Thevenin network without resistance?

I'm trying to find the Thevenin equivalent network for the circuit below.

But whenever I solve the equations, I always end up with V(t)=2V_a - 3i_s and can't get any expression showing V(t)~i(t) relationship. There's no way Thevenin resistance does not exist, but I can't get the grip of where i(t) appears..

I've tried short-circuit(V=0)/open-circuit(i=0) method along with putting test voltage between the terminals. The problem is that I can't get any expression that relates i(t) to V(t). Can somebody help me and find the v-i relationship for this circuit?

Thank you for reading my question.

Thevenin voltage is obviously 10va(t) as the voltage source is directly connected to the open circuit terminals.

It is also obvious that the thevenin resistance is zero as any load connected to the terminals will be driven directly by the voltage source (which is taken to be ideal an therefor doesn't have any internal resistance).

The rest of the circuit doesn't have any influence on the connected load.

• 10Va(t) is a dependent voltage source.
– Chu
Commented Sep 16, 2017 at 11:49
• I think you're right.. My calculation also showed V(t)=10V_a (t). So the dependent voltage source was the one that was taking care of all the loads in the circuit, and therefore Thevenin resistance was shown as 0 to me.. I now get the grip. Thank you! Commented Sep 16, 2017 at 15:21
• No problem. Please upvote/accept my answer. Somebody for some reason downvoted me. Commented Sep 16, 2017 at 18:23

Work out the open circuit voltage, $V_{TH}$, by nodal analysis at node $v_a(t)$.

Then work out the short circuit current (the dependent voltage source is short-circuited), $I_{SC}=\infty$, hence $R_{TH}=0$.

• Hey, I tried it before, but really couldn't solve for short circuit current(I_sc). Some shows that I_sc = 0 or expressions containing I_sc just cancel out, leaving nothing. Commented Sep 16, 2017 at 15:19
We first calculate the voltage at node 2, following the above statement:$V_2=10V_2\frac{R_1}{R_1+R_2}+I_1(R_1||R_2)$. Solving for $V_2$ gives $V_2=I_1\frac{R_1R_2}{R_2-9R_1}$. With the schematic values, we have $V_2=-0.75\;V$. The Thévenin voltage is simply $V_{th}=10I_1\frac{R_1R_2}{R_2-9R_1}=-7.5\;V$. As indicated in the above, the short-circuit current is infinite because you short a perfect voltage source:
In this case, the Thévenin resistance is simply $R_{th}=0\;\Omega$.