Engineering Solving phasor circuit with unknown dep. current src

[x86]

Homework Statement

Homework Equations

V=IR

The Attempt at a Solution

I try to simplify the circuit by combining the capacitor and 2ohm resistor in parallel, Z = 1-j. Then I add the inductor to get Z = 1. Knowing the current across the resistor, 2(0d) I find the voltage across the 5 ohm resistor (2(0d)). So 2/5(0d).

The real part is 0.4 (correct). The imaginary part is 0 (incorrect)??

The answer for the imaginary part is 0.4 But I don't get this in my attempt.

 

[gneill]

I try to simplify the circuit by combining the capacitor and 2ohm resistor in parallel, Z = 1-j. Then I add the inductor to get Z = 1. Knowing the current across the resistor, 2(0d) I find the voltage across the 5 ohm resistor (2(0d)). So 2/5(0d).

Combining the capacitor and 2 Ohm resistor is a good start. But then adding the inductor's impedance to it at that point is not so good.

You see, you may know the current through the resistor thanks to having its potential difference, but there is also the current through the capacitor to consider. What must the current be through the combined impedance (cap || resistor) in order for the potential difference across it to end up being 4V @ 0°?

 

[x86]

Combining the capacitor and 2 Ohm resistor is a good start. But then adding the inductor's impedance to it at that point is not so good.

You see, you may know the current through the resistor thanks to having its potential difference, but there is also the current through the capacitor to consider. What must the current be through the combined impedance (cap || resistor) in order for the potential difference across it to end up being 4V @ 0°?

If I'm not misaken, after combining the resistor and capacitor, I know the voltage through them both is V1. Then by dividing by their combined impedence, I get the current through these two elements, which I draw a black box.

So now I know the current through this black box. I combine the impedence of the black box and inductor, multiplying by the current, to get the voltage. Now I know the voltage across the current source and the 5 ohm resistor (they are in parallel).

 

[gneill]

If I'm not misaken, after combining the resistor and capacitor, I know the voltage through them both is V1. Then by dividing by their combined impedence, I get the current through these two elements, which I draw a black box.

So now I know the current through this black box. I combine the impedence of the black box and inductor, multiplying by the current, to get the voltage. Now I know the voltage across the current source and the 5 ohm resistor (they are in parallel).

Yes, that would work fine.