Complex numbers in electrical engineering

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Complex numbers in electrical engineering

In mathematics, it is common to denote the imaginary part with the letter i. In electrical engineering, however, the letter i is reserved for the electric current. To avoid confusion, we use j for the imaginary part in the following: Z = x + j y

Examples for the application of the complex numbers in electrical engineering now in the following:


Example 1: RL Series Circuit 

Series connection of ohmic and inductive resistance

Series connection of ohmic and inductive resistance

What you see is a series connection of an ohmic resistor of 120 Ω and an inductive resistor XL of 80 Ω. Determine mathematically and graphically the complex resistance Z of the series circuit.

Solution:

Complex numbers applied to a series connection of R and XL

Complex numbers applied to a series connection of R and XL


Complex numbers applied to a series connection of R and XC

Complex numbers applied to a series connection of R and C

Example 2: RC Series Circuit 

Determine the complex resistance for the series circuit of anohmic resistor and a capacitor as shown (R = 680 Ω, C = 4.7 μF). The frequency is 50 Hz.

Complex numbers applied to a series connection of R and C example

Complex numbers applied to a series connection of R and C example


Example 3: Parallel RL Circuit

Determine the impedance Z of a parallel circuit of resistance R = 120 Ω and inductive reactance XL = 80 Ω.

Solution 1:   Determine the impedance via the conductances

Complex numbers - Parallel RL Circuit - determine the impedance via the conductances

Complex numbers - Parallel RL Circuit - determine the impedance via the conductances

G  =     1        =   8,333 mS
.       R       120 Ω

BL  =     1        =   - j12,5 mS
.        XL       j80 Ω

⇒  Y = 8,333 mS - j 12,5 mS

To get to the impedance, we convert the Cartesian form into the Euler´s formula, so that we can form the reciprocal:

lYl = sqrt (G2  + BL2 )  = sqrt (8,3332 + 12,52) mS = 15,023 mS

φ = arctan-1 (BL / G)  = arctan-1 (12,5 / 8,333) = - 56,31°

⇒  Y = 15,023 e-j56,31° mS

Z   =         1                                 = 66,565 e j56,31° Ω
.       Y           15,023 e -j56,31° mS

Solution 2:  Calculate without detour

Complex numbers - Parallel RL Circuit - how to determine the impedance

Complex numbers - Parallel RL Circuit - how to determine the impedance

To the extent of our inaccuracy, this agrees with the result from solution 1!


Example 4: Parallel RC Circuit

Determine the impedance Z of a parallel circuit of resistance R = 120 Ω and capacitive reactance XC = 80 Ω.

Complex numbers - Parallel RC Circuit - how to determine the impedance

Complex numbers - Parallel RC Circuit - how to determine the impedance

Determine the impedance via the conductances

Complex numbers - Parallel RC Circuit - how to calculate the impedance

Complex numbers - Parallel RC Circuit - how to calculate the impedance