Is the Thevenin resistance correct ? Also does it matter whether the load is connected in parallel or series?

Question

So i figured out that the voltage was 9 volts by using mesh currents .

My Thevenin resistance however came out to be 60/16 or 3.75 ohms . I short circuited the voltage sources and open circuited the current source . After that I added the resistors on the right (6 & 4) in series and then added the left 6 ohm resistor in parallel. It came out like this

Rth = 10*6 / 10 + 6

Also I noticed that the load was connected in parallel with the resistors , will that change the Thevenin circuit ?

Answer 1

Looking left from the A-B terminals to determine equivalent source circuit, when you open 3A source and short the 18V source, then the two 6 ohms are in series, and their branch is in parallel with the the 4 ohms. (6+6) || 4 = 3. Your described calculation would be for a equivalent resistance looking out from the two terminals of the current supply (but ignoring the 1 ohm load), which doesn't make sense for this question.

Answer 2

Starting with the part of your question that asks whether the load is connected in series or parallel, on the left below you see some boxed system with two ports, A and B, and the load is clearly connected in parallel with it:

^{simulate this circuit – Schematic created using CircuitLab}

Upper right I connect the two in series, but then lower right I complete the loop with the red wire. Comparing left (parallel) with lower right, what's the difference? How else could you connect the load? The terms "series" and "parallel" don't make much sense unless there are other elements in the circuit for context.

To find the Thevenin equivalent, you consider the boxed region with nothing connected, and then with its terminals connected directly together:

^{simulate this circuit}

With the terminals open (above left), you measure the potential difference between them. Then with the two terminals connected directly together, you measure the current through that connection. Remembering that a voltmeter has infinite resistance, and an ammeter has zero resistance, your experiment to find the Thevenin equivalent of the boxed part looks like this:

^{simulate this circuit}

Therefore your own problem becomes two individual tasks. The first is to measure \$V_{AB}\$, when there's nothing connected to A or B:

^{simulate this circuit}

The second is to measure \$I_{AB}\$, the current through a wire directly connecting A to B:

^{simulate this circuit}

As a hint, during the short-circuit current calculation, clearly no current flows through R3, since it's in parallel with 0Ω, so R3 can be disregarded, removed altogether.

Essentially you have results from two distinct conditions:

• You now know \$V_{AB}\$ when \$I_{AB}=0\$

• You also know \$I_{AB}\$ when \$V_{AB}=0\$

This permits you to use superposition to find a circuit that would behave exactly the same, but consists of only a single voltage source \$V_{TH}\$ and single resistance \$R_{TH}\$, the so-called "Thevenin equivalent circuit". Without elaborating on how superposition yields the following conclusions, using your calculated/measured \$V_{AB}\$ and \$I_{AB}\$ from before, it boils down to this:

$$ V_{TH} = V_{AB} $$

$$ R_{TH} = \frac{V_{AB}}{I_{AB}} $$

Finally you can replace everything in the box with those two elements, connect the load between A and B, and proceed with the (now simplified) calculations for current through and voltage across that load:

^{simulate this circuit}

If it helps, it's pretty clear that now you do indeed have two resistors in series, the load and \$R_{TH}\$ which (now that I think about it) may be what you were referring to in the first place. Sorry.