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Impedance and Admittance. Objective of Lecture Derive the equations for equivalent impedance and equivalent admittance for a series combination of components.

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Presentation on theme: "Impedance and Admittance. Objective of Lecture Derive the equations for equivalent impedance and equivalent admittance for a series combination of components."— Presentation transcript:

1 Impedance and Admittance

2 Objective of Lecture Derive the equations for equivalent impedance and equivalent admittance for a series combination of components. Derive the equations for equivalent impedance and equivalent admittance for a parallel combination of components. Chapter 9.7 Fundamentals of Electric Circuits Chapters 12.6, 13.5, 15.3, 15.5, 16.2, 16.4, 17.1, and 17.4 Principles of Electric Circuits Chapters 9.6, 10.2, 10.4, 11.5, 12.2, 12.4, 13.1, and 13.5 Electric Circuit Fundamentals

3 Ohm’s Law in Phasor Notation V = I ZV = I/Y I = V/ZI = V Y

4 Series Connections Using Kirchhoff’s Voltage Law: V 1 + V 2 – V s = 0 Since Z 1, Z 2, and V s are in series, the current flowing through each component is the same. Using Ohm’s Law: V 1 = I Z 1 and V 2 = I Z 2 Substituting into the equation from KVL: I Z 1 + I Z 2 – V s = 0V I (Z 1 + Z 2 ) = V s

5 Equivalent Impedance: Series Connections We can replace the two impedances in series with one equivalent impedance, Z eq, which is equal to the sum of the impedances in series. Z eq = Z 1 + Z 2 V s = Z eq I

6 Parallel Connections Using Kirchoff’s Current Law, I 1 + I 2 – I S = 0 Since Z 1 and Z 1 are in parallel, the voltage across each component, V, is the same. Using Ohm’s Law: V = I 1 Z 1 V = I 2 Z 2 V/ Z 1 + V/ Z 2 = I S I S (1/Z 1 +1/Z 2 ) -1 = V

7 Equivalent Impedance: Parallel Connections We can replace the two impedances in series with one equivalent impedance, Z eq, where 1/Z eq is equal to the sum of the inverse of each of the impedances in parallel. 1/Z eq = 1/Z 1 + 1/Z 2 Simplifying (only for 2 impedances in parallel) Z eq = Z 1 Z 2 /(Z 1 + Z 2 )

8 Shorthand for Parallel Connections An abbreviated means to show that Z 1 is in parallel with Z 2 is to write Z 1 ǁ Z 2.

9 If you used Y instead of Z In series: The reciprocal of the equivalent admittance is equal to the sum of the reciprocal of each of the admittances in series In this example 1/Y eq = 1/Y 1 + 1/Y 2 Simplifying (only for 2 admittances in series) Y eq = Y 1 Y 2 /(Y 1 + Y 2 )

10 If you used Y instead of Z In parallel: The equivalent admittance is equal to the sum of all of the admittance in parallel In this example: Y eq = Y 1 + Y 2

11 Example 1

12 Impedance Z R = 10  Z L = j  L = j(100)(10mH) = 1j  Z eq = Z R + Z L = 10 +1j   (rectangular coordinates) In Phasor notation: Z eq = (Z R 2 + Z L 2 ) ½  tan -1 (Im/Re) Z eq = ( ) ½  tan -1 (1/10) =  5.7 o  Z eq = 10.1  5.7 o  Impedances are easier than admittances to use when combining components in series.

13 Solve for Current Express voltage into cosine and then convert a phasor. V1 = 12V cos (100t + 30 o – 90 o ) = 12V cos (100t – 60 o ) V1 = 12  -60 o V

14 Solve for Current I = V/Z eq = (12  -60 o V)/ (10.1  5.7 o  ) V = 12  -60 o V = 12V e -j60 (exponential form) Z eq = 10.1  5.7 o  = 10.1  e j5.7 (exponential form) I = V/Z eq = 12V e -j60 /(10.1 e j5.7 ) = 1.19A e -j65.7 I = 1.19A  o I = V m /Z m   V -  Z )

15 Leading/Lagging I = 1.19A e -j65.7 = 1.19  o A V = 12V e -j60 = 12  -60 o V The voltage has a more positive angle, voltage leads the current.

16 Example 2

17 Admittance Y R = 1/R = 1   Y L = -j/(  L) = -j/[(300)(1H)] = -j 3.33 m   Y C = j  C = j(300)(1mF) = 0.3j   Y eq = Y R + Y L + Y C = j   Admittances are easier than impedances to use when combining components in parallel.

18 Admittances In Phasor notation: Y eq = (Y Re 2 + Y Im 2 ) ½  tan -1 (Im/Re) Y eq = (1 2 + (.297) 2 ) ½  tan -1 (.297/1) Y eq = 1.04  16.5 o  -1 It is relatively easy to calculate the equivalent impedance of the components in parallel at this point as Z eq = Y eq -1. Z eq = Y eq -1 = 1/1.04  o  =  o 

19 Solve for Voltage Convert a phasor since it is already expressed as a cosine. I = 4A cos(300t - 10 o ) I = 4  -10 o A

20 Solve for Voltage V = I/Y eq V = I m /Y m   I -  Y ) V = (4  -10 o A)/ (1.04  16.5 o  -1 ) V = 3.84V  o V = IZ eq V = I m Z m   I +  Z ) V = (4  -10 o A)(0.959  o  -1 ) V = 3.84V  o

21 Leading/Lagging I = 4  -10 o A V = 3.84V  o Current has a more positive angle than voltage so current leads the voltage.

22 Equations Equivalent ImpedancesEquivalent Admittances In Series: Z eq = Z 1 + Z 2 + Z 3 ….+ Z n Y eq = [1/Y 1 +1/Y 2 +1/Y 3 ….+ 1/Y n ] -1 In Parallel: Z eq = [1/Z 1 +1/Z 2 +1/Z 3 ….+ 1/Z n ] -1 Y eq = Y 1 + Y 2 + Y 3 ….+ Y n

23 Summary The equations for equivalent impedance are similar in form to those used to calculate equivalent resistance and the equations for equivalent admittance are similar to the equations for equivalent conductance. The equations for the equivalent impedance for components in series and the equations for the equivalent admittance of components in parallel tend to be easier to use. The equivalent impedance is the inverse of the equivalent admittance.


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