Norton Equivalent

Discussion in 'Homework Help' started by xz4chx, Oct 23, 2012.

Sep 17, 2012
71
1
I need to find Norton Eqivalent by 1 kΩ resistor and then find $v_Q$
http://imageshack.us/photo/my-images/37/40362225.png/
So I first calculated the short circuit current where the 1kΩ resistor was.
Since $v_Q$ in the situation would be 0V because of the short dependent source would be gone shorting out the 3.3kΩ resistor and 680Ω resistor. And then I would have
$-12V/4700Ohms + i_{sc} + .005A = 0$
$i_{sc}=-2.45[mA]$
After that I need to calculate $R_{th}$ which is $v_T/i_T$
So I put a 1V test source and $i_T$ in the direction of the voltage rise. The 1V test source is - toward bottom and + towards top of circuit.
So then this gives me
$-12V/4700Ohms-i_T+1.5v_Q = 0$
Since $v_Q$ is in parallel with my test source but in opposite direction $v_Q=-v_T$
So I end up with
$-12V/4700Ohms-i_T+1.5(-1) = 0$
So $i_T = -1.503 [A]$
Divided by $v_T$ gives me $R_{th}$ which I get to equal .667Ω
So I set up the Norton Equivalent and then need to solve for $v_Q$
which I am not 100% on but I was thinking it would just be the voltage of the Norton Equivalent. So I changed it to a Thevenin Equivalent and got
$v_Q = -1.6317[mV]$

Then answer in the textbook is $v_Q= -8.53[V]$

I want to know where I went wrong as my answer is no where near the right answer.

I sorry I don't have any pictures of my work, it was hard to take a good picture of it.

2. Jeffrey Samuel New Member

Oct 25, 2012
4
0
Your Rth calculation is flawed for sure as the resultant is about the 4.7K||3.98K which can never be in the lower end to the best of my knowledge