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  #1  
Old 01-31-2010, 04:02 PM
Fraser_Integration Fraser_Integration is offline
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Default Thevenin Equivalent of a Voltage Divider circuit

Hi there.

The attached is probably really simple, but I've never thought about Thevenin being used for something like a voltage divider, would just like to confirm my steps:

I would think V(th) would be the voltage at the output terminals found by usual voltage divider equation: 10V, and for R(th) I would think that if you shorted the voltage source (that isn't there I know, but bare with me) then the two resistors would be in parallel, so effective resistance of: 22k ohms?

Thanks for looking.
Fraser
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Old 01-31-2010, 04:11 PM
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hgmjr hgmjr is offline
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You are correct.

hgmjr
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Old 01-31-2010, 07:52 PM
Fraser_Integration Fraser_Integration is offline
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thanks a lot.

you can delete the thread to de-clutter if you like.
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Old 01-31-2010, 08:44 PM
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hgmjr hgmjr is offline
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No problem. We can let it stand. Other members/guests may benefit from it.

Knowing how to solve for Thevenin's Voltage and Resistance is the first step. To really understand Thevenin's Theorem, it is important to realize how to apply it analyzing a circuit.

hgmjr
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Old 01-31-2010, 09:08 PM
thyristor thyristor is offline
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Quote:
Originally Posted by Fraser_Integration View Post
Hi there.

I would think V(th) would be the voltage at the output terminals found by usual voltage divider equation: 10V, and for R(th) I would think that if you shorted the voltage source (that isn't there I know, but bare with me) then the two resistors would be in parallel, so effective resistance of: 22k ohms?

Thanks for looking.
Fraser
Possibly a better way of understanding the Thevenin equivalent is to consider that the circuit is inside a closed box so that you cannot see the components and that all you have access to are the Vout terminals.

All you can then measure is the open circuit voltage and the short circuit current. If you could then construct a circuit that gave the identical open circuit voltage and the identical short circuit current, then you would have built an equivalent circuit.

So the open circuit voltage is (as you correctly stated)
15 x R2/(R1 + R2) = 15 x 66/99 = 10v

If we short the output terminals, then the short circuit current is 15/R1 = 15/33 = 0.45A

Therefore there appears to be a resistor inside the box of 10/0.45 = 22K fed by a 10v source.

In algebra, the o/c voltage is Vcc x R2/(R1 + R2) and the s/c current is
Vcc/R1

Therefore the equivalent resistance is
Voc/Isc = Vcc x R1R2 / Vcc(R1 + R2)

which is R1R2/(R1 + R2) or the two resistors in parallel.

So the closed box is equivalent to a 10v source with a 22k series resistor
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Old 01-31-2010, 11:25 PM
Fraser_Integration Fraser_Integration is offline
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Thank you thyristor, that explanation would read better than my made up voltage source.
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