﻿ Equivalent Impedance In RLC Circuits

Equivalent Impedance In A RLC Circuit

Strings (SiPjAjk) = S7P3A31     Base Sequence = 12735     String Sequence = 12735 - 3 - 31

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Expressions Of Pj Problems
Equivalent Impedance In A RLC Circuit
Math

Determine the equivalent impedance of the circuit shown in figure 4.36, given that:
vs(t) = 636cos(3,000t + π/12)
R1 = 3.3 kΩ; R2 = 22 kΩ
L = 1.90 H; C = 6.8 nF.

The string:
S7P3A31 (Force - Pull).
The math:

Pj Problem of interest is of type force. The resistance to flow or motion is a foce problem (force - pull).
When the signals from the active elements of a circuit are sinusoidal excitations (e.g. AC circuits), the concept of resistance is replaced by the concept of impedance. In essence, impedance may be viewed as a complex frequency in the sense that inductors and capacitors are frequency-resistors whose resistances (called reactance) are functions of the frequency of the sinusoidal excitation. Nonetheless, the fundamental DC theories (Ohm's and Kirchoff's laws) remain applicable.

R1 is in series with L:
So, Z1 = R1 + jωL
Where jωL is the reactance of the inductor.
So, Z1 = 103(3.3 + j5.7)
Phasor form for Z1 :
Z1 = 103(6.59)<1.046.

R2 is in series with C:
So, Z2 = R2 + 1/(jωC)
Where 1/(jωC) is the reactance of the capacitor.
So, Z2 = 103(22 - j49)
Phasor form for Z2:
Z2 = 103(53.71)< -1.1495
Z1 + Z2 = 25.3 - j43.3.
Phasor form for Z1 + Z2:
103(50.15)<-1.052.
Z1 is parallel to Z2:
So, Zeq = (Z1)(Z2)/(Z1 + Z2)
So, Zeq = 106(6.59<1.046)(53.71< -1.1495)/103(50.15<-1.052)
So, Zeq = 7.05<0.94 kΩ

The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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