Skip to content
Question 47 of 50

Q.Obtain an expression for average power dissipated in a series LCR circuit.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2025Subjective· 4mImportance★★★★★
94% · 47/50 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The average power in an AC circuit is not simply VrmsIrmsV_{rms}I_{rms} but includes the power factor cos⁡φ\cos\varphi accounting for the phase lag between current and voltage.

In a series LCR circuit driven by v=V0sin⁡ωtv = V_0\sin\omega t, the current is i=I0sin⁡(ωt−φ)i = I_0\sin(\omega t - \varphi), where φ\varphi is the phase angle between voltage and current, tan⁡φ=(XL−XC)/R\tan\varphi = (X_L-X_C)/R.

Instantaneous power:

p=vi=V0I0sin⁡ωt sin⁡(ωt−φ)p = vi = V_0 I_0 \sin\omega t\,\sin(\omega t - \varphi)

Using sin⁡(ωt−φ)=sin⁡ωtcos⁡φ−cos⁡ωtsin⁡φ\sin(\omega t - \varphi) = \sin\omega t\cos\varphi - \cos\omega t\sin\varphi:

p=V0I0[sin⁡2ωtcos⁡φ−sin⁡ωtcos⁡ωtsin⁡φ]p = V_0 I_0\left[\sin^2\omega t\cos\varphi - \sin\omega t\cos\omega t\sin\varphi\right]

Averaging over a full cycle: ⟨sin⁡2ωt⟩=12\langle\sin^2\omega t\rangle = \tfrac12 and ⟨sin⁡ωtcos⁡ωt⟩=0\langle\sin\omega t\cos\omega t\rangle = 0, so

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.