IMPEDANCE

In the last topic about the phasor relationship of circuit elements and this is the continuation. We already know about the formulas of the three passive elements- resistor, inductor and capacitor.  (Table 1)

In the Impedance column, the Z stands for the frequency-dependent quantity (or the Impedance) and this Z is measured in Ohms (Ω).

Note: Solving for the equivalent impedance is just equal to rules in the resistance.

For example

Find  i(t) and v(t) in the circuit below:

Solution:

from the voltage source 10cos4t, we should convert it to phasor form:

Vs = 10<0º

ω = 4

The first thing to do is to convert the capacitance into a ohms value using the formula from the table 1.

Z = R + 1/jωC

Z= 5 + 1/j4 x 0.1    Ω                    note:            (j= √(-1) )

Z = 5 -2.5   Ω

to solve for current I, use ohms law

I = Vs/Z

I = 10<0º/(5-j2.5)

multiply it by the conjugate of 5-j2.5 both numerator and denominator .

I = 10(5 +j2.5) / 25 + 6.25)

I = 1.6 + j0.8

I = 1.789 < 26/57º A.

convert the I into time domain

i(t) = 1.789cos(4t + 26.57º ) A.  

The voltage across the capacitor is:

V= IZc

I/jωC  =  1.789 < 26/57º A/(j4 x 0.1)

Convert the denominator into a phasor form that is equal to 0.4<90º.

V= 4.47<-63.43º V

convert V into time domain :

v(t) = 4.47cos(4t-63.43º)  V