Diode Analysis and Thevenins Theorem

[CoolDude420]

Homework Statement

Homework Equations

The Attempt at a Solution

The correct answer is 0.25V. I'm getting 0.3335V[/B]

 

[gneill]

Check your Thevenin voltage derivation.

 

[CoolDude420]

Check your Thevenin voltage derivation.

I tried it again, same answer.

 

[gneill]

Still incorrect. You're mucking up after the first stage when you apply the voltage division with the 500 Ohm resistor. Take the circuit one stage at a time:

First break the circuit at (1) and find the Thevenin equivalent. What do you find?

 

[CoolDude420]

Still incorrect. You're mucking up after the first stage when you apply the voltage division with the 500 Ohm resistor. Take the circuit one stage at a time:
View attachment 114680

First break the circuit at (1) and find the Thevenin equivalent. What do you find?

 

[gneill]

No. There are three resistors involved and you haven't dealt with the voltage divider. A Thevenin model consists of a voltage source with a series resistance and does not form a closed loop!

 

[CoolDude420]

No. There are three resistors involved and you haven't dealt with the voltage divider. A Thevenin model consists of a voltage source with a series resistance and does not form a closed loop!

my apologies, is this fine

 

[gneill]

No. How do you arrive at 2 kΩ? How come the original 2 V hasn't been divided by the voltage divider consisting of the two 1 kΩ resistors? This is the circuit that you are starting with for this first part:

Apply the procedure to find its Thevenin equivalent.

 

[CoolDude420]

No. How do you arrive at 2 kΩ? How come the original 2 V hasn't been divided by the voltage divider consisting of the two 1 kΩ resistors? This is the circuit that you are starting with for this first part:

View attachment 114681

Apply the procedure to find its Thevenin equivalent.

 

[gneill]

Right!

Now tack on the next stage and repeat:

What's the Thevenin equivalent for the terminals at (2)?

 

[CoolDude420]

Right!

Now tack on the next stage and repeat:

View attachment 114682
What's the Thevenin equivalent for the terminals at (2)?

Ah. I seem to have gotten it now. Thank you, not entirely sure why my method that I did initially to find Vth is wrong.

 

[gneill]

Ah. I seem to have gotten it now. Thank you, not entirely sure why my method that I did initially to find Vth is wrong.

Great.

You were doing something strange with 1.5 kΩ that I couldn't see the reasoning behind. It's best to either break the circuit up into discrete stages and proceed methodically, or apply something like mesh or nodal analysis when you want to do it all at once.

 

[CoolDude420]

Great.

You were doing something strange with 1.5 kΩ that I couldn't see the reasoning behind. It's best to either break the circuit up into discrete stages and proceed methodically, or apply something like mesh or nodal analysis when you want to do it all at once.

Okay, I've written in detail to what my approach was. I have a feeling that I'm neglecting that extension of A-B at the right and assuming no current flows? But isn't that what I'm meant to do. I mean Vth is the open circuit voltage?

 

[CoolDude420]

I think I see my mistake. I'm applying voltage division even though the current through the resistors is different.

 

[gneill]

Okay, I've written in detail to what my approach was. I have a feeling that I'm neglecting that extension of A-B at the right and assuming no current flows? But isn't that what I'm meant to do. I mean Vth is the open circuit voltage?

The problem is that you started finding the Thevenin equivalent for the first section of the circuit and did not complete the operation before you applied the partial result to the next section. Your paragraph labelled (2) assumed that the 1 V first Thevenin voltage would be expressed across the first vertical 1 k resistor and hence would be applied to the following section's voltage divider consisting of the 500 Ω resistor and 1k Ω resistor. That was an error. The 1 V should be behind a 1 kΩ series resistor (the Thevenin resistance of the first stage) that includes the 500 Ω resistor.

 

[CoolDude420]

The problem is that you started finding the Thevenin equivalent for the first section of the circuit and did not complete the operation before you applied the partial result to the next section. Your paragraph labelled (2) assumed that the 1 V first Thevenin voltage would be expressed across the first vertical 1 k resistor and hence would be applied to the following section's voltage divider consisting of the 500 Ω resistor and 1k Ω resistor. That was an error. The 1 V should be behind a 1 kΩ series resistor (the Thevenin resistance of the first stage) that includes the 500 Ω resistor.

Thanks!