Electronics Anyone?

[Mickey13]

I have been trying to get my brain around Kirchoff's Laws all day and have been given this as an exercise:

http://www.pa.msu.edu/courses/1998spring/PHY232/lectures/kirchoff/examples.html

The first example on the page is the same as mine but the first battery cell in it is facing the other way around. I have been given the B1 value of 6, the B2 value of 8 (for the battery cells), and I have been given R1, 2 and 3 values of 3, 5 and 2 but I simply cannot figure out how to determine the current at the points of R1, R2 and R3. I don't know the calculation, it explains it on that page but I can't understand it. I have a circuit diagram and analysis program which I am using to check results but I just can't get the results there.
If anyone knows a little about this subject could they explain how to work out those current values in ENGLISH!? Maybe it's because I'm tired but I'm finding it hard to get what I need from there and the other pages I have tried are ten times worse.
Maybe someone here could help me.
Thanks.

[jrbishop79]

Originally posted by Mickey13
I have been trying to get my brain around Kirchoff's Laws all day and have been given this as an exercise:

http://www.pa.msu.edu/courses/1998spring/PHY232/lectures/kirchoff/examples.html

and I have been given R1, 2 and 3 values of 3, 5 and 2
Thanks.

are those resistors supposed to be k ohms or just plain ohms?

let me piddle with it and I'll get back to you...

[Mickey13]

Just plain ohms I think, it doesn't say. Does is really matter? Surely the equation for working it out will be the same.

[Grettir]

Id say that you should probably start by redrawing the circuit so that it looks a little more like something easy to solve.like for example start from Va and go down, then when you get to the wire split draw that as if it is just Vb in series with R3, with those two series items in parallel with R2. Then would come R1 in series with Va, and in series with this parallel thing you just drew. Now doing something like node voltages seems a lot easier. btw do you know node voltages? its basically the concept that you define some nodes , usually ata point where the wires split, and then the current through a particular resistor is simply the difference of the two nodes that it is between divided by the resistance (ohm's law). and you should define one of your nodes to be ground, so now you got one less unkown. Does this make sense? hope so, if not feel free to ask more questions

[jrbishop79]

well since Va and Vb are in series aiding, then the current through R2 will add. Try this:

take Vb and R3, and R2 out of the circuit and analyze what's left with Ohm's Law. Then put Vb and R3 back in the circuit, and take out Va and R1. Add the two series currents to get the current through R2.

hope this helps.

[edit] They subtract... my mistake...

[EnigmaOne]

Hi, Mickey13,

Circuit Analysis can be a lot of fun--if you enjoy electronics merely for electronics' sake. :::grin:::

You can make the problem a trivial one by using Mesh Analysis.

Pick a reference point in each loop of the diagram...how about bottom center of each loop?

Assign a clockwise direction of current around each loop (or window) and label the left-hand current loop 'I1'. Label the right-hand current loop 'I2.'

Since I2 runs counter to I1, you'll be subtracting the current through this resistor in each of your equations ('-5I1' and '-5I2') when you write them.

Moving clockwise around loop 1 you have:

a voltage drop (+ to -) for V1 ( -6V )

a voltage rise (- to +) for R1 ( +3 * I1 )

a voltage rise (- to +) for R2 ( +5 * I1 )

and a voltage drop (+ to -) for R2 ( -5 * I2 )
because I2 runs in the opposite direction, so we have to account for that as well.

Using KVL, we can write the first mesh equation as:

( 3 + 5 )I1 - 5I2 = -6V
8I1 - 5I2 = -6V

Moving clockwise around loop 2 you have:

a voltage rise (- to +) for R2 ( +5 * I2 )

a voltage drop (+ to -) for R2 ( -5 * I1 )
Again, because I1 runs in the opposite direction, so we have to account for it.

a voltage rise (- to +) for R3 ( +2 * I2 )

and a voltage drop (+ to -) for V2 ( -8V ).

KVL gives us the second mesh equation as:

( 5 + 2 )I2 - 5I1 = -8V
7I2 - 5I1 = -8V

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~

Re-ordering terms gives us the following set of equations:

I1: 8I1 - 5I2 = -6V
I2: - 5I1 + 7I2 = -8V

Cranking these through determinants will give you:

I1 = 2A
and
I2 = 4.4A

(If I've had enough coffee this morning, that is. It's been years since I've messed with CA, and I'm going from dim memory here.)
BTW: Don't mess around with Branch-Current analysis methods. BC is only a stepping stone to learning Mesh Analysis; which is the more generalized--thus useful--form.
HTH

[edit]
I just looked back over your original definitions, so I'll clarify just a bit further:
I1 above is the loop current for the left-hand loop.
I2 above is the loop current for the right-hand loop.

Obviously, I1 is the current through R1 (2A), and
I2 is the current through R3 (4.4A).

The current through R2 will be the algebraic sum of I1 and I2, which is:
I2 - I1 = 2.4A.