Thevenin equivalent circuit with external loop

Question

I am tasked with the following exercise:

I am still finding Thevenin equivalent circuits challenging, and this example looks a bit daunting since it seems to have an external loop.

Hopefully, I made the correct assumption (though I could very much be wrong) that I only needed to find the equivalent circuit of the OPEN LOOP of this circuit. Namely, creating a thevenin equivalent circuit of this:

^{simulate this circuit – Schematic created using CircuitLab}

Which led me to getting the following Thevenin Equivalent circuit:

^{simulate this circuit}

Was my assumption correct? If not, how do I account for this? For example, this external circuit would make finding the open circuit voltage no different, I would think.

Answer 1

To find Vth

Currents flow only in close loops. You opened two terminals in the circuit. It means, no current will flow through the lower 2 ohm resistor. So you can discard it. Draw the circuit in an understandable manner then.

You can any method like KVL, nodal voltage, superposition theorem etc to find Vth.

Use KVL, assuming I is the current in the circuit: $$V_1 - IR_1 - IR_2 - V_2 = 0$$

On solving \$I = -66.66 mA\$

Therefore - $$V_{th} = 2.1 + IR_2 \approx 2 V$$

To find Rth

Go back to the original ckt. Short all voltage sources. Open all current sources. The result is:

Rth = R3 + R1||R2 = 8/3 ohms.

Hence the thevenin eq. ckt -

There is no "open circuit current" source as you drew in the ckt.

Answer 2

Some hints to find the Thevenin equivalent circuit.

1. You must deactivate any independent source in your circuit (bring the generated quantity to 0). This means your two batteries become short circuits, i.e. piece of ideal wire.

2. The circuit is now made up only of resistors. The equivalent resistance is the Thevenin equivalent resistance.

3. In the original circuit determine the voltage at the terminals. That is the value of the Thevenin equivalent source.

For this last point, note that with nothing connected to the terminals, there is no current through the lower 2Ω resistor, hence there is no voltage drop across it. This means that the open circuit voltage is just the voltage across the upper terminal and the common node of the two batteries.