Potential divider problem

[KrisBlueNZ]

Yes thanks for the response, it absolutely makes sense. One other thing though if I may ask, if we pretend R2 isn't there, why are we calculating total resistance of R1 and R3 as though they were in parallel, afterall when we pretend that R2 is not there, aren't R1 and R3 in series and total up to 400 ohms instead of 75 ohms??

That's how Thevenin equivalents work. Here's a simple example that should make it clear:

. . . . . . . . . . . 2k . . . . . . . . . . 2k
+10V -----------\/\/\/\/\/--------------\/\/\/\/\/------------ 0V
. . . . . . . . . . . . . . . . . . |
. . . . . . . . . . . . . . . . . . ------ point "X"

There are two 2k resistors connected in series across a 10V power source. They act as a voltage divider, so point "X" will sit at +5V (assuming nothing else is connected to point "X" to affect the voltage).

This circuit has a "Thevenin equivalent" - it is equivalent to a +5V voltage connected through a resistance to point "X". This resistance is calculated as the PARALLEL combination of the two resistances. 2k in parallel with 2k is 1k.

The Thevenin equivalent of the circuit above is:

. . . . . . . . . . 1k
+5V -----------\/\/\/\/\/--------- point "X"

 

[24Volts]

So, its the resistance as seen from point X !

Got it!

One last question KrisBlueNZ.

If my 2nd voltage was +5VDC instead of -5 VDC, Vnode would be 6.818 VDC right?

Thanks KrisBlueNZ

 

[(*steve*)]

24Volts, you need to sum the currents, so its i1 + i2 + i3 = 0

It's equal to zero because KCL says that if you look at any node, the sum of the currents flowing into that node equal the sum of currents flowing out of that node. If you sum all of those currents you get zero.

Now to do this, you MUST decide that if currents flowing into the node are positive, then currents flowing out are negative.

In Harald's diagram, you will note that he has drawn all the currents as flowing toward the node. This means they MUST sum to zero. So you simply add them up.

That's pure KCL and it doesn't get you all the way to a solution, in fact it leaves you with an equation that has several unknowns.

What Laplace does (in a transformative manner ) is to use the node method, which in this case describes the currents in terns of a single unknown. That unknown is the voltage at the node. These can be summed, and you have an equation with a single unknown that can be immediately solved.

The only caveat (and this applies to many of these calculations) is that you need to ensure that you get your signs right. In this case, it is fairly clear by inspection that the actual current flow for i2 is going to be in the direction opposite from the arrow. This has no effect on the validity of the process (in fact we expect it) but is a trap for young players in getting the math right.

The node method does require more discipline in getting your signs right because an error can be harder to spot.

 

[Laplace]

The node method does require more discipline in getting your signs right because an error can be harder to spot.

Actually, I use the node method because I'm no good at discipline. Node equations do not deal with currents, only node voltage. So with no current there is no concern with the direction of current flow (unless the circuit includes current sources which then does require some discipline).

Without current sources every term in the node equation that sums to zero is of the form:

(Vnode - Vothernode)/Zbetweennodes + (Vnode......

What makes it simple is that, in the equation for each node, every term begins with (Vnode - ........
That makes it simple enough for my undisciplined mind to get it right most of the time.

 

[24Volts]

Steve,

I don't get it, you say:

24Volts, you need to sum the currents, so its i1 + i2 + i3 = 0

and KrisBlueNZ says:

So the total current is:
59.1 - 45.5 - 13.6 = 0 mA as a final check.

So, steve you must mean that the current coming into a node must equal 0 by taking the highest current coming in the node and subtracting all other currents... no? KrisBlueNZ's way surely affirms the result being correct!!

Anyways I sort get both ways (KrisBlueNZ and Laplace) ways!

Pyrohas, sorry for messing up your post, it's just that I found your question interesting and really wanted to learn this. So thanks again!

24v

 

[KrisBlueNZ]

24Volts, it's a matter of the polarities of the currents. Currents that feed INTO the node are positive; currents that feed OUT of the node are negative. In this case, i2 (45.5 mA) and i3 (13.6 mA) both feed OUT of the node. So my formula could be rewritten as:
59.1 + (-45.5) + (-13.6) = 0 mA as a final check.
I considered writing it like that at the time. I guess I should have!

 

[24Volts]

24Volts, it's a matter of the polarities of the currents. Currents that feed INTO the node are positive; currents that feed OUT of the node are negative. In this case, i2 (45.5 mA) and i3 (13.6 mA) both feed OUT of the node. So my formula could be rewritten as:
59.1 + (-45.5) + (-13.6) = 0 mA as a final check.
I considered writing it like that at the time. I guess I should have!

ah! okay, wow, there's a lot of detail in a circuit like this!

I get it now. Thanks KrisBlueNZ.

24v