Question

Derive an expression for the impedance of a series LCR circuit connected to an AC power supply.

Answer 1

The alternating emf (e) source (power supply) is connected to a key K, a capacitor of capacitance C, a resistor of resistance R, and an inductor of inductance L to produce a closed series circuit.

       LCR series circuit

We consider the resistor, capacitor, and inductor to be perfect. These carry the same current, i = i₀ sin ωt at all times since they are connected in series.

The current is in phase with the voltage across the resistor, e_R = Ri. The current is led by `π/2` rad by the voltage across the inductor, e_L = X_Li, and lags behind by `π/2` rad by the voltage across the capacitor, e_C = X_Ci. The phasor diagram (Fig.) illustrates this.

Phasor diagram for an LCR series circuit

From this figure, `e_0^2 = e_R^2 + (e_L - e_C)^2`

= `R^2i_0^2 + (X_Li_0 - X_Ci_0)^2`

= `i_0^2 [R^2 + (X_L - X_C)^2]`

∴ `e_0 = i_0 sqrt(R^2 + (X_L - X_C)^2) = i_0Z`, where

 `Z = e_0/i_0 = sqrt(R^2 + (X_L - X_C)^2)` is the effective resistance of the circuit. It is called the impedance.