Question

Derive an expression for electric power in an LCR a.c. series circuit.

Answer 1

Let E be alternating e.m.f. applied to an LCR circuit and is given by

E = E₀ sin ωt

If a.c. lags behind the applied e.m.f. by Φ, then

I = I₀ sin (ωt - Φ)

Instantaneous power,

or Average power = Apparent power x power factor

Average power is also called true power.

∴ True power = Apparent power x power factor

∴ Power factor = cos Φ

= \(\frac{\text{ True power }}{\text{ Apparent power }}\) ............(2)

In practie, the true power is expressed in kilowatt (kW) and apparent power in kilowatt ampere (kVA).

Special cases

(i) If a.c. circuit contains only resistance (R)

Here Φ = 0, so cos Φ = cos 0° = 1

∴ P_av = E_vI_v cos 0°

or P_av = E_vI_v

(ii) If a.c. circuit contains only pure inductance (L)

Here Φ = π/2

So Pav = E_vI_v cos π/2 = 0 [∵ cos π/2 = 0]

i.e. Average power consumed in pure inductive circuit in a.c. is zero.

(iii) If a.c. circuit contains pure capacitance (C)

Here Φ = -π/2

∴ P_av = E_vI_v cos (-π/2) = 0 [∵ cos -π/2 = -cos π/2 = 0]

i.e. Average power consumed in pure capacitive circuit in a.c. is zero.

(iv) If a.c. circuit contains LCR circuit

Here 

tan Φ = \(\frac{\omega L-\frac{1}{\omega C}}{R}\)

From the figure, we have

cos Φ = \(\frac{R}{\sqrt{R^2+(\omega L- \frac{1}{\omega C})^2}} = \frac{R}{Z}\) .............(3)

where Z = \({\sqrt{R^2+(\omega L- \frac{1}{\omega C})^2}}\) .............(4)

Z is the impedence of LCR circuit

∴ P_av = E_vI_v cos Φ = \(\frac{E_vI_vR}{Z}\)

or P_av = \(\frac{E^2_vR}{Z^2}\)  [∵ I_v = \(\frac{E_v}{Z}\)]

or P_av = \(\frac{Er^2R}{{{R^2+(\omega L- \frac{1}{\omega C})^2}}}\)