# Thevenin's equivalent circuit

• Engineering

## Homework Statement

Find the voltage and the current through R5

http://i.imgur.com/bL9AFVo.jpg?1 ## Homework Equations

Thevenin's theorem.

## The Attempt at a Solution

http://i.imgur.com/Lz9k8l2.jpg?1 * To find Vth:
1- Open circuit R5.
2- Find the voltage at A and B by determining the current through R1, R2, R3 and R4 (Their Req=(R1+R4)//(R2+R3)=368.4 ohms
3- The current is equal to 0.24 A, hence the Vab=Vth=30V

http://i.imgur.com/Lz9k8l2.jpg?1

* To find Rth (=Rab):
1- Open circuit R5 and short circuit the voltage source.
2- (R1+R4)//(R2+R3)=368.4 ohms

My problem is here, does the wire (caused by short circuiting the source) cause short on (R1+R4)?

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phinds
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You need to make your figures just a little more fuzzy ... I can almost read these.

gneill
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My problem is here, does the wire (caused by short circuiting the source) cause short on (R1+R4)?
Shorting the battery joins the top rail to the bottom rail making them all one node. That means R1 and R4 each have one lead that shares that one node... what about their other leads?

Edited.

Sorry.

Shorting the battery joins the top rail to the bottom rail making them all one node. That means R1 and R4 each have one lead that shares that one node... what about their other leads?
Their other leads are connecting together with B, does that make them in series?

gneill
Mentor
Their other leads are connecting together with B, does that make them in series?
Let's see... two components whose leads share two nodes... nope not series... what's the other choice? I see, do I have to consider B as an element?
I know that when two elements shares two extraordinary nodes (A node connects more than two elements), they're are in parallel; and when two elements shares one ordinary node (A node connects only two elements), they're in series.

gneill
Mentor
I see, do I have to consider B as an element?
Not if the components in question are in parallel; It doesn't matter what else connects to the nodes that they share. Besides, in this case you're looking at the case where the terminals A and B are open.

I know that when two elements shares two extraordinary nodes (A node connects more than two elements), they're are in parallel; and when two elements shares one ordinary node (A node connects only two elements), they're in series.
Not sure about the terminology "extraordinary" and "ordinary" nodes. Must be something new. I was taught about "essential" and "non-essential" nodes, a non-essential node being a node embedded in a serially connected branch.

So this problem is similar to this one? gneill
Mentor
Very similar indeed.

So my Rth would be: (R1//R4)+(R2//R3)=405.7 ohms.

Can I assume - in similar cases - that there is an "imaginary" voltage source to compute the total resistance?

gneill
Mentor
So my Rth would be: (R1//R4)+(R2//R3)=405.7 ohms.

Can I assume - in similar cases - that there is an "imaginary" voltage source to compute the total resistance?
You can if you like, but if the resistors can be reduced by the usual parallel/serial combination methods it's not needed.

Thank you for your quick responds :)

Would you explain the difference in these two cases? (Where I am asked to evaluate the equivalent between a and b)

http://i.imgur.com/eHYrn47.jpg?1 gneill
Mentor
Thank you for your quick responds :)

Would you explain the difference in these two cases? (Where I am asked to evaluate the equivalent between a and b)

http://i.imgur.com/eHYrn47.jpg?1
Looks like you've spotted the difference already. The second one has a direct wire running from a to b so its equivalent resistance is zero (short circuit as you state). You've made a calculation error for the equivalent resistance of the first case; the method is right but your result is not.

Yes, it should 10 not 5.

But for this kind of problems, how should I start? where terminals a and b are not on the same side.
http://i.imgur.com/rs6IAmJ.jpg?1 gneill
Mentor
Yes, it should 10 not 5.

But for this kind of problems, how should I start? where terminals a and b are not on the same side.
http://i.imgur.com/rs6IAmJ.jpg?1
All of a given node is the same place electrically, so you can slide connections to a given node to anywhere on the node (the contiguous conducting path) and not change the circuit. So nestle the a and b terminal up to either side of the 5 Ohm resistor. Ta da! Same side. If the terminals in question are truly separated by a lot of circuitry though, look to simplifying other portions of the circuit as you can. At times there may be more intuition than canned rules involved, and experience is the best guide. There's no one standard method that applies to all cases.

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Okay, so if I assumed a voltage source between a and b, then there will be an essential between 8, 20 and 30. How could 8 and 20 be in series?
http://i.imgur.com/ICCTPwn.jpg?1 gneill
Mentor
Okay, so if I assumed a voltage source between a and b, then there will be an essential between 8, 20 and 30. How could 8 and 20 be in series?
http://i.imgur.com/ICCTPwn.jpg?1
The 8 and 20 Ohm resistors are not in series due to the presence of the 30 Ohm's connection at their junction.

What's the relationship between the 20 and 30 Ohm resistors?

They're in parallel (c and b are common)

gneill
Mentor
They're in parallel (c and b are common)
Yes. So what strategy does that suggest to you?

I get it now, 20 and 30 are in parallel, which results in a series resistance with 8, and finally parallel with 5 which leads us to 4 ohms equivalent resistance.

Last question: If I want to determine the current in resistance, and that resistance is a part of Wye or Delta, would the current passing through the resistance be the same after the converting? and how do I know my resistance after the converting?

gneill
Mentor
I get it now, 20 and 30 are in parallel, which results in a series resistance with 8, and finally parallel with 5 which leads us to 4 ohms equivalent resistance.

Last question: If I want to determine the current in resistance, and that resistance is a part of Wye or Delta, would the current passing through the resistance be the same after the converting?
Nope. Once you make a circuit transformation involving a given component that component is "lost" to you for further analysis because it becomes a new beast involving a mix of other components.

and how do I know my resistance after the converting?
Presumably you would apply the appropriate Y-Δ or Δ-Y transform formulas to obtain the values.

If you're looking for current values through a particular component, after making obvious and easy circuit simplifications it's often easier to just employ another analysis method (KVL, KCL, loop, nodal, mesh...) to find the value.

Sincere thank you, gneil, you've opened my mind to this field :)

Really appreciate it. :)