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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. |
Answer» 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)`. |
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