Phasor Relationship For Circuit Elements

Impedance and Admittance

Impedance Z – of a circuit is the ratio of the phasor voltage to the phasor current, measured in Ohms(Ω).

               Z=V/I                    or           V=ZI

Admittance Y – is the reciprocal of impedance, measured in siemens(s).

               Y=1/Z = I/V

                              Impedances and admittances of passive elements.

Elements	Impedance	Admittance
R (Ω)	Z=R	Y=1/R
L (H)	Z= jωL	Y=1/jωL
C (F)	Z=1/jωC	Y=jωC

That three elements(R,L,C) only the Resistances are real and the rest were imaginary and the reactances. The Capacitor is the positive reactance while the Inductor is the negative reactance.

Example:

We can transform the imaginary elements into an impedance by the use of shown equations.

1H

Z= jωL = j1x10 = j10

1F

Z=1/jωC = 1/j10x1 = -j0.1

In order to proceed and get the unknown values, we must first identify if what concept should be done in this problem. Since, circuit analysis was all about analyzing the different character of a circuit in any ways, we will use the basic law which we’ve been through in a few months “The Kirchhoff’s Law” was already discuss here in the past topics. We will just recall some important matters.

Kirchhoff’s Law has two parts, the Kirchhoff’s voltage Law and Kirchhoff’s current Law.

Kirchhoff's voltage law, states that the algebraic sum of all the voltages around a closed circuit equals zero. 

Kirchhoff's current law, states that the algebraic sum of all the currents entering and leaving a node is equal to zero. 

Looking at the circuit, the KVL must be applied since the sum of all the voltage drops in a closed circuit will equal the voltage source if only we’ll combine all of the impedances in order to have a single loop. And here we can apply the time domain converted into phasor domain.

Z= 1 + (1/j10 + 1/-j0.1 + 1/1)^-1 = 1.01010 – j0.1 = 1.015∠ -5.653

We can solve now the unknown which is the current using Ohm’s Law;

I=V/R = 2∠0/ 1.015∠ -5.653

I = 1.9704∠5.653 = 1.9704cos(10t+5.65) A