Hi again,
Well, first we might want to clear something up here.
If we take a resistor of R1=1k and connect it in series with two other resistors R2=1k and R3=1k, we have to total of 3k.
Now if we replace R3 with a 2k resistor we have a total of 4k. We have 4k partly because R1=1k and it stayed that way even when we changed the value of R3.
Now if we put that 1k back so R3=1k again but this time replace R1 with a voltage source of 2 volts. We have v=2 and R total = 2k so the current is 1ma, and if we look at the voltage source we see 2v with 1ma so 2/0.001=2k, so the voltage source looks sort of like 2k.
Now we replace R3 with 2k again, keeping the voltage source of 2v in place of R1. Now we have a voltage source 2v and two resistors R2 is 1k and R3 is 2k so the total resistance is 3k and 2/3000=6.666e-4 amps. Now 2v/6.666e-4 amps equals 3k, so NOW the voltage source looks like a 3k resistance.
So which is it...is the voltage source like a 2k resistor or like a 3k resistor?
Now in a another experiment, we use two separate voltage sources V1=1v and V2=2v, and some resistors.
Connecting the circuit up, we measure the voltage at some node and find that the voltage is 2.4 volts.
Now we short out V1, and find the voltage at that node is 1.9 volts, and unshort that source and instead short the other source V2 and find the node measures 0.5 volts. Now we add the two voltages, 1.9+0.5 and we get 2.4 volts, the same voltage that we get when we had both sources unshorted. We do this with 100 other combinations of resistors and voltages sources and we always find that the voltage with two sources equals the sum of the voltages with each source taken alone, with the other voltage source shorted out.
So the conclusion is that each source contributes exactly the same amount as if the other source was zero impedance.
Thus we say that the impedance of a voltage source is zero.
Maybe what you are after is the impedance of the circuit with the source removed?
Also, i see no source impedance or load impedance, which would affect the outcome.
Does this all make sense to you?
Ramuna,
You haven't shown the relative polarity between the L2-L3 pair, nor between the L4-L5 pair. You could use the dot convention:
https://en.wikipedia.org/wiki/Dot_convention
or some other means.
I assume you want the impedance across the A-B terminals when the source V is not present. In other words, you want the ratio V/I, where I is the current supplied by the source V; is this the case?
Hi again,
Well first we need to get something else straight...
When you say:
"Down the rabbit hole with MrAl!",
does that mean i am down here all by myself or does that mean that you are down here with me? <ha ha>
Ok so it appears that you want to redefine circuit analysis. So you say that you want to state that a voltage source impedance is equal to the external network impedance rather than zero. This makes some sense because if we have a voltage source 2v and 2k resistor we have 1ma and 2v/1ma is 2k, so the voltage source appears to have a resistance (impedance) of 2k. But what happens when we change the 2k resistor to 4k is we get 0.5ma, so now the voltage source looks like a 4k resistance. So your theory works out
So what you have to do now is show how this viewpoint can help us analyze a circuit because after all if it does not help us then it is of no use.
For example, we have a 2v voltage source and a 1v voltage source. The negative terminals are connected together, and a resistor of 1k is connected between the positive terminals. Show how you solve this circuit for the most necessary quantities using your new theory. We'll take a look at your solution and go from there.
BTW you've made this very interesting
Th
Thank you Mr Electrician Sir! You have correctly identified what I did wrong. I have uploaded the circuit diagram with the transformer poles drawn using the dot convention. Based on this, I think you will agree that my (eq4) on my worksheet page 2of4 should be -L4*i4 + M2*i1 - V = 0. And yes I do want the ratio V/I where I is the current flowing between B and A. Please confirm that with the correction to my eq4, I should correctly derive expressions for i1 - i4 and that the required impedance is V /(i3 + i4).View attachment 80673
Thank you MrAl for your kind remarks
The aggregation of voltage sources, under the infinite impedance voltage source convention, follows the rules of vector addition. In the example you give, both voltage sources have a common ground, and the potential difference between their free terminals is 1 volt. This then is the voltage across the 1k resistor. Consequently the current running through it is 1 mA by the usual application of Ohm's law.
[FONT=Courier New]
CIRCUIT 1:
v
+----R1----+----R2----+
| |
E1 E2
| |
+----------+----------+
|
o
GND
CIRCUIT2:
v
o----R1----+
|
E1
|
GND o----------+
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