FERGUSON I.T. ELECTRONICS STUDY Lab#1:  Ex#4.   ===================================

   Now lets explore the effects, on the remaining curcuit,
of changing the value of R5 which is a variable resistor.
We will see the effect of making R5 really huge and then very small.
Plug in  R1,R2,& R3=1 ohms,  R4, R5, & R6=4,  R7, R8, & R9=5 ohms
... then observe the effects of plugging in
 R1,R2,& R3=10 ohms & R5=.05 ohms

If R1,2,3 were lamps, what would be the effect of varying R5? Why?
Hint: Current flows through the lest resistive branch of any curcuits.
How did current through R5 & R1,2,3 change?

R1= R2= R3= R4= R5= R6= R7= R8= R9=!!

I1 current = Itot - I5 - I4
Rs = R1 + R2 + R3 = ohms. Rt1 = (R4 x Rs)/(R4 + Rs) =(x)/(+) = ohms. (black section) Rg = R6 + R7 + Rt1 = ohms. (green section) Rt2 = (R5 x Rg)/(R5 + Rg) = (x)/(+) = ohms. (black & green sec.)
Rt = R8 + R9 + Rt2 = ohms. (blue branch resistance)

Itot = 12volts / Rt = amps
(Answer: Itotal current =.9293) when R5 was 4 ohms.

I5 = Itot x current divider (Rg/[Rg+R5])= amps !!!

In this present example . . after changing R1,2,3 & R5 ..
I1 (thru R1) << I5
I1 current = Itot - I5 - I4 = .9241 amps - .6897 amps - I4 amp!!!
but what is I4 ???
I7 = Itot - I5 = .9241amps - .6897 amps = +.2344 amps !!!
I4 = I7 x current divider = .2344 amps x 9/(9+4) = +.16 amps !!! so ... I1 = I7 - I4 = .2344 - .16 = +.0744 amps !!!!

checking ... when you add'em back together at node A
I4 + I1 + I5 = .16 + .0744 + .6897 = .9241 amps !!

The current through R5, after changing it to .05 ohms and R1,2,3 to 10
Itot - I5 = .0018 amps !!!!
How does I5 compare to Itot and I7 ? Notice that the current thru the branches with the most resistance had the lowest currents
because the voltage and current for a resistor are indirectly proportional to eachother,
so when one goes up the other one goes down and vice versa.
When you finish tryings the higher R5, send an email to zjev06@gmail.com.
Give a summary of the results of each and tell me what you have learned.

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