Equivalent resistance in a Wye formation

[Frankenstein19]

Homework Statement

Find the Thévenin resistance for each of the circuits shown by zeroing the sources.

Homework Equations

The Attempt at a Solution

I replaced the current source with an open circuit and I'm left with a wye formation but I don't know how to calculate the equivalent resistance

 

[gneill]

After suppressing (opening) the current source, what closed paths are left for current to flow via the output terminals?

 

[Frankenstein19]

After suppressing (opening) the current source, what closed paths are left for current to flow via the output terminals?

None? It'll just look like a T

 

[gneill]

None? It'll just look like a T

Yes, but suppose you were to apply a voltage (or current) source across the two output terminals. What path or paths could current follow? Remember, a "circuit" implies a closed path (loop).

Edit: Remember that in finding the Thevenin equivalent, you want to find the circuit's equivalent resistance "looking into" the output terminals. What load resistance would a test source "see" if connected there?

 

[Frankenstein19]

Yes, but suppose you were to apply a voltage (or current) source across the two output terminals. What path or paths could current follow? Remember, a "circuit" implies a closed path (loop).

Sorry if I'm nor using correct terminology but through the 20 an 10 ohm resistor?

 

[gneill]

Sorry if I'm nor using correct terminology but through the 20 an 10 ohm resistor?

Correct! The 5 Ω resistor, being disconnected at one end, is unable to pass a current through it. So the only path available is the one you've identified.

So then, what's the equivalent resistance?

 

[Frankenstein19]

Correct! The 5 Ω resistor, being disconnected at one end, is unable to pass a current through it. So the only path available is the one you've identified.

So then, what's the equivalent resistance?

30! Haha bless your soul for spoon feeding this to me. But I'm still a little confused as to why I'd suppose there was a source connected to the output terminals. I wasn't doing that with previous exercises. (I'll try it though to confirm)

 

[gneill]

30! Haha bless your soul for spoon feeding this to me. But I'm still a little confused as to why I'd suppose there was a source connected to the output terminals. I wasn't doing that with previous exercises. (I'll try it though to confirm)

The idea is that the network presents a certain resistance at the output terminals. A resistance can pass current through it. Hence, if you imagine that there is a source connected across it, then only the paths that can pass current can be involved in the net resistance.

When you come across more complex networks with controlled sources (rather than fixed voltage and current sources) that you cannot suppress, one of the techniques that you'll have to employ is to apply a fixed source to the terminals and then analyze the resulting current and voltage that results through and across the terminals in order to use Ohm's law to determine the "load" resistance: $R = V/I$.