VCI Xc⋅=Note the magnitude of the voltage across the inductance is larger than the source voltage magnitude. Also, the angle is 90oahead of the voltage across the resistance.arg VL(2922.62deg=VL231.17volt=VL213.38788.911j+volt=VLI Xl⋅=arg VR(2967.38-deg=VR46.234volt=VR17.78242.677j-volt=VRI R⋅=c. To find the voltage across each element, multiply current times impedance:arg I( )67.38-deg=I9.247amp=Note: "arg" is the angle of a complex numberI3.5568.535j-amp=IVZ=Z512j+ohm=ZRXl+Xc+=Xc13j-ohm⋅=Xl25j ohm⋅=R5 ohm⋅=b. The current is the voltage divided by the impedance. Since the three elements are in series we can add them directly:We can let the voltage be our reference, so the angle is zero
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