The best way to visualize the Thevenin equivalent is to assume you are physically looking at the circuit from the end you are calculating from. In this case you are looking in from the far right. (When I first studied this there would actually be a picture of an eye looking at the circuit from the one end.)

When looking in from the far right end what equivalent voltage and resistance will you see?

First you see the 5k which is in series with the output. Further in you have an 8k across the circuit to ground. Now temporarily forget about the voltage supply (actually replace it with a short), and the 3k is seen as being in parallel with the 8k. So the Thevenin equivalent resistance (as seen from the far right), is 5k + (8k // 3k) = 7.18k.

Now calculate the actual voltage that will be "seen" at the far right. This is just a simple voltage divider with a 3k and 8k. So 22v/(3k+8k)x8k = 16v. (The 5k is still there but it does not alter the voltage as there is no initial current flow.)

So at the end the equivalent Thevenin circuit is just a single 16v source with a single series connected 7.18k resistor.

The best way to visualize the Thevenin equivalent is to assume you are physically looking at the circuit from the end you are calculating from. In this case you are looking in from the far right. (When I first studied this there would actually be a picture of an eye looking at the circuit from the one end.)

When looking in from the far right end what equivalent voltage and resistance will you see?

First you see the 5k which is in series with the output. Further in you have an 8k across the circuit to ground. Now temporarily forget about the voltage supply (actually replace it with a short), and the 3k is seen as being in parallel with the 8k. So the Thevenin equivalent resistance (as seen from the far right), is 5k + (8k // 3k) = 7.18k.

Now calculate the actual voltage that will be "seen" at the far right. This is just a simple voltage divider with a 3k and 8k. So (22v/(3k+8k))x8k = 16v. (The 5k is still there but it does not alter the voltage as there is no initial current flow.)

So at the end the equivalent Thevenin circuit is just a single 16v source with a single series connected 7.18k resistor.

The best way to visualize the Thevenin equivalent is to assume you are physically looking at the circuit from the end you are calculating. In this case you are looking in from the far right. (When I first studied this there would actually be a picture of an eye looking at the circuit from the one end.)

When looking in from the far right end what equivalent voltage and resistance will you see?

First you see the 5k which is in series with the output. Further in you have an 8k across the circuit to ground. Now temporarily forget about the voltage supply (actually replace it with a short), and the 3k is seen as being in parallel with the 8k. So the Thevenin equivalent resistance (as seen from the far right), is 5k + (8k // 3k) = 7.18k.

Now calculate the actual voltage that will be "seen" at the far right. This is just a simple voltage divider with a 3k and 8k. So 22v/(3k+8k)x8k = 16v. (The 5k is still there but it does not alter the voltage as there is no initial current flow.)

So at the end the equivalent Thevenin circuit is just a single 16v source with a single series connected 7.18k resistor.

The best way to visualize the Thevenin equivalent is to assume you are physically looking at the circuit from the end you are calculating from. In this case you are looking in from the far right. (When I first studied this there would actually be a picture of an eye looking at the circuit from the one end.)

When looking in from the far right end what equivalent voltage and resistance will you see?

First you see the 5k which is in series with the output. Further in you have an 8k across the circuit to ground. Now temporarily forget about the voltage supply (actually replace it with a short), and the 3k is seen as being in parallel with the 8k. So the Thevenin equivalent resistance (as seen from the far right), is 5k + (8k // 3k) = 7.18k.

Now calculate the actual voltage that will be "seen" at the far right. This is just a simple voltage divider with a 3k and 8k. So 22v/(3k+8k)x8k = 16v. (The 5k is still there but it does not alter the voltage as there is no initial current flow.)

So at the end the equivalent Thevenin circuit is just a single 16v source with a single series connected 7.18k resistor.