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I am currently learning how ideal sources may be replaced with linear sources to make the application of other circuit theory possible. Unfortunately, I don't understand how the following conversion is possible. This is the original circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

And my prof. converts it into the following, unfortunately very quickly and with little commentary:

schematic

simulate this circuit

with: $$ V_1^* = V \frac{R_3}{R_1+R_3} $$ $$ R_1^* = \frac{R_1R_3}{R_1+R_3} $$ $$ V_2^* = V \frac{R_4}{R_2+R_4} $$ $$ R_2^* = \frac{R_2R_4}{R_2+R_$} $$

Ultimately, the question in this exercise is what voltage-drop occurs over \$R_5\$, but my question relates more to the conversion above, and less to the accomplishment of said goal. I only mention it, since it may allow some simplifications which would not be possible, would we seek information about other specific nodes or currents, for example.

Let's only consider \$V_1*\$ and \$R_1*\$ since the conversion for the other voltage-source is equivalent.

The voltage itself of course stems from a voltage-division between \$R_1\$ and \$R_3\$. Yet I don't quite see how this comes about, since, going from the cathode and through \$R_3\$, our remaining voltage should at first glance be smaller, the larger \$R_3\$ is. A second glance gives me even more trouble, since I don't see a way for current to travel from the upper cathode to the upper anode while going through the rest of the circuit (since it would ultimately have to be at node A again before going through \$R_1\$ and into the anode.

When calculating the inner resistance, the resistors \$R_1\$ and \$R_3\$ are clearly considered to be parallel. But, considering one of the voltage-sources, I see no way that current could travel from cathode to anode without going through both resistors.

I assume that in both cases my problem is that I consider only one of the sources at a time, while I should consider them both at once. They supply the same voltage, and are oriented so that no potential-difference occurs between their cathodes, which means that there is no potential-difference between their anodes as well. In this exercise they actually stem from a single source, which got split into these two to make a simplification of the circuit possible.

So, why can we make the above replacements, and why are the relations the way they are? Am I right that I should consider both sources at the same time? Some explanation and/or pointing to relevant theory would be highly welcome.

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    \$\begingroup\$ Another view is to convert a voltage source and its series resistor into its Norton equiv. That puts two resistors in parallel, which can now be combined to yield a simpler Norton, which can now be converted back to a Thevenin equiv. Or has been discussed, just convert each source and resistor divider pair into its equiv Thevenin. Same thing either way. \$\endgroup\$
    – jonk
    Jan 17 at 10:41
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For general notes on the thevenin-theorem, you may consult Christianidis Vasilleios answer.

In the case of my exercise, and the solution provided by my professor, the procedure was related, but a little different, and I want to explain it now:

Thevenin's theorem states that we can replace any bipolar circuit with only linear elements by an equivalent circuit, containing only one voltage source and one inner resistance.

Let's name the node between the two voltage-sources C, and cut out the upper voltage-source, together with resistors \$R_1\$ and \$R_3\$. This is the cut-out circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

This is clearly a bipole, and only contains linear elements, so we may form an equivalent circuit, containing only one voltage source and one resistor.

For the resistor, let's name it \$R_1^*\$, short the voltage source. This will lead to: $$ R_1^* = R_1||R_3 = \frac{R_1R_3}{R_1+R_3} $$

For the voltage, note that whatever load \$R_L\$ we may put between A and C, the voltage-drop across it will equal the voltage-drop across \$R_3\$. (Since \$R_L\$ is in parallel to \$R_3\$. $$ R_L = R_3 $$ There will also be a voltage-drop across \$R_1\$. Here the voltage-divider rule comes into play: $$ \frac{U_3}{U} = \frac{R_3}{R} $$ or: $$ \frac{U_L}{U} = \frac{R_3}{R_1+R_3} $$ This leads to: $$ U_L = U \frac{R_3}{R_1+R_3} $$ which will be the equivalent voltage for our cut-out circuit.

Plugging this new circuit back into the original one will give:

schematic

simulate this circuit

The same may be done with the lower voltage-source together with the respective resistors \$R_2\$, \$R_4\$ leading to the result of the question.

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    \$\begingroup\$ Yes, this is exactly what you have to do. It soon becomes second nature to notice that R1 and R3 are forming a voltage divider across V1 to drive a load, and so that load will effectively be driven by a reduced voltage in series with an effective source resistance which is the two voltage divider elements in parallel. I'd accept nthis one \$\endgroup\$
    – Neil_UK
    Jan 17 at 10:51
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I found my notes for thevening from college:

It needs two steps to be done:
1)Start from the original circuit, make some modifications, find the Rth
2)Start from original circuit again, make some modifications, find the Vth

In order to find the Thevening Resistor:
If you have a Resistor between you A,B, you delete it (so it becomes an open circuit, ant in this spot, at the final circuit, there will be your Vthevening.

If you have power sources, shorτ them (Your 'V') and If you have current sources, you would make it an open circuit and calculate the equivalent resistor (Rthevening).
So your schematic In order to find Rth becomes like this:

schematic

simulate this circuit – Schematic created using CircuitLab

In order to find the Vth:
You begin from your original circuit again, and you Remove the R between A,B so it becomes an open circuit, and you calculate your Vth there. Now, im stuck here, i do not remember how you would calculate the Vth if the circuit is complicated, but if, lets say, you had two Voltage sources in series, you would add them.

schematic

simulate this circuit

Another example from my notes at college (it's in Greek though) and I don't remember the steps really clear, it might help you:

enter image description here

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    \$\begingroup\$ Oh yes, the thevenin-resistance.. Indeed, now I see: each of the upper and lower resistors together with their accompanying voltage-sources form a bipole, and I can simplify them to their thevenin-equivalent. The equivalent resistance is then clear. I must look now into finding the equivalent voltage, I will comment again at that point. \$\endgroup\$ Jan 17 at 9:20
  • \$\begingroup\$ I read my notes again, i edited my answer, it must be more clear now. Check it again. But i dont remember much from Thevening unfortunately. This is all i can help. \$\endgroup\$ Jan 17 at 9:34
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    \$\begingroup\$ Also check: youtube.com/watch?v=zTDgziJC-q8 It might help. \$\endgroup\$ Jan 17 at 9:38
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    \$\begingroup\$ Yes, the organic chemistry tutor is always helpful ;) You pointed me in the right direction, yet your procedure is different from what my prof. did, and so you didn't quite answer the exact question. Your answer is still very helpful, don't get me wrong. I personally think that it is better without the handwritten exercise in Greek though (but that is only my opinion). I posted my own answer now, and will wait for any community-response in the form of voting to decide which answer to accept. \$\endgroup\$ Jan 17 at 10:16

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