Thevenin's equivalent help!!

Hello,

I need some help with this circuit. I can't seem to find the Thevenin's equivalent.
once I remove the load resistor (RLoad), R3 throws me.

The answer should be Vthev = 10.5V and Rthev = 150 Ohms.

i can get the Rthev value but not the Vthev voltage.

Any help would be much appreciated

thanks.

Try superposition theorem.

Q1) With the load resistor removed, how much current is flowing in R3?

Q2) With that current flowing in R3, what is the voltage at the output in terms of the voltage at the left side of R3?

Q3) What is the voltage on the node to the left of R3 if you ignore R3?

hello, could someone be kind enough to show me how to calculate this via super position as I cant seem to do it.

I can work out the currents when the 13V supply is shorted however dont know how to calculate the current with the 8V shorted.

thanks

manish1991 said:

hello, could someone be kind enough to show me how to calculate this via super position as I cant seem to do it.

I can work out the currents when the 13V supply is shorted however dont know how to calculate the current with the 8V shorted.

thanks

Click to expand...

What is it that you are doing differently between the two cases? They are topologically identical situations.

You need to show your work! How else can we figure out where you are going astray?

hey sorry here is my working.

like i said before the 8v being in the middle branch is throwing me.

this how far i have gotten:

13V shorted

when the 8V is shorted, the resistance seen by the 13V is:
[(100+220)||100]+100 = 176.2
Thus the current leaving the 13V source will be 13/176.2 = 73.8ma

What I did was replace the load with a short when doing the superposition.

When I found the Rth, I removed the load and shorted both voltage sources.

The current through R3 was the only current that interested me as that would be the maximum current with a shorted load. So I followed the superposition rules and calculated the current in R3.

Vth was simply Rth * the current through R3.

The way I learned how to do it is to open circuit the network (remove the load resistor), then calculate the open circuit voltage of 10.5V.

Then place a short across the load resistor, and calculate the current through the short 0.07A

Divide the open circuit voltage by the short circuit current to find the equivalent resistance. 10.5/0.07 = 150Ω

So the network can be replaced by a voltage source of 10.5V in series with a 150Ω resistor. (Thev equiv)

Either circuit (original or Thev equiv) should put the same voltage across 220Ω.

Attached is the proof of the pudding... Note that V(5) = V(6)

oh I see

Thanks for the help guys

Really appreciate it

manish1991 said:

hey sorry here is my working.

like i said before the 8v being in the middle branch is throwing me.

this how far i have gotten:

13V shorted

when the 8V is shorted, the resistance seen by the 13V is:
[(100+220)||100]+100 = 176.2
Thus the current leaving the 13V source will be 13/176.2 = 73.8ma

Click to expand...

How is this "13V shorted"? It seems more like "13V NOT shorted, 8V shorted".

You got the current from the 13V source (with the 8V shorted) just fine, but what you want is the current in the load, which is only a portion of that 73.8mA. You can use current division to determine that.

Now, it's important that you understand that the circuit topology you are working with if you leave the 8V source in place and short the 13V supply instead. Redraw the circuit with the 13V branch (i.e., the 13V source and the 100Ω resistor above it) in the middle and the 8V branch (i.e., the 8V source and the 100Ω resistor above it) on the left side. Do you see that you haven't changed anything electrically but that you now have the same problem to solve?

Even if you use this approach to find the current in the 220Ω load resistor, it hasn't gotten you much closer (some, but not much) to finding Vth. The two primary ways to find Vth are (1) directly, by removing the load and calculating the open circuit voltage, which IS Vth; or (2) indirectly by determining the Req and the short circuit current and multiplying the two.

You can use any method you want to do the analysis for either approach. I suspect it would be very good practice for you to use all of the major approaches (node voltage, mesh current, superposition) for both approaches to see how they work and that they yield the same result at the end of the day.

As for the "best" approach to use for this particular circuit, I would say that treating the circuit as a voltage divider is the simplest, which is basically using node voltage analysis.

You might go back and try to answer the questions I asked in Post #3. You might be in a better position to answer them now.