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Power dissipated in an L-C-R series circ...

Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` is

A

`(epsilon^(2)sqrt(R^(2)+(omega L-(1)/(omega C))^(2)))/(R )`

B

`(epsilon^(2)[R^(2)+(omega L-(1)/(omega C))^(2)])/(R )`

C

`(epsilon^(2)R)/(sqrt(R^(2)+(omega L-(1)/(omega C))^(2)))`

D

`(epsilon^(2)R)/([R^(2)+(omega L-(1)/(omega C))^(2)])`

Text Solution

Verified by Experts

The correct Answer is:
D
Average power, `P = V_("rms")I_("rms")cos phi`
Here, `Z = sqrt(R^(2)+(X_(L)-X_(C ))^(2)), cos phi = (R )/(Z)`
But `I_("rms")=(V_("rms"))/(Z). Therefore P= V_("rms")^(2)(R )/(Z^(2))`
`therefore P = V_("rms")^(2) (R )/({R^(2)+(X_(L)-X_(C ))^(2)})=(epsilon^(2)R)/([R^(2)+(omega L-(1)/(omega C))^(2)])`
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Knowledge Check

  • In L-C-R series AC circuit,

    A
    If `R` is increased, then current will decrease
    B
    If `L` is increased, then current will decrease
    C
    If `C` is increased, then current will increase
    D
    If `C` is increased, then current will decrease

  • In an L-C-R series, AC circuit at resonance

    A
    the capacitive reactance is more than the inductive
    B
    the capacitive reactance equals the inductive reactance
    C
    the capactive reactance is less than the inductive reactance
    D
    the power dissipated is minimum

  • An L-C-R series circuit , connected to a source E, is at resonance. Then,

    A
    the voltage across R is zero
    B
    the voltage across R equals applied voltage
    C
    the volatage across C is zero
    D
    the voltage across C equals applied voltage