- #1

- 136

- 0

Does anyone know what Im on about? I cant think of how you could do this, but then again, I dont really understand Thevenin all too well.

Cheers.

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- Thread starter morry
- Start date

- #1

- 136

- 0

Does anyone know what Im on about? I cant think of how you could do this, but then again, I dont really understand Thevenin all too well.

Cheers.

- #2

SGT

Does it help?

- #3

- 1,681

- 3

flow is connected to a flowmeter via a section of finite diameter pipe.

B) A second pump which maintains a constant flow rate and can develop any

pressure at the head is connected to itself with another short pipe section.

A pressure gauge measures the pressure at across this pipe.

For any type of pump and pipe section in situation A, you can find another

pipe and pump of type B that will have an identical pressure and flow

rate

A or B. You won't know from measuring the flow and pressure at the service

pipe whether you are geting your fluid from a type A or type B system.

- #4

- 136

- 0

Hmm, Im still a bit confused, but I kinda get it. Thanks guys.

- #5

- 1,309

- 19

What exactly are you having trouble with in understanding Thevenin equivalent circuits?

- #6

- 1,497

- 4

This mechanical analogy may not be physically correct, but the idea is the same.

Suppose you pick any I-beam in a skyscraper, and want to analyze all forces, shears and stresses that are acting on it. If you could reduce all that junk to a single force and shear you could model whats going to the I-beam.

Similarly with resistors (I-Beam)

Suppose you have a glob of interconnected resistors with voltage and current sources. If you pick any resistor in the circuit, you can model whats going on between the two nodes across you resistor with a single voltage source (thevenin) and another resistor (thevenin).

If you then take your resistor that you picked and connect it to the thevenin voltage source and the thevenin resistor, it will give you the same effect if your resistor was connected to whole circuit.

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