4.6 Series $L-C-R$ circuit

  Alternating Current

Consider an inductor of self inductance $L$, capacitor of capacitance $C$ and a resistor of resistance $R$ is connected in series to an AC power supply as shown in the figure.

For phasor diagram we can write,

Phasor diagram of resistor $+$ Phasor diagram of inductor $+$ Phasor diagram of capacitor $=$ Phasor diagram of circuit

So, voltage $V$ can be written as,

$$V = {V_L} + {V_C} + {V_R}$$$$V = i{X_L}\left( {\widehat j} \right) + i{X_C}\left( { - \widehat j} \right) + iR$$or $$V = i\left( {R + \widehat j\left[ {{X_L} - {X_C}} \right]} \right)$$or $$V = iZ$$

where,

$$Z = R + \widehat j\left[ {{X_L} - {X_C}} \right]$$ or $$Z = R + \widehat j\left[ {\omega L - \frac{1}{{\omega C}}} \right]$$

The modulus of impedance can be written as,

$$\left| Z \right| = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $$ or $$\left| Z \right| = \sqrt {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} $$

The angle by which the potential difference leads the current is, $$\tan \phi = \left( {\frac{{{V_L} - {V_C}}}{{{V_R}}}} \right)$$$$\tan \phi = \left( {\frac{{{X_L} - {X_C}}}{R}} \right)$$$$\tan \phi = \left( {\frac{{\omega L - \frac{1}{{\omega C}}}}{R}} \right)$$

Note:

1. If ${X_L} > {X_C}$ then $\tan \phi $ is positive. Therefore, $\phi $ is positive. Hence, the voltage leads the current by a phase angle $\phi $. Thus, the AC circuit is inductor dominated circuit.

2. If ${X_L} < {X_C}$ then $\tan \phi $ is negative. Therefore, $\phi $ is negative. Hence, the voltage lags the current by a phase angle $\phi $. Thus, the AC circuit is capacitors dominated circuit.

3. If ${X_L} = {X_C}$ then $\tan \phi $ is zero. Therefore, $\phi $ is zero. Hence, in such a case current and voltage are in phase with each other. Thus, the AC circuit becomes pure resistor circuit as impedance becomes equal to the resistance of the circuit.