Transfrom scheme into eq. scheme with one EMF and one resistor.

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I don't know English well(I'm from russian speaking countries) but I try to explain the problem(and practise my english too)

Homework Statement


Transfrom scheme into eq. scheme with one EMF and one resistor.


Homework Equations


Where is the series configuration of resistors and where is the parallel configuration?

The Attempt at a Solution


R1=7 Ohms
J1= 1 A
E1= 8 V
E2= 4 V
R2=5 Ohms
R3=1 Ohm
R4=3 Ohms
 
Attachments
  • Circuit schematic with sources E1, E2 and J1 and resistors R1 to R4, with terminals a and b on the right side.
    file.JPG
    10.6 KB · Views: 432

Discussion



There does not appear to be any immediate opportunities to combine resistances in either series or in parallel. At each junction there is more than two components attached, so serial configurations are not available, and while some components share a node at one end, at the other they have another component situated between their connections, so no parallel configurations.

What is the goal for this problem? If you are looking to find the Thevenin equivalent for the circuit, then when you disable the current and voltage supplies to find the Thevenin resistance some opportunities may arise to combine resistances in series and parallel.
 


I need to transform this scheme to scheme(pic.2)

I don't know what can I do with(pic.1)

gneill said:
when you disable the current and voltage supplies
how can I do it?
 
Attachments
  • Circuit schematic with voltage sources E1 and E2, current source J1, resistors R1 to R4, and terminals a and b on the right.
    file.JPG
    11.1 KB · Views: 453
  • Circuit schematic: a voltage source E in series with resistor R, connected to terminals labeled a and b, with the source's arrow pointing up.
    temp.JPG
    3.4 KB · Views: 451


Okay, you're looking to find the Thevenin equivalent for the circuit in pic. 1. That means finding the open circuit voltage at a-b, and the Thevenin resistance "looking into" a-b. Have you covered Thevenin equivalents in your coursework or textbook? If not, there's plenty of references online if you do a search.
 


How can I delete J1(or substitute it) from the circuit?
 


Why do you want to delete or substitute it?

You should be trying to find the output voltage at a-b, and then the equivalent resistance at a-b with all the sources suppressed (which means shorting voltage supplies and removing current supplies).
 


gneill said:
Why do you want to delete or substitute it?
Because I don't know how to
gneill said:
find the output voltage at a-b.
 


You could write KVL equations for the three loops. Assign an unknown voltage across J1, and as an additional equation, note that J1 is related to the currents in the two loops that it is a member of.
 
Attachments
  • Circuit schematic with three loop currents I1, I2 and I3, EMF sources E1 and E2, a current source J1, resistors R1 to R4, and terminals labelled a and b on the right.
    Thev3.jpg
    17.2 KB · Views: 471


KVL
(I1-I2)*R2+R4(I1-I2)=E1
I2R3+R2(I2-I1)-J1=0
I3R3+R4(I3-I1)+J1=-E2

I Can find I1,I2,I3.I can find amperage(google translator) in every branch.But what it gives to me?
 


With I2 you can find the voltage across R3. That's the same as the voltage at a-b, and is your Thevenin equivalent voltage.

Next you need to find he Thevenin resistance by suppressing all the sources and finding the equivalent resistance at a-b.
 


gneill said:
Next you need to find he Thevenin resistance by suppressing all the sources and finding the equivalent resistance at a-b.

What do I need to do with U3=I3*R3=1.6*1=1.6V?
 


The voltage across R3 will be the voltage of the source in the Thevenin equivalent, the E in your second pic.

But there's a bit of a problem with the value that you calculated; I don't get the same value. Can you check your algebra? If you can't find a problem, then perhaps you should post your loop equations.
 


I found a mistake in my posts' equatations.In the third equatation I3*R1!
I got this values(I1,I2,I3) in mathcad
 
Attachments
  • Mathcad worksheet solving three circuit loop equations with Find(I1,I2,I3) giving the column vector 7, 6, 8/5; variable definitions R1 := 7, J1 := 1, E1 := 8, E2 := 4, R2 := 5, R3 := 1, R4 := 3 shown above.
    filee..jpg
    12.7 KB · Views: 501


Here are the equations that you wrote:

LOOP 1: (I1 - I2)R2 + R4(I1 - I2) = E1

LOOP 2: I2*R3 + R2(I2 - I1) - J1 = 0

LOOP 3: I3*R1 + R4(I3 - I1) + J1 = -E2

First, note that J1 is a current, not a voltage. So it shouldn't appear alone in a voltage equation; the units don't match.

In your first equation, the two currents passing through R4 should be I1 and I3.

In your second equation, you should have a variable for the VOLTAGE across the current source J1. (Perhaps call it EJ1. It is an unknown that can be solved for).

Since an additional unknown is introduced (EJ1), an additional equation is required. In this case it is the fact that J1 = I3 - I2.
 


builder_user said:
EJ1=-2
i1=4
i2=3
i3=1?

Nope.:frown:

Try these equations in Mathcad:

E1 - (I1 - I2)*R2 - (I1 - I3)*R4 = 0

(-I2 + I1)*R2 - I2*R3 - EJ1 = 0

-E2 - (I3 - I1)*R4 + EJ1 - I3*R1 = 0

I3 - I2 = J1
 


15/8 1/2 3/2 35/8

EJ1 direction is opposite to the direction of J1?
 


gneill said:
Here's a hint. I1 should be 1A.

Strange.Mathcad can't make mistakes.

But I can.I found mistakes in the equatations.
New result is ------ 1 -3/8 5/8 29/4

And so U3 is 3/8 V?
 
Last edited:


-3/8 V. Watch the sign of the current. :smile:

Now, on to the equivalent resistance!
 


builder_user said:
How can be voltage less than zero?

The voltage you measure depends upon your reference point. Here I'm assuming that you want the voltage at point a, on the schematic diagram, with respect to point b. Point a is at a lower potential than point b. If you were to put a voltmeter across a-b with the voltmeter's ground lead connected to point b, then you would read a negative voltage.
 


gneill said:
The voltage you measure depends upon your reference point. Here I'm assuming that you want the voltage at point a, on the schematic diagram, with respect to point b. Point a is at a lower potential than point b. If you were to put a voltmeter across a-b with the voltmeter's ground lead connected to point b, then you would read a negative voltage.

I see.So what's the next step?
 


The next step is to determine the Thevenin resistance. That would be the resistance of the network from the point of view of a-b with all sources suppressed. Replace all voltage sources with short circuits, and remove all current sources.
 


gneill said:
Replace all voltage sources with short circuits.

Replace...Like this?
 
Attachments
  • Left: a circle with a vertical line through it and an upward chevron arrow, a current source symbol. Right arrow points to a rectangular loop with a resistor on the left side and a circle with an upward chevron arrow on the right side.
    exp.JPG
    5.5 KB · Views: 443


The new circuit is pic.1 or pic.2?I think there're not differences between them.
 
Attachments
  • Circuit schematic with four resistors labelled R1 to R4 and terminals a and b; R2 and R4 form a series branch in parallel with R3 on the left, connected to R1 below and to terminal b.
    news.JPG
    5 KB · Views: 403
  • Circuit schematic with four resistors labelled R2, R3, R4 and R1, arranged in two vertical branches on a rectangular frame, with terminals a and b on the right side.
    scheme3.JPG
    6.6 KB · Views: 358
Last edited:


Pic 2. is your new circuit. E1 was short-circuited (replaced with a wire). There is a significant difference between the circuits due to this short-circuit. Can you spot what it is?
 


there is no resistance and no voltage drop across the short?
Is there a current?
 
Last edited:


Correct. So R2 and R4 are shorted out; they can be removed, leaving only the shorting wire.
 

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