Before starting the simulation, guess: Both voltage sources have the same aplitude and frequency. How much current flows through the resistor?
You can't tell, because it is dependent on the phase.
That is why we use complex numbers to express AC voltages:
U = Amplitude*exp(j*2pi*Frequency*Phase),
With j being the imaginary unit: j*j = -1.
Now lets calculate the current through the resistor:
I = (U2-U1)/R = (10V*exp(j*2Pi*1000*0°) - 10V*exp(j*2Pi*1000*30°))/R
Looks complicated, but it really is not. I will replace 2pi*1000 with f for frequency and simplify a bit:
I = (10V-10V*exp(j*f*30°))/R = 5.18V*exp(j*f*-150°)/R = 5.18mA*exp(j*f*.150°)
We can now easily see that we have an amplitude of 5.18mA with a phase of 150°.
This could be calculated without the use of complex voltages, but those calculations become really really nasty when you have a real network of voltages. Most modern calculators can calculate complex values, so basically no work for you.
