Question

A Review | Constants An L – R-C series circuit has voltage amplitudes Vư = 180 V, Vc = 120 V, and VR = 160 V. At time t the instantaneous voltage across the inductor is 80.0 V. Part A At this instant, what is the voltage across the capacitor? Express your answer with the appropriate units. ? Å Value * o Units VC= Submit Request Answer

Answer 1

The voltage across the resistor (R), inductor (L), and capacitor (L) in series R-L-C circuit is given by

                                      

VR= IR sin wt

                                      
Vi = IX sin (wt +90°)
            ;
Xc = iw

                                      
VC = IXc sin (at – 90°)
            ;
Xc = iwc

where $\omega$ is frequency, and t is time.

Given,

                                     

Vimar = IX = 180 volt

                                     

Vomar = IXc = 120 volt

                                     

VRmar = IR = 160 volt

at a certain time t, the voltage across the inductor is 80 volt.

Therefore, we can write

                                      

80 = 180 sin(wt +90°)

                               or,  

coswt = 80/180

                               or,  

coswt = 4/9

             and,                  

sin wt = V1 -coswt = V1 -(4/9)2 =

So at that moment of time

Part A:

the voltage across the capacitor is

                                       

Vc = 120 sin (wt - 90°) = 120 coswt = 120 X 4/9 = 53.33 volts

Part B:

the voltage across the capacitor is

                                       

VR= 160 sin wt = 160 X V65/9 = 143.33 volts