# Fisika Study Center

## Complex Impedance in Alternating Current

physics.fisikastudycenter.com,- discussing alternating current circuit (AC circuit) using complex impedance to find the currents, angles on series or series-parallel circuit. Question 1

Given an AC electric circuit contents a resistor R and an inductor L as shown below: Where
R = 2 Ω
L = 2 mH
and the voltage source of ν = 120 sin 5000 t volts
Determine the equivalent impedance of this circuit, express the answer in rectangular form of complex impedance!

Discussion

Rectangular form of impedance :

Z = R + j ( XL − XC )

with
R = resistance (Ω)
XC = capasitive reactance (Ω)
XL = inductive reactance (Ω)

We have no capacitance in above circuit so XC = 0
and the inductive reactance
XL = ω L
XL = (5000)(2x10−3)
XL = 10 Ω

Z = R + j ( XL − XC )
Z = (8 + j 10) Ω

Take a note that expressing Z in complex impedance will be usefull in an advance circuit containing series and parallel as a combination, as shown in question 4, for the simple circuit as question 1, its easy enough to solve by using ordinary ways without complex impedance. Question 1, 2 and 3 will introduce us to the complex impedance forms.

Question 2
In below circuit R = 10 Ω, C = 5 μ F and ν = 200 sin 10000t volts Express the circuit impedance in rectangular form of complex impedance!

Discussion
Z = R + j ( XL − XC )
XL = 0
XC = 1/ω C
XC = 1/(10000)(5x10−6)
XC = 20 Ω

Z = R + j ( XL − XC )
Z = 10 − j 20 Ω

Question 3
Express the impedance of below circuit in rectangular form of complex impedance R = 8 Ω
L = 32 mH
C = 800 μF
ν = 120 sin 125 t volts

Discussion
Z = R + j ( XL − XC )
XL = ΩL
XL = (125)(32x10−6)
XL = 4 Ω

XC = 1/ω C
XC = 1/(125)(800 x 10−6)
XC = 10 Ω

Z = R + j ( XL − XC )
Z = 8 + j ( 4 − 10 )
Z = 8 − j 6 Ω

Question 4
Given an alternating current circuit contents R, L and C as the figure below At frequency of ω = 400 rad/s and ν = 120 /0o volts,
a) find the equivalent impedance of the circuit, express in complex
b) the total current of circuit
c) the current of resistance R = 5 Ω
d) the current of L = 25 mH

Discussion
a) equivalent impedance of this circuit:

R1 = 8Ω
R2 = 5Ω
XL = ΩL
XL = (400(25x10−6)
XL = 10 Ω → + j 10 in the polar form we get: b) Total current c) Curent through L → IL or I2 in the figure above d) Curent through R = 5 Ω → I5 or I1 in the figure above Prepared by fisikastudycenter.com
Literatures
-Theory and Problems of Electric Circuit, Joseph A Edminister, McGraw-Hill
-Soal UN Fisika SMA 2008

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