Homework Help: Use of Thevenin's theorem in Steady State Analysis

1. Jul 25, 2011

rudderauthori

1. The problem statement, all variables and given/known data
R1 = 29 ohm
R2 = 23 ohm
f = 60hz
L = 20mH
C = .1uF
Is = 27A
K = 12
See picture for circuit.

2. Relevant equations
KCL, KVL, is = Is cos (wt), V = IZ

3. The attempt at a solution

I first did a source conversion to volts then tried to love using KCL and the nodal method. When I would try to get an answer it wouldn't work. The problem mostly being I don't know how to handle the dependent voltage source in the middle when I would set up my equations.

I1 = (Vs + K*I1)/(R1+j*w*L)
I2 = (K*I1-Vo)/(R2)
Io = Vo/(-j(1/wC))

783V-R1I1-j*w*L-K*I1 = 0 ??

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2. Jul 25, 2011

Zryn

For Question 6 and 7, instead of doing a source conversion on the current source and resistor, then using nodal, would you consider using mesh analysis?

In the left most loop we can say the mesh current is equal to the current source value of 27A.

Can you write an equation using this information to solve for the mesh current (i1) in the second equation? This would then given you the dependent source voltage, which you could use in a voltage divider to determine the real and imaginary components across the capacitor (which is the same as across the terminals x and y).

For Question 8 and 9, in finding the Thevinin Equivalent Impendance of a circuit, are you familiar with the process of "Short Circuit the voltage (including dependent voltage) sources and Open Circuit the current (including dependent current) sources and combine everything thats left (all the series and parallel combinations) until you have one impedance"?

3. Jul 25, 2011

Staff: Mentor

You have to be a bit subtle about writing your loop equations when there's a controlled source involved.

Note that in the case of this circuit the controlled source is effectively isolating the loops to its left from the loops to its right (in terms of KVL equations). No matter what happens at X-Y, no matter what current is pushed through R2, the voltage between the ground node and the top of the k*i1 source will be fixed by that k*i1 source. Current flowing through the k*i1 source will not affect its voltage! Further, the open circuit voltage at the x-y terminals will be given by k*i1 in concert with the voltage divider comprising R2 and C.

So, just looking at the leftmost two loops, the loop with the Is source has a fixed mesh current of Is. So you're really down to one loop to be concerned with. Call the mesh current in the central loop I2. What then is the current, I1, flowing through R1 in terms of I2 and Is?