Hello!

After a gap of 1 year, I am starting my Graduate studies & I am already stuck with Q2 on 4 which deals with Thevenin. I have brushed up the topic but I am stuck at ONE place:

My initial approach was to convert the Voltage Source to Current Source OR converting Current to Voltage Source. My initial thought process was that the Voltage Source was connected to a Series Resistance of 2kohms. Hence, it can be converted into a Current Source of 3A with 2kohms Parallel resistance. If this idea is correct, then we can get a cumulative current source of 5A and I can go ahead.

However I am not sure if this idea is correct.

Need help guys.
Please let me know if this idea is correct. .If NOT please suggest a method, and I will try it out.

Thanks in advance

 

My initial approach was to convert the Voltage Source to Current Source OR converting Current to Voltage Source.

IMO, it is not a good approach. I don't see how can you do this here.

My initial thought process was that the Voltage Source was connected to a Series Resistance of 2kohms.

No, they are not in series.

Please let me know if this idea is correct.

It is not correct.

If NOT please suggest a method, and I will try it out.

• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points

http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

 

To add to above regarding 2, Rth.
Remove the 6 kOhm reistor from the circuit.
Another way to find Thevenin Equivalent Resistance, is to replace independent voltage source with a short, replace independent current source with open. This way you will have 3 resistors, 2k and 4k are in parallel and this combination is in series with 2k, find the equivalent resistance, this equivalent resitance is Thevenin Equivalent Resistance of the circuit.

 

Remove the 6 kOhm reistor from the circuit.

The 6k resistor IS the output. One would normally remove the load.

 

The statement of the problem is somewhat ambiguous. If we are being asked to find the Thev. Equiv. of everything to the left of the 6K resistor, then remove the 6K resistor, and compute Vo (I get 10.6667V) . Then place a short circuit across Vo, and compute the current that would flow through the short (I get 0.0032A).

The Thev Equiv would be a 10.66667V source in series with a 10.66667/0.0032 = 3.33333K resistor. If you connect the 6K to either the original circuit, or to the Thev Equiv the voltage at Vo is 6.85714V, proving that the Thev. Equiv. is...

 

Can someone tell me why the problem stated the 6k resistor IS the output if we are going to remove it as if it is the load?

IF the 6k is part of the output and NOT THE LOAD ... the supply is capable of 6.86 V @ 3.2 mA
IF the 6k IS THE LOAD ... the supply is capable of 10.67 V @ 3.2 mA

The wording of the problem is paramount when dealing with academic exercises. With 25 marks on the line, you decide. Also look carefully as the Vo terminals are clearly to the right of the 6k resistor, making it NOT THE LOAD.

With the statement in the problem that the "6k resistor is the output", I would use it in the calculation. There is no ambiguity when you consider the output terminals being to the right of the 6k resistor and you just short the output terminals to determine the maximum current the supply can provide.

 

Ok. So, I checked with my professor & he has asked me to go ahead taking 6kohms as the load resistor. I could calculate Rth easily. While calculating Vth, I decided to use KVL. Can somebody please tell me how to find VTH using KVL?

 

Guys, I have solved it. .Thank you for your help

 

Glad you got it.

I hate ambiguous questions. I take enough tests annually with too much fluff in the question, that one has to read the question carefully to identify the stem of the question, disregarding all the superfluous BS, and eliminating the distractors.

In ancient times, bleeder resistors improved regulation. It was not efficient. They probably don't discuss bleeders anymore. LOL.