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Thevenin Voltage And Resistance
#1
Hi all.  I have a problem with Thevenin's Theorem.

To get Thevenin's Resistance  you take out the load and measure resistance at A and B.  Or theoretically  You short the voltage source, and calculate the total resistance between A and B.  I did this starting at the left with  R1 and R6 in parallel in series to R2, this in parallel to R7, etc...  I got 5.96 K ohms.  I think this is right.

The problem I am having is getting the Thevenin Voltage.  With a meter you measure the voltage between A and B with load out.  Without a meter I am stuck.  Can some one give me a pointer to find the voltage out of this series parallel network?

Thevenin's Theorem


[img][Image: ThevelinProblem.jpg][/img]
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#2
r5 and 9 in parallel, result in series with r4, then just repeat that for each rung in the ladder till done.
--------------- ---- --- -- -  -
If things weren't meant to be modified, they would not come with wires attached.
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#3
(2016-03-28, 09:59 PM)darrin Wrote: To get Thevenin's Resistance  you take out the load and measure resistance at A and B.  Or theoretically  You short the voltage source, and calculate the total resistance between A and B.  I did this starting at the left with  R1 and R6 in parallel in series to R2, this in parallel to R7, etc...  I got 5.96 K ohms.  I think this is right.

This is special ladder because 1/(1/6+1/3) = 2k. Plus 4k = 6k again: no matter how many steps, Thevenin resistance is 6k.

(2016-03-28, 09:59 PM)darrin Wrote: The problem I am having is getting the Thevenin Voltage.  With a meter you measure the voltage between A and B with load out.  Without a meter I am stuck.  Can some one give me a pointer to find the voltage out of this series parallel network?
1/

There is several methods to do this. Since you are learning Thevenin, they want you use Thevenin to simplify the circuit by eliminating one loop at a time.

Step1: analyse V1, R1, R6 - rest of the circuit is load. Thevenin resistance for step1 is 2k, Thevenin voltage for step1 is V1/(6k+3k)*3k = V1/3. You can replace V1,R1,R6 with V1/3 + 2k in series.

Step2 ... N: Because we have 2k + 4k = 6k in series with V1/3, it's going to be the same step as before, divides voltage by 3. Resulting Thevenin voltage of this circuit will be V1/(3^N) where N is number of steps - in your case of 4 - Thevenin voltage is V1/81
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#4
Thankyou Roman and Jon for the advice.

Roman your case about the thevenins resistance makes total logical sence, thankyou.
AS for the voltage, I do not know what you mean V1/(6k+3k)*3k = V1/3, wouldnt that equal V1/27K
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#5
Division and multiplication have same precedence in math, so V1/(6k+3k)*3k = (V1/(6k+3k))*3k

Does this help: ?

Code:
.       3k      V1
. V1 * ----- = ----- * 3k
.      6k+3k   6k+3k

Btw you should always check units. When needing volts, Volts / (Ohms * Ohms) makes no sense. But (Volts / Ohms) * Ohms = Volts does.

Did you understand loop elimination by replacing each loop from left to right by new power source and resistor in 4 steps ?


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#6
Hi guys,
Keep running with this...it is helping more than you know.
Math has been my weakness and I need to spend more time working things out to confirm anticipated results.
Thanks for posting in the Learning section!
Student Bob D Smile
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#7
Thankyou very much Roman. Just did what you said. Everything is correct and I understand this now.
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