krit.club logo

Alternating Current - Power in AC Circuits

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

β€’

Instantaneous power in an AC circuit is given by the product of instantaneous voltage v=Vmsin⁑(Ο‰t)v = V_m \sin(\omega t) and instantaneous current i=Imsin⁑(Ο‰t+Ο•)i = I_m \sin(\omega t + \phi).

β€’

The average power dissipated in an AC circuit over a complete cycle depends not only on the voltage and current but also on the phase difference Ο•\phi between them.

β€’

The term cos⁑ϕ\cos \phi is known as the Power Factor of the circuit. Its value ranges from 00 to 11.

β€’

In a purely resistive circuit, Ο•=0\phi = 0, so cos⁑ϕ=1\cos \phi = 1. The power dissipation is maximum.

β€’

In a purely inductive or purely capacitive circuit, the phase difference Ο•=Ο€2\phi = \frac{\pi}{2}. Consequently, the power factor cos⁑ϕ=0\cos \phi = 0, and no power is dissipated. This lead to the concept of Wattless Current.

β€’

The power is only dissipated across the Resistor (RR) in an LCR circuit; the average power consumed by an ideal inductor or an ideal capacitor is zero.

β€’

At resonance in an LCR circuit, XL=XCX_L = X_C, which means Ο•=0\phi = 0 and cos⁑ϕ=1\cos \phi = 1. The circuit behaves as a purely resistive circuit and power dissipation is maximum (P=VrmsIrmsP = V_{rms} I_{rms}).

πŸ“Formulae

Pinst=vβ‹…iP_{inst} = v \cdot i

Pavg=VrmsIrmscos⁑ϕP_{avg} = V_{rms} I_{rms} \cos \phi

Vrms=V02,Irms=I02V_{rms} = \frac{V_0}{\sqrt{2}}, \quad I_{rms} = \frac{I_0}{\sqrt{2}}

PowerΒ Factor=cos⁑ϕ=RZ\text{Power Factor} = \cos \phi = \frac{R}{Z}

Z=R2+(XLβˆ’XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Pavg=Irms2RP_{avg} = I_{rms}^2 R

Iwattless=Irmssin⁑ϕI_{wattless} = I_{rms} \sin \phi

πŸ’‘Examples

Problem 1:

A series LCR circuit with R=80 ΩR = 80 \, \Omega, XL=100 ΩX_L = 100 \, \Omega, and XC=40 ΩX_C = 40 \, \Omega is connected to a 200 V,50 Hz200 \, V, 50 \, Hz AC supply. Calculate the power factor and the average power dissipated in the circuit.

Solution:

  1. First, calculate the impedance ZZ: Z=R2+(XLβˆ’XC)2=802+(100βˆ’40)2=802+602=6400+3600=100 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + (100 - 40)^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = 100 \, \Omega.

  2. Calculate the Power Factor: cos⁑ϕ=RZ=80100=0.8\cos \phi = \frac{R}{Z} = \frac{80}{100} = 0.8.

  3. Calculate IrmsI_{rms}: Irms=VrmsZ=200100=2 AI_{rms} = \frac{V_{rms}}{Z} = \frac{200}{100} = 2 \, A.

  4. Calculate Average Power: Pavg=VrmsIrmscos⁑ϕ=200Γ—2Γ—0.8=320 WP_{avg} = V_{rms} I_{rms} \cos \phi = 200 \times 2 \times 0.8 = 320 \, W.

Explanation:

The power factor is the ratio of resistance to impedance. The average power is the product of the effective voltage, effective current, and the power factor. Alternatively, Pavg=Irms2R=22Γ—80=320 WP_{avg} = I_{rms}^2 R = 2^2 \times 80 = 320 \, W gives the same result.

Problem 2:

Show that the average power consumed by a pure inductor over one complete cycle of AC is zero.

Solution:

For a pure inductor, the current ii lags the voltage vv by a phase angle of Ο€2\frac{\pi}{2}. If v=Vmsin⁑(Ο‰t)v = V_m \sin(\omega t), then i=Imsin⁑(Ο‰tβˆ’Ο€2)=βˆ’Imcos⁑(Ο‰t)i = I_m \sin(\omega t - \frac{\pi}{2}) = -I_m \cos(\omega t). Pavg=1T∫0Tvβ‹…i dt=1T∫0T(Vmsin⁑ωt)(βˆ’Imcos⁑ωt) dtP_{avg} = \frac{1}{T} \int_0^T v \cdot i \, dt = \frac{1}{T} \int_0^T (V_m \sin \omega t) (-I_m \cos \omega t) \, dt Pavg=βˆ’VmIm2T∫0Tsin⁑(2Ο‰t) dtP_{avg} = -\frac{V_m I_m}{2T} \int_0^T \sin(2\omega t) \, dt. Since the integral of a sine function over a full period is zero, Pavg=0P_{avg} = 0.

Explanation:

Mathematically, the phase difference Ο•\phi is 90∘90^\circ. Using the formula Pavg=VrmsIrmscos⁑90∘P_{avg} = V_{rms} I_{rms} \cos 90^\circ, since cos⁑90∘=0\cos 90^\circ = 0, the power dissipation is zero.

Power in AC Circuits - Revision Notes & Key Formulas | CBSE Class 12 Physics