Question

In a series LCR circuit connected to an a.c. source of voltage v = v_msinωt, use phasor diagram to derive an expression for the current in the circuit. Hence, obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum

Answer 1

The phasor diagram for the series LCR circuit is

The phasor relation for the voltages is
V_L + V_R + V_C = V

Because VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (V_C + V_L) which has a magnitude |v_{Cm -} v_Lm|.

Because V is represented as the hypotenuse of a right triangle whose sides are V_R and (V_C + V_L), the Pythagorean theorem gives

`v_m^2=v_(Rm)^2+(V_(Cm)-V_(Lm))^2`

`:.v_m^2=(i_mR)^2+(i_mX_c-i_mX_L)^2`

`:.v_m^2=i_m^2[R^2+(X_C-X_L)^2]`

`:.i_m=v_m/sqrt(R^2+(X_C-X_L)^2)`

This is the expression for the current in the circuit.

The instantaneous power in the circuit is

P = vi = (v_msinωt)(i_msin(ωt+Φ))

`:. P=(v_mi_m)/2(sinomegat)(sin(omegat+phi))`

`:.P=(v_mi_m)/2(cosphi-cos(2omegat+phi))`

The average power over a cycle is given by the average of the two terms in the RHS of the above equation.
It is only the second term (cos(2ωt+)) which is time-dependent. Its average is zero. Therefore, we get

`P=(v_mi_m)/2cosphi=v_m/sqrt2xxi_m/sqrt2cosphi`

∴ P = VI cosΦ

It can also be written as  P = I²ZcosΦ

At resonance X_C − X_L = 0 and Φ = 0. Therefore, cosΦ = 1 and power is P=I²Z=I²R.

That is, maximum power is dissipated in a circuit at resonance.