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Q.Derive an expression for power dissipated in LCR series AC circuit and explain power factor. OR Explain principle, construction and working of A.C. generator.

Haryana BsehBSEH Intermediate Board 2025Subjective· 5mImportance★★★★★
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The average power dissipated in a series LCR circuit depends on the phase angle between current and voltage through the power factor cosφ = R/Z, and is zero for a purely reactive (L or C only) circuit.

Setup: In a series LCR circuit driven by v=vmsin⁡ωtv = v_m\sin\omega t, the current lags/leads the voltage by phase angle φ:

i=imsin⁡(ωt−ϕ)i = i_m\sin(\omega t - \phi)

where im=vm/Zi_m = v_m/Z, Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2} is the impedance, and tan⁡ϕ=XL−XCR\tan\phi = \dfrac{X_L - X_C}{R}.

Instantaneous power:

p=vi=vmsin⁡ωt⋅imsin⁡(ωt−ϕ)p = vi = v_m\sin\omega t \cdot i_m\sin(\omega t - \phi)

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

p=vmim[sin⁡2ωtcos⁡ϕ−sin⁡ωtcos⁡ωtsin⁡ϕ]p = v_mi_m\left[\sin^2\omega t\cos\phi - \sin\omega t\cos\omega t\sin\phi\right]

Averaging over a full cycle: ⟨sin⁡2ωt⟩=1/2\langle\sin^2\omega t\rangle = 1/2 and ⟨sin⁡ωtcos⁡ωt⟩=0\langle\sin\omega t\cos\omega t\rangle = 0, so

Pˉ=vmim2cos⁡ϕ=vm2⋅im2cos⁡ϕ=VrmsIrmscos⁡ϕ\bar{P} = \frac{v_mi_m}{2}\cos\phi = \frac{v_m}{\sqrt2}\cdot\frac{i_m}{\sqrt2}\cos\phi = V_{rms}I_{rms}\cos\phi

Power factor: The term cos⁡ϕ\cos\phi is called the power factor. Using the impedance triangle, cos⁡ϕ=R/Z\cos\phi = R/Z, so equivalently

Pˉ=Irms2R\bar{P} = I_{rms}^2 R …

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