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

A

ε²R / [R² + (L\omega - 1/C\omega )²]

B

ε²√[R² + (L\omega - 1/C\omega )²] / R

C

ε²[R² + (L\omega - 1/C\omega )²] / R

D

ε²R / √[R² + (L\omega - 1/C\omega )²]

Step-by-Step Solution

In an L-C-R series circuit, power is dissipated only in the resistor, as the average power over a cycle for pure inductors and capacitors is zero. The average power dissipated is given by P=I2RP = I^2 R, where II is the rms current . The rms current is given by I=ε/ZI = \varepsilon / Z, where ε\varepsilon is the rms emf and ZZ is the impedance. The impedance Z=R2+(XLXC)2=R2+(Lω1Cω)2Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (L\omega - \frac{1}{C\omega})^2} . Substituting II into the power equation: P=(εZ)2R=ε2RZ2=ε2RR2+(Lω1Cω)2P = \left( \frac{\varepsilon}{Z} \right)^2 R = \frac{\varepsilon^2 R}{Z^2} = \frac{\varepsilon^2 R}{R^2 + (L\omega - \frac{1}{C\omega})^2}.

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