For the L-R-C series circuit described in the above Illustration 5.7, describe the time dependence of the instantaneous current and each instantaneous voltage.

Explanation

we found the amplitude of the current and voltages. Now we have to find the expressions for the instantaneous values of the current and voltages. As we learned, the voltage across a resistor is in phases with the current but the voltages across an inductor or capacitor are not. We also learned in this section that `phi` is the phase angle between the source voltage and the current.
The current and all the voltage oscillate with the same angular frequency.
Let us choose
`E=E_(0)sin omega t implies E=50 sin(10,000 t)`
then `I_I_(0) sin (10,000 t+ phi)`,
where `phi= -53^(@)=-(53xxpi)/(180) rad = -0.93 rad`
`implies I=0.10 sin (10,000 t-0.93)`
Current and voltage are in phase in resistor so
`V_(R)=V_(R0) sin (10,000t-0.93)=30 sin(10,000 t-0.93)`
In inductor voltage voltage leads current by `pi//2` so ,
`V_(L)=V_(L0) sin [10,000t-0.93+(pi//2)]=60 cos(10,000 t-0.93)`
In capcitor voltage lags currents lags currents by `(pi//2)` so
`V_(C)=V_(C0) sin[10,000t-0.93+(pi//2)]`
`=-20 cos (10,000t-0.93)`.