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**Electric Circuit. Norton/Thevelin Equivalents**

**Unsolved Problem**

I used a node voltage equation to try and get Vth and the I in the short-circuit to get the Rth but it didn't work.

**Unsolved Problem**

I used a node voltage equation to get Vth and I in the short-circuit. Got Rth and used in the in Vth^2 / 4 RL (with RL = Rth)

My answer was wrong. Is the method correct ?

**Solved this my self**

**Removed to make space for another problem**

For me, I find that the Norton equivalent is going to be 12A since the independent current source forces that ampage. But apparently I'm wrong. (I get feedback from a web form when I submit the answer). Any one can tell me whats wrong with my thought process ? Thank you.

Solution:

After doing a source transformation and adding up the resistors. You can get the correct the Norton current value.

**Solved**

I thought that if I did a source transformation getting a 36000V source would solve the problem because there isn't any other voltage source in the circuit. I was wrong. Any hints on this one ?

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