Resistance Inductance and Capacitance in Series

The three elements Resistance Inductance and Capacitance in Series are shown in Figure (A) joined in series across an A.C. supply of R.M.S. voltage V.

Let V_R = I_R = voltage drop across R —in phase with I

V_L = I.X_L = voltage drop across L —leading I by π/2

V_C = I.X_C = voltage drop across C —lagging I by π/2

In voltage triangle of Figure (B), OA represents V_R, AB and AC represent the inductive and capacitive drops respectively. It will be seen that V_L and V_C are 180° out of phase with each other i.e. they are in direct opposition to each other.

Subtracting BD (= AC) from AB, we get the net reactive drop AD = I (X_L − X_C)

The applied voltage V is represented by OD and is the vector sum of OA and AD

(impedance)² = (resistance)² + (net reactance)²
or Z² = R² + (X_L− X_C)² = R² + X²

where X is the net reactance (Figure (A), (B) & (C)).
Phase angle φ is given by tan φ = (X_L − X_C)/R = X/R = net reactance/resistance

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Power factor is

Hence, it is seen that if the equation of the applied voltage is v = V_msin ωt, then equation of the resulting current in an R-L-C circuit is given by i = I_msin (ωt ± φ)

The + ve sign is to be used when current leads i.e. X_C > X_L.
The − ve sign is to be used when current lags i.e. when X_L > X_C.

In general, the current lags or leads the supply voltage by an angle φ such that tan φ = X/R.

Using symbolic notation, we have (Figure (D)), Z = R + j (X_L − X_C)

Numerical value of impedance

Its phase angle is Φ = tan⁻¹ [X_L − X_C/R]

Z = Z ∠ tan⁻¹[(X_L − X_C)/R] = Z ∠ tan⁻¹(X/R)

If V = V ∠ 0, then, I = V/Z

Summary of Results of Series AC Circuits

Type of Impedance	Value of Impedance	Phase angle for current	Power factor
Resistance only	R	0°	1
Inductance only	ωL	90° Lag	0
Capacitance only	1/ωC	90° Lead	0
Resistance and Inductance		0 < φ < 90° lag	1 > p.f. > 0 lag
Resistance and Capacitance		0 < φ < 90° lead	1 > p.f. > 0 lead
R-L-C		between 0° and 90° lag or lead	between 0 and unity lag or lead