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answer>>>>>>>>>>>>>>>>>.........................................

Soma Mesh , 7 Years ago
Grade 12
anser 1 Answers
Raman Parjapat
First of all, it is from Alternating Current chapter in Physics.
Given in the question, L=20mH     C=120uF    R=60ohm    t=60s
AC source of 24V/50Hz.
Time period,T =1/50 =20ms
The given is much larger than time peroid,
Hence we can write the average dissipated power
R2=602=3600ohm
XL=\large \omega L=2\large \Pi \upsilonL
     =2×3.14×50×20×10-3=6.14ohm
Xc=1/\large \omegaC
    =1/2×3.14×50×120×10-6=26.54ohm
Z2=R2+(XL−Xc)2
    =3600+(6.14-26.54)2
     =3600+416.16
    =4016.16ohm2
         
 
Average power  dissipated,Pav=Vrms×Irms×\large \cos \phi =\frac{R}{Z}
                                                  =Vrms×Vrms×R/(Z×Z)
                                                   =(Vrms)2×R/Z2
                                                    =24×24×60/4016.16
                                                     =8.60W
 
Last Activity: 6 Years ago
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