# Voltage across current-controlled-current-source

1. May 9, 2014

### kexanie

1. The problem statement, all variables and given/known data
Draw the Thevenin and Norton equivalent circuits for diagram 1, labeling the elements and terminals.
ZR = 5 Ω, ZL = j5Ω, Vs = 3∠30°, the current flowing throught the resistance is Ix, and the current flowing through the dependent current source is 0.5Ix.

2. Relevant equations

KVL and KCL in Steady-State Sinusoidal Analysis.
Ohm's law in Steady-State Sinusoidal Analysis.

3. The attempt at a solution

Because this circuit contains a dependent source, I cannot find the Thevenin resistance by zeroing the sources and combining the impedances in series. Thus, I try to find the open-circuit voltage.

But what I don't understand is, by using KVL, Vs = VL + VId + VR. How can I derived the voltage across the dependent current source?

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2. May 9, 2014

### Staff: Mentor

You should be able to see that this circuit must have some kind of load attached at a-b or else it will not be a correct circuit. Can you spot the contradiction that would arise if the terminals are left open?

Once you've satisfied yourself that this is the case, decide what kind of "load" you want to place on the circuit to facilitate analysis.

EDIT: Actually, it occurs to me that there is a valid solution obtainable for the open circuit voltage provided that a particular current value and voltage across the current source occurs. The value for the current can be arrived at by considering the "contradiction" I mentioned above and thinking through the implications.

3. May 10, 2014

### kexanie

If the teriminals are left open, the current flowing through the resistor is Ix, and for the dependent source it is 0.5Ix. However, since they are in series, the current should be the same.

Hence Ix = 0 ? There is no current flowing through the loop? And the voltage between terminal a and b is just Vs?

Actually my instructor told me that the dependent current source will be "deactivated" in this case but I did not get it.

4. May 10, 2014

### Staff: Mentor

Yes, that is correct and well constructed logic.

And now you do