Thevenin voltage for a simple circuit

Question

Trying to find the Thevenin equivalent voltage for the circuit on the picture below. This voltage is supposed to be found at the round dots to the very right in the circuit.

According to the answer sheet the answer is supposed to be 16V for the Thevenin voltage.

However I keep getting the wrong answer and I don't understand why, I'm pretty new to this so there might be something very basic I'm overlooking. I've tried several variants, but still keep getting the wrong answer, should I overlook a resistor or something? Any help highly appreciated.

I found the \$R_{th} = 7.18 k\Omega\$

\$R_{parallel} = 2.18 k\Omega\$

\$Uth = 22V \times \dfrac{5+8}{5+2.18} = 39.8 V \$ but should be \$16 V\$

Answer 1

We ignore the R10 because effectively there is an infinite resistor across the two output terminals. Which means you have the same voltage drop at the output as is dropped across R9 which is 16volts 22*(R9/(R9+R8))

(Just think of it as another voltage divider of an infinite resistor and R10. The infinite resistor gets all the voltage across it with nothing dropped across R10.)

The memory aid of an infinite resistor across any open terminals, makes those out-lying dummy resistors (Like R10) easier to understand.

Answer 2

You don't need to find Rth. Note that R10 is immaterial since the Thevenin voltage is found with the output open circuited so no current flows in R10. Without R10, the Thevenin voltage consists of a voltage divider, R8 and R9, and a source voltage of 22 volts. Just calculate the output voltage based on that configuration.

Answer 3

I think you are confusing Thevenin equivalent circuit with voltage and resistance.

Thevenin equivalent circuit consists of an equivalent voltage source in series with thevenin equivalent resistance, and optionally load resistance / element which you assumed as load.

In your case, the equivalent voltage is really the one appearing across R9 because R10 is not in a closed path and so no current flows through it.

Answer 4

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.

Answer 5

^{simulate this circuit – Schematic created using CircuitLab}

In the above circuit (which represents your circuit) you can see that since no current goes through \$R_4\$, you simply measure voltage of \$V_1\$ (because \$ V_2 = V_1\$).

And we have: \$ V_1 = \frac{R_1}{R_1+R_2} Vs \$

In reality \$ V_1 \$ is \$ V_{th} \$ because that is the voltage you will see when you insert a very big resistance (load resistance) on your side.

Answer 6

The circuit is open at R10, so R10 is not affect the circuit. There is only one voltage source in this circuit . Therefore Uth = Vs = 22V R9 and R8 are in series. Therefore to find voltage across R9 (which I think you are looking for) , just need to use voltage divider rule.

V9= 22 x 8/(8+3) = 16V

Answer 7

Just ignore R10 (R8 and R9 are going to form a voltage divider of exactly 16V). As R10 in this case would be in series with voltage divider and voltage doesn't go through as current does, the voltage measured after R10 is going to be the same 16 V.