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
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.
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.
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
G = 1 = 1 = 8,333 mS
. R 120 Ω
BL = 1 = 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 = 1 = 66,565 e j56,31° Ω
. Y 15,023 e -j56,31° mS
Solution 2: Calculate without detour
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.
Determine the impedance via the conductances