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.
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.
