For the following circuit:

I’m solving for the Rth at node 2. In the solution guide, it shows that the Rth = (10k || 10k). I don’t understand how this is so, by following the resistence equivalent rules, these two resistance should be in series.

I’d appreciation for the clarification.

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1 Answer

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I assume for node 2 you’re ignoring everything to the right of it?

Anyways, to calculate the Thأ©venin equivalent, you look at short-circuited current and open-circuit voltage. If you just want resistance (if you have VTh and ITh, then R = V/I), replace voltage sources with shorts and current sources with opens.

For just resistance, if we replace the 10 V DC source with a short, then we can see that it’s just two 10 kخ© resistors in parallel just like your guide shows. To check it, we could find VTh and ITh.

VTh = 5 V, a simple voltage divider between equal resistors. ITh = 1 mA (10 V, 10 kخ©).

Rth=VthIth=5 V1 mA=5 kΩ=10 kΩ∥10 kΩR_{th} = \frac{V_{th}}{I_{th}} = \frac{5\mathrm{\ V}}{1\mathrm{\ mA}} = 5\mathrm{\ k\Omega} = 10\mathrm{\ k\Omega} \parallel 10\mathrm{\ k\Omega}

Just what I needed, thanks!

– Willtd

Sep 14 ’14 at 20:43

I was waiting for this sort of matter. Thank you very much for the

post.