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Chapter 24: Application of Complex Numbers to Series AC Circuits

24.1 Introduction

AC circuits may be analysed by using complex numbers for simplified phasor diagrams. This can solve complicated circuits.

24.2 Series AC Circuits

Pure Resistance

In pure resistance the circuit in polar equation is given by:

Z = (VR∠0°) / (IR∠0°) = R

Pure Inductance

In pure inductance the current lags the applied voltage by 90°:

Z = (VL∠90°) / (I∠0°) = XL∠90° = jXL

Where XL is inductive reactance, XL = ω L = 2\pi fL ohms.

Pure Capacitance

The voltage lags the current by 90° in this circuit:

Z = (VC∠-90°) / (IC∠0°) = XC∠-90° = -jXC

Where XC is 1/ω C. A notable equation converts the imaginary representation:

-jXC = -j / ω C = 1 / jω C

R-L Series Circuit

The current is said to be lagging even though the phase is +90° as the current is behind the voltage:

V = VR + jVL
Z = R + jXL

R-C Series Circuit

The relations for the voltage and impedance triangles are:

V = IZ   and   VC = IXC
Z = R – jXC

R-L-C Series Circuit

The voltage and total impedance are defined as:

V = VR + j(VL – VC)
|Z| = √(R2 + (XL – XC)2),   φ = tan-1((XL – XC) / R)
Z = R + j(XL – XC) = |Z|∠φ

General Series Circuit

In an a.c. circuit containing several impedances connected in series, say Z1, Z2, Z3 … Zn, the total equivalent impedance is given by:

ZT = Z1 + Z2 + Z3 + … + Zn

Chapter 25.3: Parallel AC Networks

For a circuit containing parallel impedances Z1, Z2 and Z3, the potential difference is the same and equal to the supply voltage V:

I = I1 + I2 + I3

If ZT is the total impedance, then:

1 / ZT = 1 / Z1 + 1 / Z2 + 1 / Z3

In general for impedances connected in parallel, the total admittance YT is:

YT = Y1 + Y2 + Y3 + … + Yn

For Two Impedances Connected in Parallel

ZT = (Z1 Z2) / (Z1 + Z2)

Some useful current equations:

V = I ZT = I ((Z1 Z2) / (Z1 + Z2))
I1 = I (Z2 / (Z1 + Z2))
I2 = I (Z1 / (Z1 + Z2))