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An inductor `20 mH`, a capacitor `100muF` and a resistor `50Omega` are connected in series across a source of emf `V=10sin314t`. The power loss in the circuit isA. `1.13W`B. `0.79W`C. `2.74W`D. `0.43W` |
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Answer» Correct Answer - B `V_(0)=10V,omega=314 rad//s` `X_(L)=omegaL=(314)(20xx10^(-3))=6.280` `X_(C)=(1)/(omegaC)=(1)/(314xx100xx10^(-6))=31.84Omega` and `R=50Omega` `Z=sqrt((X_(c)-X_(L))^(2)+R^(2))` `=sqrt((31.84-6.28)^(2)+(50)^(2)=56Omega` Power loss `P=(i_(rms))^(2).R=(i_(0)^(2)R)/(2)` Hence `P=((0.18)^(2)xx50)/(2)=0.81W` |
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