Thevenin equivalent circuits help?

I need to calculate Rt, which is the net resistance when thesource is deactivated. I have no clue why Rt is 3 ohms, shouldn't it be 1.2 ohms (3*2/(2+3))

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Reply 1

Joinedup

Original post by ErniePicks

I need to calculate Rt, which is the net resistance when thesource is deactivated. I have no clue why Rt is 3 ohms, shouldn't it be 1.2 ohms (3*2/(2+3))

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well would you agree that the resistance of this

Thevenin problem1.png

is zero?

Reply 2

ErniePicks

Original post by Joinedup

well would you agree that the resistance of this

Thevenin problem1.png

is zero?

Why is the resistance zero?

Reply 3

Joinedup

Original post by ErniePicks

Why is the resistance zero?

it's a circuit with a conductor in parallel with a 2 ohm resistor, all the current goes through the conductor and none goes through the 2 ohm resistor... i.e. if you've got a zero resistance current path putting a resistor in parallel with it doesn't make the overall resistance go up.

or another way of looking at it

conductors have zero resistance, which is to say they're equivalent to zero ohm resistors

it's a 'resistor' of zero ohms in parallel with a resistor of 2 ohms

the rule for parallel circuits is (1/R₁ + 1/R₂) ^-1

which tends towards zero as either R₁ or R₂ tend to zero

Reply 4

ErniePicks

Original post by Joinedup

When you say conductor do you mean the cell/power source? I'm confused, i've done plenty of circuit calculations with a resistor in parallel to the cell and i'd always take into account then when calculating the resistance and it would be correct.
What is so special about this situation? I've never encountered anything like this in all the circuit calculations i've done, is there a special name for it or anything?

Reply 5

ErniePicks

for example

In this the 20 ohms resistor is parallel to the cell but current still flowed through it and all i had to do was add it into the 5 ohm resistor with the product/sum rule and then add the other resistor for the total resistance

Reply 6

atsruser

Original post by ErniePicks

for example

When you calculate Thevenin equivalent values, you have to replace all voltage sources with short circuits i.e. with resistances of 0 ohm. You thus need to calculate the resistance of 2 ohm || 0 ohm =

\frac{2 \times 0}{2+0} = 0

ohm. This is then in series with the 3 ohm resistance.

Reply 7

Joinedup

Original post by ErniePicks

you're trying to get the resistance between the 2 terminals on the right hand side of the picture. if you put a pd across those terminals there's be a current.

this is the conductor that's in parallel with the 2 ohm resistor
Thevenin problem2.png

probably it's a bit weird because you've substituted some component in order to do your circuit analysis

Reply 8

ErniePicks

Original post by atsruser

\frac{2 \times 0}{2+0} = 0

ohm. This is then in series with the 3 ohm resistance.

So for the example i posted just above yours would the 20 ohm resistor have no current going through it?

Reply 9

atsruser

Original post by ErniePicks

So for the example i posted just above yours would the 20 ohm resistor have no current going through it?

Your question is a little odd, since we aren't really worried about actual current flows when calculating Thevenin equivalents.

However, if you created a circuit as you have shown, and you applied a voltage across the terminals, then current *would* flow through the 20 ohm resistor - in this case, it is in parallel with the 5 ohm resistor, as opposed to the previous case where it was in parallel with the voltage source which you replaced with a short circuit (i.e. with a 0 ohm resistor).

Reply 10

Joinedup

Original post by atsruser

I found this stuff a bit of a cognitive jump when I started and thinking about the current that would flow if you threw away the original circuit, built a circuit like that and put a ohmmeter on it was how I dealt with it.

but when you substitute a voltage source out they aren't real circuits and they're not supposed to be the same as the original circuit - it's just a step you go through in order to do the analysis... but they do obey the same rules you already know for series or parallel resistances... and at the end, if you followed the rules you come out with a correct answer.

Reply 11

ErniePicks

Thanks for the help guys, i've moved on to Norton equivalent circuits and was doing fine until i hit this snag in calculating the Rn

It says that the supply is deactivated, is this what happens when there is a constant current supply when finding the Rn for Norton equivalent? I'm confused how they got 20 ohms for Rn, guessing it has something to do with the supply being activated. I was thinking about handling it like the other question but the 10 ohm resistors are parallel with each other and not the supply.

Reply 12

Joinedup

Original post by ErniePicks

Thanks for the help guys, i've moved on to Norton equivalent circuits and was doing fine until i hit this snag in calculating the Rn

It's a similar thing to thevanin...

the difference is when doing:

thevanin you replace voltage source with a closed (or short) circuit... i.e. no potential difference because resistance = 0

norton you replace the current source with a open circuit... i.e. no current because resistance is infinte.

in this case the only current path is through the right hand and centre resistors in series so you just add them together as usual for series resistors.

Reply 13

ErniePicks

Original post by Joinedup

It's a similar thing to thevanin...

in this case the only current path is through the right hand and centre resistors in series so you just add them together as usual for series resistors.

But wouldn't the current come in through the left hand side and go through the left resistor anyways?

Reply 14

Joinedup

Original post by ErniePicks

But wouldn't the current come in through the left hand side and go through the left resistor anyways?

Where could any current come from? there's nothing there - it's a dead end for this part of the analysis.

Reply 15

ErniePicks

Original post by Joinedup

Where could any current come from? there's nothing there - it's a dead end for this part of the analysis.

But then how does the current go through the centre and right resistors? Where does it come from and how did you determine that it goes through those 2?

Reply 16

Joinedup

Original post by ErniePicks

But then how does the current go through the centre and right resistors? Where does it come from and how did you determine that it goes through those 2?

What the writing I underlined, 20 ohms and the arrow means is the resistance as seen from that side of the circuit... the resistance you'd see if you built that circuit and put a ohm meter between terminal A and terminal B - these are the terminals on the side the arrow is pointing at.