What is the power dissipated in an LCR series circuit connected to an ac source of emf ϵ ?

Question

What is the power dissipated in an LCR series circuit connected to an ac source of emf ϵ ? (A) ((ϵ )2 √R2 + (ω L - (1/ω C))2/R) (B) ((ϵ )2 [R2 (ω L - (1/ω C))2]/R) (C) ((ϵ )2 R/√R2 + (ω L - (1)ω C)2) (D) ((ϵ )2 R/[R2 + (ω L - (1)ω C)2])

Tardigrade Admin · Accepted Answer

Average power, P=Vr m sIr m scos φ Here, Z=√R2 + (XL - XC)2cos φ=(R/Z) But Ir m s=(Vr m s/Z). ∴ P=Vr m s2(R/Z2) ∴ P=Vr m s2(R/ . R2 + (XL - XC)2 . )=((ϵ )2 R/[R2 + (ω L - (1)ω C)2])

Q. What is the power dissipated in an $L CR$ series circuit connected to an ac source of emf $ϵ$ ?

$\frac{( ϵ ) ^{2} R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}{R}$

$\frac{( ϵ ) ^{2} [ R ^{2} ( ω L - \frac{1}{ω C} ) ^{2} ]}{R}$

$\frac{( ϵ ) ^{2} R}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$

$\frac{( ϵ ) ^{2} R}{[ R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2} ]}$

Q. What is the power dissipated in an LCR series circuit connected to an ac source of emf ϵ ?

R(ϵ)2R2+(ωL−ωC1​)2​​

R(ϵ)2[R2(ωL−ωC1​)2]​

R2+(ωL−ωC1​)2​(ϵ)2R​

[R2+(ωL−ωC1​)2](ϵ)2R​

Average power, P=Vrms​Irms​cosϕ Here, Z=R2+(XL​−XC​)2​cosϕ=ZR​ But Irms​=ZVrms​​.∴P=Vrms2​Z2R​ ∴P=Vrms2​{R2+(XL​−XC​)2}R​=[R2+(ωL−ωC1​)2](ϵ)2R​

Q. What is the power dissipated in an $L CR$ series circuit connected to an ac source of emf $ϵ$ ?

$\frac{( ϵ ) ^{2} R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}{R}$

$\frac{( ϵ ) ^{2} [ R ^{2} ( ω L - \frac{1}{ω C} ) ^{2} ]}{R}$

$\frac{( ϵ ) ^{2} R}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$

$\frac{( ϵ ) ^{2} R}{[ R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2} ]}$