Why does a pure resistance has a an imaginary current and voltage?

[skyline1397]

While solving the below circuit:



The solution is as shown below:

I noticed the following while analyzing it:



how a pure resistance both the voltage across it and 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?

 

[Harald Kapp]

how a pure resistance both the voltage across it and current through it have imaginary parts?

Because the circuit is not purely resistive. The imaginary part is added by the inductor in this case.
The same current must go through resistor and inductor. See how this in influences your calculations and the real and imaginary parts.

I'll also move this thread to where it belongs: homework.

 

[skyline1397]

The current is the same. But why the resistor voltage has an imaginary part??

 

[Harald Kapp]

What is the equation for a voltage across a resistor?

When you look at the voltage across and the current through the resistor only, you will see that both are perfectly in phase. Since the current has an imaginary part, so does the voltage.
Real and imaginary make sense only in combination with a reference frame that establishes a 0 ° phasor (or a purely real signal). Then you can determine the phase of other signals (voltages, currents) with reference to that 0 ° signal. When you look at a single signal, isolated from the others in the system/circuit, phase makes not much sense (if any at all).
In your example the reference signal with phase 0 ° would be the driving voltage source. Only when looking at the voltage with respect to this 0 ° reference will you "see" the imaginary part of the voltage across the resistor. You will not "see" the imaginary part when you look at the voltage and current across and through the resistor only.

 

[Ratch]

The same current passes through all components in a series circuit. Therefore, if current has "imaginary" components, all series components will "see" them. That word is a misnomer by the way. There is nothing imaginary about the shock you will receive if you put part of your body across an energized cap. The correct word to use is orthogonal.

To answer your question, an ammeter responds to the absolute value of the current. You can the calculate the voltages across each of the components by multiplying its impedance by the absolute current (150 amps). The total, added collaterally for resistance voltages and orthogonally for reactances should add up to the source voltage often called the complex voltage.
Sqrt[(150*12.8)^2 + (150*9.6)^2]=2400

Same is true for power using I squared times impedance.
Sqrt[(150^2*12.8)^2 + (150^2*9.6)^2]=360000

I hope this clears things up for you. Ratch