calculation dilemma

[Ratchit]

Hi Ratch, I greatly appreciate your input! However you use Kirchoff's laws resulting in 3 equations with 3 unknowns, which I understand. But here I specifically want to use equivalent voltage source and equivalent substitution resistance based on Thevenin theorema.
Both approaches have their merrit and should be able to be used?

I don't think you appreciate my calculations. One way is to determine the resistance from a resultant current when sourced by a given voltage. A second way is to determine the resistance from a resultant voltage when sourced by a given current. Both methods gave the same answer and comply with Thevenin's assertion that a circuit can be reduced to a voltage in series with a resistance, (685/24 ohms) in this case. That is about as Thevenin as you can get. I don't understand your reluctance to accept that.

What you have is a circuit for a Wheatstone bridge. You can collapse the resistors directly by using the delta to Y transformations. The final resistor value will be the Thevenin equivalent.

Ratch
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[earckens]

Those currents are in opposite direction, and you used 10 ohms instead of 17.14. Current through R5 should be then 4.38mA. But anyway, this current is not the total current, only what is flowing between left and right branch through R5. But this current then gives you the voltage at the left and right nodes, which then gives you the current through each of the top resistors.

The current through R5 (4.38mA) gives the voltage across R5. How can this give the voltage at the nodes (with reference to ground)?

 

[kubeek]

That is what the thevenin equivalent does. The 0.6V source with 12 ohms in series is the same thing as the 1V source with 20 and 30 ohms divider. So in post 18 the voltage on left side of R5 is the same as on the left of R8. And since the second circuit is a simple series cicuit, you should esily find what that voltage is and then apply it to the upper circuit and solve for the currents through R1 and R2.

 

[earckens]

That is what the thevenin equivalent does. The 0.6V source with 12 ohms in series is the same thing as the 1V source with 20 and 30 ohms divider. So in post 18 the voltage on left side of R5 is the same as on the left of R8. And since the second circuit is a simple series cicuit, you should esily find what that voltage is and then apply it to the upper circuit and solve for the currents through R1 and R2.

Hi Kubeek, what I do not understand is how the voltage across R8 (being 10 Ohms x 4.39mA = 0.0438V) can be referenced to ground. In the second circuit there are two voltage sources: how do you find the voltage on the left of R8, referenced to ground?
My reasoning would be that current flows from V2 (3/5V) to V3 (3/7V) so the total voltage across R6//R7, R8 and R9/R10 is the difference between the two voltage sources?
Hence the voltage at the node between R6//R7 and R8 = V2 - (4.38mA x R6//R7) = 0.6V - (0.00438 x 12 Ohms) = 0.5474V (referenced to ground) ???

 

[kubeek]

My reasoning would be that current flows from V2 (3/5V) to V3 (3/7V) so the total voltage across R6//R7, R8 and R9/R10 is the difference between the two voltage sources?
Hence the voltage at the node between R6//R7 and R8 = V2 - (4.38mA x R6//R7) = 0.6V - (0.00438 x 12 Ohms) = 0.5474V (referenced to ground) ???

That is exactly right

 

[earckens]

That is exactly right

Thanks!!!

Using the Thevenin theorema is a very helpful way to gain insight in circuit internal impedances and substitution voltage source calculation. That is why I wanted to get to the bottom of this using this procedure.

 

[The Electrician]

Thanks!!!

Using the Thevenin theorema is a very helpful way to gain insight in circuit internal impedances and substitution voltage source calculation. That is why I wanted to get to the bottom of this using this procedure.

Here's a very detailed description of the technique of using Thevenin to solve the unbalanced Wheatstone bridge, which is what you have.

http://www.hallikainen.com/rw/theory/theory6.html

 

[Tesla23]

This type of problem can also be solved with the Extra Element Theorem:
https://en.wikipedia.org/wiki/Extra_element_theorem

by inspection:
\[ Z_\infty = 50||70 = 350/12 \]
\[Z_e^\infty = 60||60 = 30 \]
\[Z_e^0 = 20||30 + 40||30 = 204/7 \]
\[Z = \frac{350}{12}\frac{10 + 204/7}{10 + 30} = \frac{685}{24} \]

 

[Ratchit]

Thanks!!!

Using the Thevenin theorema is a very helpful way to gain insight in circuit internal impedances and substitution voltage source calculation. That is why I wanted to get to the bottom of this using this procedure.

And all this time, I thought that the Thevenin method was used to reduce a complicated circuit to a simple resistance and voltage source. By doing so, it was supposed to obviate the need to perform a detailed and unnecessary analysis of the internal workings of a perplexing circuit. One learns something every day.

Ratch