So I was told that $V_x$ is equal to 10V. I tried reasoning out why $V_x$ was 10V and this is what I got:

-You can treat the Vx as an open circuit which means current does not flow through the resistor (meaning it's a short circuit) and thus from KVL we get $V_x$ = 10V

However, I am still not convinced that $V_x$= 10V. A better explanation would be greatly appreciated.

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  • $\begingroup$ Welcome to Engineering SE. You are asked to calculate $V_{th}$ for the circuit. So the first step is to calculate the open circuit voltage across $V_{AB}$. Since no current is flowing across the Circuit $V_{AB}$ is same as $V_x$ is same as 10V. Here is a good link Thévenin's theorem or Thevenin's Equivalent $\endgroup$ – Mahendra Gunawardena Oct 26 '15 at 2:08

You start by making A-B an open circuit. This implies that the current entering or leaving node A from the load is 0 Amps. This implies that the current in the two 1 $\Omega$ resistors have the same current flowing through them.

The open circuit across A-B also implies that 0 current is flowing into or out of Node B. This means 0 current is flowing through $\frac{5}{11}\Omega$ resistor, so there is 0 voltage drop across that resistor, so $V_x$ is also 10 V.

This leaves you with 10 V at $V_x$ and $0.2V_x = 2 V$ across the voltage controlled voltage source. This means the current flowing in the loop with the two 1 $\Omega$ resistors is $\frac{2 V}{2\Omega} = 1 A$. This also means the voltage between the voltage controlled voltage source and the bottom 1 $\Omega$ resistor is $12 V$.

With this information you can calculate the voltage drop across the lower 1 $\Omega$ resistor to be $1V$, which gives you a $V_{th}$ of $11 V$.


You are right in that $V_x$ can't be 10 V. Fortunately, 10V for $V_x$ simplifies the circuit considerably, making it particularly easy to analyze.

Since $V_x$ is the same voltage as the supply on the left, both ends of the 5/11 Ω resistor are at the same voltage. That means the current thru it is zero. That means the current thru anything directly in series with is must also be zero. Therefore, the current thru the 5/11 Ω resistor, the 10 V power supply, and the 1 Ω resistor at right are all 0. The voltage across any resistor with 0 current thru it must also be 0. We therefore know that point A is at 0 V.

Taking the loop with the 2 V supply and the two 1 Ω resistors in isolation, we can easily see that 1 A must be circulating, as indicated by your curved arrow. 1 A thru a 1 Ω resistor causes a 1 V drop, so the right end of this loop must be 1 V higher than the left.

Putting this back into the rest of the circuit that is known to carry 0 current just doesn't work. According to previous logic, point A must be at 0 V, but adding the +2 V of the supply and -1 V of the resistor to the 10 V at $V_x$ yields 11 V at A, which is a contradiction.

Clearly something is wrong somewhere.

  • 1
    $\begingroup$ This: "That means the current thru anything directly in series with is must also be zero." doesn't imply this: "the 1 Ω resistor at right are all 0". Why? Because you have a voltage controlled voltage voltage source and the loop there. So your second answer of 11 V is correct. $\endgroup$ – Eric Oct 26 '15 at 16:21

Thevenin Circuit

When calculating $V_{th}$ , $V_{AB}$ is open circuited as in the diagram. As a result there is no current flow across $\frac{5}{11} \Omega$. Thus $V_x = 10V$


  • $\begingroup$ But there is current in loop I. $\endgroup$ – Eric Oct 26 '15 at 16:53

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