While solving the below circuit:
The solution is as shown below:
I noticed the following while analyzing it:
Why does the voltage across a pure resistance and the current through it have imaginary parts?
What do these imaginary parts represent?
Similarly for a pure reactive load (inductive) what does the real part of the voltage Vx mean?
Best Answer
The imaginary part tells you that the current flow through the resistor is lagging the voltage applied to the series network by a certain amount. As far as the resistor is concerned, both its voltage and its current are totally in phase but, relative to the applied voltage applied across the series network of R and L, they are both lagging.
The imaginary part indicates by how much the current in both L and R (the same current) is lagging the applied voltage to that series network.
Your math is correct apart from forgetting to add 90 ° to the inductor voltage value