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calculation dilemma

Thread starter #3
I can think of at least two ways to calculate the total resistance - thevenin substitution or kirchhoff laws.
That is were I intend to use this example. Except: for Thevenin I need to calculate exactly this substitution resistance. So how do you calculate this substitution resistance for this example?
 

kubeek

Well-Known Member
#4
I would transfer R1-R3 to its equivalent, and R2-R4. Now you get a series circuit Vt13 - Rt13 - R5 - Rt24 - Vt24. From that you get current through R5 and voltages at both sides of R5. Find the rest of the currents and then plop the whole mess into ohms law and get the total resistance of the circuit.
 
Thread starter #5
I would transfer R1-R3 to its equivalent, and R2-R4. Now you get a series circuit Vt13 - Rt13 - R5 - Rt24 - Vt24. From that you get current through R5 and voltages at both sides of R5. Find the rest of the currents and then plop the whole mess into ohms law and get the total resistance of the circuit.
How do you calculate the "equivalent" of R1 and R3 when there is R5 in between?
You cannot have it in series, and not with R5 in parallel because R5 belongs to the R4 and R2 branch.
 

dknguyen

Well-Known Member
#6
I would do mesh analysis and pull the equivalent resistance out of that. R5 is practically made for that here. Parallel/series resistor simplification, thevnin, etc are shortcuts, not fundamental methods and so have their limitations.
 
Last edited:

Ratchit

Well-Known Member
#7
What is the correct way to calculate the substitution resistance of the schematic attached?
Do you mean the Thevenin resistance, also known as the output resistance between Vcc and ground?

Method 1) Insert an independent voltage source in series with R3 and ground. Calculate the output voltage Vcc. Then calculate the current of Vcc with it shorted to ground. Using the definition of resistance, the equivalent resistance is Vopen/Ishort.

Method 2) Insert a 1 amp independent source at terminals of Vcc and ground. Then calculate the voltage of Vcc. The voltage value will also be the resistance value because you are using a 1 amp current source

Ratch
 
Thread starter #8
I would do mesh analysis and pull the equivalent resistance out of that. R5 is practically made for that here. Parallel/series resistor simplification, thevnin, etc are shortcuts, not fundamental methods and so have their limitations.
Interesting. Can you give plse a shortcut or link on how to perform mesh analysis? Thks!
 
Thread starter #9
Do you mean the Thevenin resistance, also known as the output resistance between Vcc and ground?

Method 1) Insert an independent voltage source in series with R3 and ground. Calculate the output voltage Vcc. Then calculate the current of Vcc with it shorted to ground. Using the definition of resistance, the equivalent resistance is Vopen/Ishort.

Method 2) Insert a 1 amp independent source at terminals of Vcc and ground. Then calculate the voltage of Vcc. The voltage value will also be the resistance value because you are using a 1 amp current source

Ratch
Method 1: this schematic is actually the result of having this actual voltage source shorted out and the resulting schematic is what is shown in post 1.
Back to square 1: I need the equivalent resistance of what is shown in post 1.
 

Ratchit

Well-Known Member
#10
Method 1: this schematic is actually the result of having this actual voltage source shorted out and the resulting schematic is what is shown in post 1.
You showed me a schematic, I described two methods to obtain a result.

Back to square 1: I need the equivalent resistance of what is shown in post 1.
Then follow what I described.

Ratch
 

kubeek

Well-Known Member
#11
How do you calculate the "equivalent" of R1 and R3 when there is R5 in between?
You cannot have it in series, and not with R5 in parallel because R5 belongs to the R4 and R2 branch.
Assume 1V on the input if you please, it makes life a bit easier. Thevenin equivalent of R1 and R3 (with 1V applied) is a Vt13=0.6V source with Rt13=12 ohms in series. Equivalent to R2-R4 is Vt24=3/7V and Rt24=30||70ohms in series. Now you have a simple circuit with two voltage sources and three series resistors, Vt13 - Rt13 - R5 - Rt24 - Vt24. Get current through R5, and backtrack to get current through the rest of the resistors. Then calculate the total resistance from R=1V/Itotal.
 
Thread starter #12
Assume 1V on the input if you please, it makes life a bit easier. Thevenin equivalent of R1 and R3 (with 1V applied) is a Vt13=0.6V source with Rt13=12 ohms in series. Equivalent to R2-R4 is Vt24=3/7V and Rt24=30||70ohms in series. Now you have a simple circuit with two voltage sources and three series resistors, Vt13 - Rt13 - R5 - Rt24 - Vt24. Get current through R5, and backtrack to get current through the rest of the resistors. Then calculate the total resistance from R=1V/Itotal.
Lile this (attached)? By the way: Rt24 = 30||40 ohms, not 30||70?
 

Attachments

kubeek

Well-Known Member
#13
Up until point 3 it is correct. After that, calculate the voltage at points left and right of the R5 - those are the same voltage as in the original schematic. Then simply use ohms law to find the current in the top two resistors, add those two together, should yield 22.62mA and 12.41mA, so total 35.03mA, therefore total resistance of about 28.55ohms.
 

dknguyen

Well-Known Member
#14
Interesting. Can you give plse a shortcut or link on how to perform mesh analysis? Thks!
This has everything you should need:
https://en.wikipedia.org/wiki/Mesh_analysis

You assign loops currents and assume a direction for each loop. You can decide to have adjacent loops or to have a smaller loop inside a big loop. It doesn't matter. Pick what is easiest. In your schematic, remember that the +V and GND form a loop on their own.

Then trace V=IR around each loop and solve.
 

Ratchit

Well-Known Member
#15
Lile this (attached)? By the way: Rt24 = 30||40 ohms, not 30||70?
Show below are 3 solutions. The first solution is a mesh solution for the Vcc loop on the left side of the circuit. The second solution is a mesh solution for the Vcc loop on the right side of the circuit. The third solution is my method 2 suggestion from above showing a node solution with a 1 amp current source applied to the Vcc terminal. The calculated voltage value of the third method is the same as the impedance because the current source is 1 amp. Better learn how to use these methods or you will wander around in a wonderland of perpetual fog. All the methods below use 3 equations and solve for 3 unknowns.

earckens.JPG
As you can see, the impedance is 685/24 ohms.

Ratch
 
Thread starter #16
(...) After that, calculate the voltage at points left and right of the R5 - those are the same voltage as in the original schematic. (...)
Please can you explain?

Using Thevenin I understand your reasoning for Vt13, Rt12, Vt24, Rt24. But to calculate the voltages left and right of R5, and taking my reasoning for points 1 and 2 is correct ( ie: I1 = (0.6 / (12+10+17) and I2 = (3/7) / (12+10+17) ) , what schematic must be used for that calculation?
 
Thread starter #17
Show below are 3 solutions. The first solution is a mesh solution for the Vcc loop on the left side of the circuit. The second solution is a mesh solution for the Vcc loop on the right side of the circuit. The third solution is my method 2 suggestion from above showing a node solution with a 1 amp current source applied to the Vcc terminal. The calculated voltage value of the third method is the same as the impedance because the current source is 1 amp. Better learn how to use these methods or you will wander around in a wonderland of perpetual fog. All the methods below use 3 equations and solve for 3 unknowns.

View attachment 114842
As you can see, the impedance is 685/24 ohms.

Ratch
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?
 

kubeek

Well-Known Member
#18
Please can you explain?

Using Thevenin I understand your reasoning for Vt13, Rt12, Vt24, Rt24. But to calculate the voltages left and right of R5, and taking my reasoning for points 1 and 2 is correct ( ie: I1 = (0.6 / (12+10+17) and I2 = (3/7) / (12+10+17) ) , what schematic must be used for that calculation?
Sorry missed those currents before, not sure where those are in the schematic.

Voltage between R1,R3 and R5 is the same as between R6, R7 and R8. So if you know voltage at that point, you also know current through R1. Do the same for R2, add the two together and you get the total current.
 

Attachments

Thread starter #19
Sorry missed those currents before, not sure where those are in the schematic.

Voltage between R1,R3 and R5 is the same as between R6, R7 and R8. So if you know voltage at that point, you also know current through R1. Do the same for R2, add the two together and you get the total current.
If understand well, using Thevenin:

1. Take my schematic from post 1 (or from your post 18, top drawing), add a 1V source to the top (Vcc), substitute the right side with Ut24 (calculated 3/7 V = 1V x 30/(30+40) ) and Rt24 = 30||40 = 17Ohm, the left side with Ut13 (calculated 0.60V = 1V x 30/(30+20) )and Rt13 = 20||30 = 12Ohm

2. Resulting schematic = your schematic, bottom drawing in post 18

3. a. Short V3 and calculate current from V2: 0.6V / (10+12+10) = 0.0188A
b. Short V2 and calculate current from V3: (3/7V) / (10+12+10) = 0.0134A
c. Sum of both a and b = 0.0321A = total current

4. R(substitute) = 1 / 0.0321 = 31.1Ohm

Something is wrong because this is different from 28.55Ohm of your post 13
 

kubeek

Well-Known Member
#20
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.
 

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