How can I come up with a equivalent resistance?

[i55BaSH]

Greetings,

I have the following circuits (picture attached):

V1 = 9 V
R1 = 1 kΩ
R2 = 7 kΩ
R3 = 7 kΩ
R4 = 0.7 kΩ
R5 = 1.5 kΩ
R6 = 1 kΩ

And I am trying to come up with an equivalent resistance so that I can calculate the voltage ( using the voltage divider formula ) at Node.

For the picture in the left I can come with an equivalent resistance:

Req_a = R3 and R4 in series as one
Req_b = Req_a and R2 in parallel as one
Req_c = Req_b and R5 in series as one

The final formula results ( under left picture ).
I have a big problem trying to come up with this approach on right picture.
Using the voltage divider formula, I just know that the equivalent resistance should be ~ 4133.6 Ω. Can you please help me?

Best regards,

 

[danadakk]

You have the V applied to the entire network, and the current going into
it. So what is the Requiv of the entire Network ? E = I x R or R = ?

Then you have a R in series with the compound network, so now what
is Requiv of a series of two R's ? Rtotal = Rseries + Requiv .........

 

[i55BaSH]

I am sorry but, I cannot follow you.

I just want to calculate the potential difference at Node ( green in pictures ) and for that I am using the voltage divider formula with R1 and a equivalent resistance.

V_Node = Req / ( Req+R1 )

In the left picture, for the red rectangle I managed to come up with a equivalent resistance for the entire block so that I can use the voltage divider formula.

In the right picture, I added the 6th resistor and I am stunned that I cannot come with the equivalent resistance for the entire orange block.

If it is not much to ask, can you help me with baby steps? Do I need to use Nodal Analysis? or is it there a method to compute a equivalent resistance for the orange block.

 

[KlausST]

Hi,

what´s the exact question?
What values are given .. and what values do you need to calculate?

Something´s wrong with circuit#2. I guess there is a problem with one given value. Please check all numbers carefully.

Klaus

 

[i55BaSH]

Hi Klaus,

At Node I want to calculate the potential difference using the voltage divider formula:

For circuit 1:

V_node = V1 * (Rb / (Ra+Rb)) -> where Ra = R1 and Rb is the equivalent resistance of R2, R3, R4, R5 as following:

(( R3 series R4 ) || R2 ) series R5

For circuit 2:

7.24V = 9V (Rb / (1kΩ+Rb))

using the voltage divider formula and knowing that on Node I have 7.24V it resulted Rb = 4133.6 Ohm as equivalent resistance.

But my question is:

How Do I arrange R2,R3,R4,R5,R6 so that it gives me that value? ( just like I explained for circuit one)

 

[barry]

you’re going to have either do a mesh analysis or node analysis. There may be another way, but that’s how i’d approach it.

 

[FvM]

To evaluate equivalent resistance incrementally, you can apply delta-wye transformation, either to R2-R4 or R4-R6. The remaining circuit can be solved by simple parallel and series resistor calculation. https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/

 

[danadakk]

Are the currents and R values, as shown in right hand pic the "givens" ?

If so the node V is simply Vnode = 9V - 1.76 mA X R1.......

But to use V divider you need the equiv R at Vnode. Using above Vnode, and you know the total
current going into the complex part of the network, you have -

Requiv = Vnode/1.76 mA.....

But if you want to do the derivation of Requiv, FVM would be the way at getting at Requiv, has
the advantage of not needing any I's, just Vsource (9V).....

 

[FvM]

Don't expect that 1.76 mA is exact solution. It's a simulator result with limited resolution. Exact Req is 4.1053k, corresponding to 1.7629 mA.

 

[BradtheRad]

Could it be that R4 doesn't enter into the calculation at all? For R2 R3 R5 R6 the result is:

a) Sum series components R2 R5 =8500. Take reciprocal=.000117647

b) Sum series components R3 R6 =8000. Take reciprocal=.000125

c) Sum reciprocals=.000242647. Take that reciprocal=4121.21 Ω.

--- Updated Dec 28, 2024 ---

That's if R4 is absent or infinite. And even if R4 is 0Ω (straight wire), the result is 4100Ω.

 

[Easy peasy]

This is a classic Kirchhoff's law problem - write equations for each loop and solve - then check with LT spice

 

[FvM]

Could it be that R4 doesn't enter into the calculation at all?

No. As stated, you get printed resistance value of 4136.6 if you use rounded simulated current values for the calculation, but it's surely considering R4. Obviously the exercise problem is asking for an exact calculation method. That's either the normal way (post #6, #11) or delta-wye transformation if an equation system with five currents is over your head.

 

[yongmin capacitor]

你需要复习一下串联和并联的定义，然后这个问题就迎刃而解了，

，

[moderator provided translation. Please post in English in the future]
"You need to review the definitions of series and parallel, and then the problem will be solved."

 

[FvM]

You need to review the definitions of series and parallel, and then the problem will be solved

Correct for the first, but not the second circuit which can't be solved by only applying series/parallel conversions.

 

[Cjey]

It's hard for me to appreciate where the question is leading. If you were interested in a particular resistor I could thevenise it for you. However, I believe the only way to learn is by doing. check out this link

I hope it helps.