RLC Series Circuit Problem: Find V_source from V_R, V_L, V_C

[nucleawasta]

Homework Statement

In an L-R-C series circuit, the rms voltage across the resistor is 32.0 , across the capacitor it is 90.1 and across the inductor it is 51.5 .
What is the rms voltage of the source?

Homework Equations

Well there are lots of equations V_rms =V/sqrt(2)
V_R=IR
V_L=IX_L(I\omegaL)
V_C=IX_c(I/(\omegaC)

The Attempt at a Solution

So what i did was convert all the values from rms into their amplitude voltages and summed them. Then took the resultant rms of that answer. This is not correct however.

I was using kirchhoffs rules of the sum of voltages around loop =0. Does this principle hold true in this situation?

 

[GRB 080319B]

For an AC circuit, the voltage across the source is equal to the vector sum of the voltages of the components. Using the phasor diagram, you can graphically determine the direction of the source voltage, if it leads the current or not, and how to find the vector sum of the voltages. Rms and peak amplitude voltage are proportional by sqrt{2}, so you don't need to convert to peak and then back to rms, just leave everything in rms and the source voltage will be in rms. The source voltage is related to the other voltages like this:

Vs = sqrt{ Vr^2 + (VL - Vc)^2 }

 

[nucleawasta]

For an AC circuit, the voltage across the source is equal to the vector sum of the voltages of the components. Using the phasor diagram, you can graphically determine the direction of the source voltage, if it leads the current or not, and how to find the vector sum of the voltages. Rms and peak amplitude voltage are proportional by sqrt{2}, so you don't need to convert to peak and then back to rms, just leave everything in rms and the source voltage will be in rms. The source voltage is related to the other voltages like this:

Vs = sqrt{ Vr^2 + (VL - Vc)^2 }

Thank you so much! I've ben stuck on this for so long.