Voltage across current-controlled-current-source

[kexanie]

Homework Statement

Draw the Thevenin and Norton equivalent circuits for diagram 1, labeling the elements and terminals.
Z_R = 5 Ω, Z_L = j5Ω, V_s = 3∠30°, the current flowing throught the resistance is I_x, and the current flowing through the dependent current source is 0.5I_x.

Homework Equations

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

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 = V_L + V_Id + V_R. How can I derived the voltage across the dependent current source?

 

[gneill]

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.

 

[kexanie]

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.

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

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

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

 

[gneill]

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

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

Yes, that is correct and well constructed logic.

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

And now you do

 

[HalfLostAndSearching]

The voltage across the dependent current source can be derived by using Ohm's law and KCL. Since the current flowing through the dependent current source is 0.5Ix, the voltage across it can be calculated by multiplying this current by the impedance of the dependent source. This can be represented as VId = (0.5Ix)(ZL). By substituting this into the KVL equation, Vs = VL + (0.5Ix)(ZL) + VR, the voltage across the dependent current source can be included in the overall voltage equation.

To find the Thevenin and Norton equivalent circuits, we first need to find the Thevenin resistance. This can be done by removing the dependent source and shorting the output terminals. By doing so, we can calculate the equivalent resistance seen from the output terminals, which in this case would be ZR in parallel with ZL. This gives us the Thevenin resistance, RT = (ZR*ZL)/(ZR+ZL).

Next, we can find the Thevenin voltage by using KVL in the original circuit. By shorting the output terminals and considering the voltage across the dependent current source as VId = (0.5Ix)(ZL), the voltage equation becomes Vs = VId + VR. Solving for VR, we get VR = Vs - VId. This gives us the Thevenin voltage, VT = Vs - VId.

To find the Norton equivalent circuit, we simply need to convert the Thevenin voltage and resistance to their respective Norton equivalents. The Norton current, IN, is equal to the Thevenin voltage divided by the Thevenin resistance, IN = VT/RT. The Norton resistance, RN, is equal to the Thevenin resistance, RN = RT.

Thus, the Thevenin equivalent circuit would consist of a voltage source, VT, in series with a Thevenin resistance, RT. The Norton equivalent circuit would consist of a current source, IN, in parallel with a Norton resistance, RN. Both circuits would have the same output characteristics and can be used interchangeably in the original circuit.