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Consider the following R-L-C circuit in ...

Consider the following R-L-C circuit in which R=`12 Omega`. `X_(L)= 24 Omega`, `X_(C)= 8 Omega`. The emf of source is given by V =`10 sin (100pit)V`.
Find the energy dissipated in 10 min.
If resistance is removed from the circuit and value of inductance is doubled, express variation of current with time t in the new circuit.

Text Solution

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(i) Impedance,
`Z= sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt((12)^(2)+(24-8)^(2))=20 Omega`
`cos phi=R/Z = 12/20 = 3/5`
Power consumed,
`P=1/2V_(0)I_(0)cosphi=1/2V_(0).V_(0)/2cosphi`
`(1/2)(10)^(2)/(20) xx 3/5 = 1.5 W`
Energy dissipated, `E=Pt=(1.5)(1.0 xx 60)=900J`
(ii)
Impedance, `Z=X^(')L-X_(c) = 48-8=40Omega`
`therefore` Current, `I_(0) = V_(0)/X=10/40 = 1/4A`

Current lags the voltage by `pi/2`
`I=I_(0)sin(omegat-pi/2) = 1/4 sin (100pit-pi/2)`
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