version. What do you think?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

Hi,

i think you are over thinking this by a mile.

I'm very confused about the 8 mA and 2 mA symbols

The 8mA and 2mA are ideal current sources. (neglect internal resistance R=0)

12V is an ideal voltage source.

now what I need to do is find the voltage across the 1kOhm resistor by simplifying the circuit. Thevenins therorem: take any two nodes, you want to find the voltage across; the rest of the circuit can be simplified into a voltage source and an equivalent resistor.

For this question what I have to do is find Vo by simplifying the rest of the circuits into a voltage source and a equivalent resistor.

Edit: To simplfy the circuit I can use KCL, KVL and superpositioning.

Let me know if you cant solve it, I will post this in a Physics forum.

Thanks,

C25

*Last edited by careless25 (2011-12-01 02:58:38)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi

I'm having a re-think based on what you've said. But posting elsewhere might be a good idea. This is a long way from my subject area.

If I come up with anything I'll post it.

Good luck with the problem.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

Hi,

I have posted the question here, if you wanted to follow it:

http://forum.allaboutcircuits.com/showthread.php?t=62776

C25

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