Hoping someone can help.....This has to do with a simple THEVENIN CIRCUIT with only 3 components.(1) AN AC SOURCE VOLTAGE AT 40 VOLTS ZERO DEGREES (2) AN INDUCTOR XL=30 (3)A RESISTOR 10 OHMS All three are connected end to end in series....WHAT I WANT TO FIND OUT IS THE THEVENIN VOLTAGE and the THEVENIN IMPEDANCE CAN YOU GIVE ME THE ANSWER IN BOTH POLAR AND RECTANGULAR FORM AND HOW YOU DO THE CALCULATIONS *THE LOAD OF THE CIRCUIT WOULD COME FROM ACROSS THE RESISTOR*
Take a look at the information contained in the AAC tutorials at the link below. http://www.allaboutcircuits.com/vol_1/chpt_10/7.html/thevenin Once you have reviewed the material, you can take a shot at a solution and then post your effort here. hgmjr
We have deleted your latest post, as it duplicated this one. If you can indicate what satge of your calculations are causing problems, someone can probably help you to see the solution. But we can't help without some indication of your work.
To find the thevenin voltage you have to remove the load so that it becomes open circuit, but when I remove any load it's still a closed loop; or is it that you have to remove the load resistor rather than an actual load...I am confused......I SENT THE SECOND QUEST BECAUSE THERE WAS AN ERROR IN THE FIRST ONE..The AC source voltage is LESS THAN ZERO DEGREES AND NOT JUST ZERO DEGREES AS WAS STATED IN MY FIRST THREAD.
Hi, Maybe I need to learn something. I always thought the AC source was the standard to measure phase shift against, so it had to assume a zero value. If you have two elements in a series circuit, removing either one opens the current path, resulting in an open circuit.