This the simple circuit that I am building for a variety of training exercises. I do not have a 12v supply, I have 18v. So I thought that I could use a voltage divider (another training exercise for me to build one).

Back to my questions. Would the readings for the above circuit be the same if the voltage source were a 12v battery or a voltage divided circuit furnishing 12v ?
I hope that made sense cause it's hard to put into words.
Does 12v battery = 12 divided voltage when I hang my resistor/diode load onto either of them.
From my other thread it seemed that my resistor/diode load has an impact on the voltage divided circuit since it was parallel(?)to a portion of it.
Or I could be full of pumpkin seeds as is normal for me

Thanks again

 

What really matters is the value of the series resistor and the wattage rating.
With 18V supply, the resistor has to drop 8V @ 5mA
R = 8000/5 = 1.6kΩ
Power dissipated = 40mW
Hence a ¼W resistor is ok.

If the zener diode is rated at 1W, you can safely increase the current to 50mA.
R = 160Ω
Power dissipated = 400mW
You want a resistor rated for 1W.

If you keep the 400Ω resistor, your zener regulated circuit is good for 8/400 = 20mA with the 18V supply.
Use a ¼W or ½W resistor.

 

Thank you MrChips, but with enormous respect, i am really looking for an answer to my question.
The numbers that you gave me are good and appreciated but I want to go through the exercises that you just did, myself.
I need to be able to solve for these and this circuit has is my test bed.

 

If you have an 18 v source and you want to use it and a voltage divider to replace the 12 volt source and the resistor, then...

You will need to create a source/voltage divider such that the thevenin equivalent is 12 volts/400 ohms.

So you have
Vout = 18 * RL/(RS + RL)
12 = 18*RL/(RS + RL)

and the parallel combination of the resistors = 400

RS*RL/(RS + RL) = 400

You will need to solve those simultaneously.

 

If you have an 18 v source and you want to use it and a voltage divider to replace the 12 volt source and the resistor, then...

You will need to create a source/voltage divider such that the thevenin equivalent is 12 volts/400 ohms.

So you have
Vout = 18 * RL/(RS + RL)
12 = 18*RL/(RS + RL)

and the parallel combination of the resistors = 400

RS*RL/(RS + RL) = 400

You will need to solve those simultaneously.

So then, it appears that if I had a 12v battery instead of the voltage divider I could just plug the components into the breadboard and begin taking meter readings. There would be no need to do fancy calculations.
Correct ?

 

If you want a voltage divider to deliver 5mA, make sure that the voltage divider can deliver at least 50mA.

 

So then, it appears that if I had a 12v battery instead of the voltage divider I could just plug the components into the breadboard and begin taking meter readings. There would be no need to do fancy calculations.
Correct ?

Correct, but I thought you wanted to do it with and 18 volts source for the educational value. Solving those two equations simultaneously is not difficult. The first one simplifies to RL = 2RS, then sub 2RS for each occurance of RL in the second. You will get RL = 1200 and RS = 600.

So you have your 18 volt battery, a series 600 ohm resistor, and a shunt 1200 ohm resistor. The Zener is in parallel to the 1200 ohm resistor. The resulting circuit will preform exactly like the original circuit.

 

Correct, but I thought you wanted to do it with and 18 volts source for the educational value. Solving those two equations simultaneously is not difficult. The first one simplifies to RL = 2RS, then sub 2RS for each occurance of RL in the second. You will get RL = 1200 and RS = 600.

So you have your 18 volt battery, a series 600 ohm resistor, and a shunt 1200 ohm resistor. The Zener is in parallel to the 1200 ohm resistor. The resulting circuit will preform exactly like the original circuit.

Thank you Ylli, that is what I wanted to know. Silly me thought that I could use the divider in place of the battery and everything else would remain the same. HAH ! I mean 12v = 12v, right.
You cannot imagine how much time I spent in trying to figure out why my readings were all over the place.
Now I know that the stupid electrons were doing exactly what I told them, blowing the meter fuse, and smoking resistors.
This exercise has been an eye opener, along with the circuit simplification I saw in the other thread.
I will proceed to solve the problem as well as fiddle around with other components and see if I can predict the meter readings.
Sheesh, I am still on problem 1

 

Long time ago I learnt that, stupid, electrons are not. look somewhere else...

 

Take a look at this voltage divider circuit:

The current through R1 and R2 is 18/1800 = 10mA.

Now add a 10V zener across R2:

Current through R1 is 8/600 = 13.3mA
Current through R2 is 10/1200 = 8.3mA
Hence the current through the zener is 13.3 - 8.3 = 5mA.

Now remove R2.

Current through the zener is 8/600 = 13.3mA

Hence, what is the purpose of R2 except to waste 8.3mA?
With R1 = 600Ω alone plus the 10V zener, you have a 10V supply that is good for delivering 0-13.3mA.

You don't need R2.