A Thevenin/Norton Equivalent

  • Thread starter siylence
  • Start date
  • #1
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Homework Statement


This was a timed assignment that I never got an answer on, nor was I given the correct answer once the assignment was finished. I understood Thevenin/Norton (or so I thought) until I got a Volt source that was PARALLEL to the first resistor and everything fell apart.
The question is in the picture:
thevenin.jpg



Homework Equations





The Attempt at a Solution



I am honestly stumped on this one. My textbook has no examples of a problem like this, nor did my professor give one like this in our notes. I feel like there was a curveball thrown.

I ended up getting something like 23.5 Ohms for Rth but that was from calculating the 10 Ohm and 4 Ohm resistors together as if they were in parallel. Then adding that to the 40 Ohm as if it were in series, then taking that resistance and adding it in parallel with the 52 Ohm resistor.

I'm an idiot.
 

Answers and Replies

  • #2
vk6kro
Science Advisor
4,081
40
The 52 ohm resistor isn't affecting the output because it is in parallel with the supply which is assumed to be perfect. So, you don't need to include it.

So, you find the Thevenin equivalent using just the 10 and 40 ohms then add the 4 ohms to the resistance Rth, because it is in series.
 
  • #3
7
0
Sorry this took me so long to reply, but are the answers:
Vth = 120 V
Rth = 24 Ohms
In = 5 A?
 

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