Question

A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.

Answer 1

Let an alternating Emf E = E₀ sinωt is applied to a series combination of inductor L, capacitor C and resistance R. Since all three of them are connected in series the current through them is same. But the voltage across each element has a different phase relation with current.

The potential difference V_L, V_C and V_R across L, C and R at any instant is given by

V_L = IX_L, V_C = IX_C and V_R = I_R

Where I is the current at that instant.

X_L is inductive reactance and

X_C is capacitive reactance.

V_R is in phase with I. V_L leads I by 90° and V_C lags behind I by 90°

In the phases diagram,

V_L and V_C are opposite to each other. If V_L > V_C then resultant (V_L − V_C) is represent by OD. OR represent the resultant of V_R and (V_L − V_C). It is equal to the applied Emf E.

`E^2 = V_R^2 + (V_L -V_C)^2`

`E^2 =I^2 +[R^2+(X_L -X_c)^2]`

`or I =E/sqrt (R^2 + (X_2 -X_c)^2)`

The term  `sqrt(R^2 +(X_2  - X_c))` is called impedance Z of the LCR circuit.

`Z = sqrt(R^2 +(X_2 -X_c)^2) =sqrt(R^2 +(L omega-1/(comega))^2)`

Emf leads current by a phase angle Φ

`tan phi = (V_L -V_C)/R = (X_L - X_c)/R =(Lomega -1/(comega))/R`

When resonance takes place

`omegaL= 1/(omegac)`

Impedance of circuit becomes equal to R. Current becomes maximum and is equal to `E/R`

`omega_0 = 1/sqrt(LC)`

`f_0 = omega_0/(2pi) = 1/(2pisqrt(LC))`

This is the condition for resonance.

When at resonance f = f₀ the current in the circuit is maximum and hence impedance of the circuit is maximum for values of f less than or greater than f₀ comparatively small current flames in the circuit.