Power dissipated in an LCR series circuit connected to an a.c. source of emf e is :

Question

Power dissipated in an LCR series circuit connected to an a.c. source of emf e is : (A) (ε2 √R2+(L ω-(1/C ω))2/R) (B) ε2[R2+(L ω-(1/C ω))2] (C) (ε2 R/√R2+(L ω-(1)C ω)2) (D) (ε2 R/[R2+(L ω-(1)C ω)2])

Tardigrade Admin · Accepted Answer

Average power, P=E text r.m.s I text r.m.s cos φ Z=√R2+(XL-XC)2, cos φ=(R/Z) But I textr . m . s =(E text r . m . s /Z), ∴ P=E text r . m . s 2 ⋅ (R/Z2) ∴ P =E text r . m . s 2 (R/ R2+(XL-XC)2 ) =(ε2 R/[R2+(L ω-(1)C ω)2])

Q. Power dissipated in an $L CR$ series circuit connected to an a.c. source of emf $e$ is :

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

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

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

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

Q. Power dissipated in an LCR series circuit connected to an a.c. source of emf e is :

Rε2R2+(Lω−Cω1​)2​​

ε2[R2+(Lω−Cω1​)2]

R2+(Lω−Cω1​)2​ε2R​

[R2+(Lω−Cω1​)2]ε2R​

Average power, P=Er.m.s ​Ir.m.s ​cosϕ Z=R2+(XL​−XC​)2​,cosϕ=ZR​ But Ir . m . s ​=ZE r . m . s ​​, ∴P=E r . m . s 2​⋅Z2R​ ∴P=E r . m . s 2​{R2+(XL​−XC​)2}R​ =[R2+(Lω−Cω1​)2]ε2R​

Q. Power dissipated in an $L CR$ series circuit connected to an a.c. source of emf $e$ is :

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

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

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

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