CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

For a given series LCR circuit, list-I shows instantaneous voltage across inductor, capacitor, resistance and source and list-II shows values corresponding to quantites of  list-I in SI units.

Match List-I with List -2  when current in circuit is half of its maximum value for the first time.

List- 1List - 2I)VLP)10II)VcQ)12III)VRR)16IV)ϵS)17.3T)16U)17.3
  1. I S II  U III  P IV
  2. I P II  U III  S IV Q
  3. I P II  P III  Q IV
  4. I S II  U III  Q IV


Solution

The correct option is A I S II  U III  P IV
i=i02   θ=30ϵ=ϵ0sinθ=20×12=10 Vi=i0sinθ=1×12=0.5 AVR=iR=0.5×20=10 A

VL=Ldidt=Li0ωcosωt=20×103×1×103×32=17.3 V
Since XC=XL, the circuit is purely resistive. So,
VC=VL=17.3 V
ϵ0=20sin37=10 V

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image