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**Home ⇒ Overview Courses ⇒ Electrical engineering ⇒ Complex numbers ⇒ Complex numbers in electrical engineering**

**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 X_{L} of 80 Ω. Determine mathematically and graphically the complex resistance Z of the series circuit.

Solution:

**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 X_{L} = 80 Ω.

Solution 1: Determine the impedance via the conductances

G = 1 = 1 = 8,333 mS

. R 120 Ω

B_{L} = 1 = 1 = - j12,5 mS

. X_{L} 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 (G^{2} + B_{L}^{2} ) = sqrt (8,333^{2} + 12,5^{2}) mS = 15,023 mS

φ = arctan^{-1} (B_{L} / 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 X_{C} = 80 Ω.

Determine the impedance via the conductances