Network Theorems - Alternating Current examples - J. R. Lucas by rku10038

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									Network Theorems - Alternating Current examples - J. R. Lucas
In the previous chapter, we have been dealing mainly with direct current resistive circuits in
order to the principles of the various theorems clear. As was mentioned, these theories are
equally valid for a.c.

Example 1                                                    j20 Ω
                                   I1    20 Ω              63.66 mΗ I2

                                                   I
                                        19.89 µF
         E1 = 100∠0o V                                                        E2 = 70∠-30o V
                                        −j160 Ω




For the circuit shown in the figure, if the frequency of the supply is 50 Hz, determine using
Ohm’s Law and Kirchoff’s Laws the current I in the 160 Ω capacitor.
Impedance of capacitor and inductor at 50 Hz are
         j XL = j 2π × 50 × 63.66 × 10−3 = j 20 Ω
and      - j XC = 1/( j 2π × 50 × 19.89 × 10−6) = − j 160 Ω
Using Kirchoff’s current law
         I = I1 − I2
Using Kirchoff’s voltage law
         100∠0o = 20 I1 – j160 (I1 – I2)               ⇒       5 = (1-j8) I1 +j8 I2             ...... (1)
and      70∠-30 = – j 20 I2 – j 160 (I1 – I2) ⇒
                  o                                                   o
                                                               7∠-30 = – j 16 I1+ j 14 I2       ...... (2)
multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives
35 - 28∠-30o = (7 – j 56 + j 64) I1 + 0                        ⇒          10.751 + j 14 = (7 + j 8)I1
                            17.65∠52.48 o
i.e.     I1       =                       = 1.660∠3.67o A
                            10.63∠48.81o


substituting in (1),
         j8 I2 = 5 – (1-j8) × 1.660∠3.67o
i.e.     8∠90o I2 = 5 – 8.062∠-82.87o × 1.660∠3.67o = 5 – 13.383∠-79.20o = 2.492 + j13.146
              13.38∠79.27 o
i.e.     I2 =               = 1.673∠-10.73o A
                  8∠90 o


Thus the required current I is = 1.660∠3.67o – 1.673∠-10.73o
                                    = 1.657 + j 0.106 – 1.644 + j 0.311 = 0.013 +j 0.417
                                    = 0.42∠88.2o A
The problem could probably have been worked out with lesser steps, but I have done it in this
manner so that you can get more familiarised with the solution of problems using complex
numbers.


Network Theorems- a c examples – Professor J R Lucas   1                                   November 2001
    Example 2
    Let us solve the same problem as earlier, but using Superposition theorem.
                                                      j20 Ω
                            I1     20 Ω             63.66 mΗ I2

                                            I
                                 19.89 µF
   E1 = 100∠0o V                                                         E2 = 70∠-30o V
                                 −j160 Ω




    This circuit can be broken into its two constituent components as shown.
                                          j20 Ω                                        j20 Ω
                 I1A 20 Ω               63.66 mΗ I2A             I1B 20 Ω            63.66 mΗ I2B

E1 = 100∠0o V                                                                      IB
                                 IA                                          IB
                      19.89 µF                             +          19.89 µF
                                                                                                        E2 = 70∠-30o V
                      −j160 Ω                                         −j160 Ω



    Using series parallel addition of impedances, we can obtain the supply currents as follows.
    Equivalent        Zs1 = 20 + (-j160)//j20,                 Zs2 = j20 + 20//(-j160)
                                      − j160 × j 20                         20 × (− j160)
                           = 20 +                   ,             = j20 +
                                         − j140                              20 − j160
                           = 20 + j 22.857,                       = j20 + 19.692 – j 2.462 = 19.692 + j17.538
                           = 30.372∠48.81o Ω,                     = 26.370∠41.69o Ω
                                 100∠0 o                                70∠ − 30 o
    source current I1A =                    ,              −I2B =
                              30.372∠48.81o                           26.370∠41.69 o
                           = 3.293∠-48.81o,                       = 2.655∠-71.69o,
    Using the current division rule (note directions of currents and signs),
                                                  j 20                                         20
                 IA = 3.293∠− 48.81o ×                 ,       IB = 2.655∠− 71.69 o ×
                                                − j140                                      20 − j160
                                                                       53.10∠− 71.69 o
                    = 0.470∠131.19o,                              =                      = 0.329∠11.18o
                                                                      161.25∠ − 82.87  o


    Using superposition theorem, the total current in
             I    = 0.470∠131.19o + 0.329∠11.18o = -0.310 + j 0.354 + 0.323 + j0.064
                  = 0.013 + j 0.419 = 0.42∠88.2o A
    which is the same answer obtained in the earlier example.




    Network Theorems- a c examples – Professor J R Lucas   2                                      November 2001
 Example 3
 Let us again consider the same example to illustrate Thevenin’s Theorem.
                                                          j20 Ω
                         I1      20 Ω                   63.66 mΗ I2
                                                P

                                            I
                               19.89 µF
E1 = 100∠0o V                                                             E2 = 70∠-30o V
                               −j160 Ω

                                                Q


 Consider the capacitor disconnected at P and Q.
                                                                       100∠0 o − 70∠ − 30 o
 Current flowing in the circuit under this condition =
                                                                           20 + j 20
             100 − 60.62 + j 35 52.69∠41.63o
           =                   =             = 1.863∠ − 3.37 o
                 20 + j 20       28.28∠45 o


 ∴Thevenin’s voltage source = 100∠0o – 20 × 1.863∠-3.37o = 62.80 + j 2.19 = 62.84∠2.00o
                                                                      20 × j 20
 Also, Thevenin’s impedance across Q = 20//j20 =                                = 14.142∠45o = 10 + j 10
                                                                      20 + j 20
 ∴Thevenin’s equivalent circuit is
                              10 + j 10 Ω                          From this circuit, it follows that
                                                    P
  Eth = 62.84∠2.00o V                                                     62.84∠2.00 o     62.84∠2.00 o
                                                I                  I =                  =
                                                        19.89 µF         10 + j10 − j160 150.33∠ − 86.19 o
                                                        −j160 Ω
                                                                       = 0.418∠-88.2o A

                                                    Q              which is again the same result.


 Example 4
 Let us again consider the same example to illustrate Norton’s Theorem.
                                                          j20 Ω
                         I1      20 Ω                   63.66 mΗ I2
                                                P

                                        I
                               19.89 µF
E1 = 100∠0o V                                                             E2 = 70∠-30o V
                               −j160 Ω

                                                Q
 Consider the capacitor short-circuited at P and Q.
                                                         100∠0 o 70∠ − 30 o
 The Norton’s current source is given as                        +           = 5 –1.75 –j3.031
                                                           20       j 20
          = 3.25 – j 3.031 = 4.444∠-43.00o

 Network Theorems- a c examples – Professor J R Lucas        3                                   November 2001
                               1    1
Norton’s admittance =            +     = 0.05 − j 0.05 S
                               20 j 20                                                          I
                                                                    4.444∠-43.0o                        −j160
or same as Thevenin’s impedance 10 + j10 Ω
The Norton’s equivalent circuit is as shown in the figure.
The current through the capacitor can be determined using the current division rule.
                                       10 + j10                        14.142∠45o
         I = 4.444∠−43.0o ×                         = 4.444∠−43.0o ×
                                    10 + j10 − j160                  150.33∠ − 86.19 o
              = 0.418∠-88.2o A


Example 5
Using Millmann’s theorem find the current in the capacitor.
                                                                 j20 Ω
                                     I1    20 Ω                63.66 mΗ I2
                                                           S

                                                       I
                                          19.89 µF
             E1 = 100∠0o V                                                        E2 = 70∠-30o V
                                          −j160 Ω

                                                           N


                          1                 1                        1
                             ⋅ 100∠0 o +        ⋅0 +                    ⋅ 70∠ − 30 o
         VSN    =
                  ∑ Y.V = 20             − j160                    j 20
                   ∑Y                  1
                                         +
                                              1
                                                   +
                                                                     1
                                      20 − j160                    j 20

                      5 + 0 − 1.75 − j 3.031   3.25 − j 3.031   4.444∠− 43.00 o
                =                            =                =
                    0.05 + j 0.00625 − j 0.05 0.05 − j 0.04375 0.06643∠ − 41.19
                = 66.89∠-1.81o V
∴        I      = 66.89∠-1.81o /(-j160) = 0.418∠88.19o A


Example 6
Determine the delta equivalent of the star connected network shown.

                                                                                YAB
    A                      S                    B              A                                    B
                                  j20 Ω
               20 Ω

                 −j160 Ω                               ≡              YCA                 YBC


                           C
                                                                                  C



Network Theorems- a c examples – Professor J R Lucas       4                                 November 2001
                       1    1
                         ×
                       20 j 20             1             1
         YAB     =                =                =            ,             ∴ ZAB = 17.5 + j 20 Ω
                   1     1      1   j 20 − 2.5 + 20 17.5 + j 20
                     +        +
                   20 − j160 j 20
                        1     1
                            ×
                     − j160 j 20             1              1
         YBC     =                 =                  =           , ∴ ZBC = 160 – j 140 Ω
                   1      1      1   160 + j 20 − j160 160 − j140
                     +        +
                   20 − j160 j 20
                      1     1
                        ×
                     20 − j160              1              1
         YCA     =                =                  =             , ∴ ZCA = –140 – j 160 Ω
                   1      1     1   − j160 + 20 − 160 − 140 − j160
                     +        +
                   20 − j160 j 20
Example 7
Determine the star equivalent of the delta connected network shown.

                     17.5+j20 Ω
  A                                         B              A              S                   B

                                                               ZA                  ZB

− 140 − j160Ω                     160−j140 Ω           ≡             ZC


                                                                          C
                         C

                          (17.5 + j 20)(−140 − j160)         26.575∠48.81o × 212.603∠ − 131.19 o
         ZA =                                              =
                     17.5 + j 20 − 140 − j160 + 160 − j140              37.5 − j 280

                     5650∠ − 82.38 o
                 =                    = 20.00∠ − 0.01o = 20 Ω (same as original value in Ex 6).
                     282.5∠ − 82.37 o


                           (17.5 + j 20)(160 − j140)         26.575∠48.81o × 212.603∠ − 41.19 o
         ZB =                                              =
                     17.5 + j 20 − 140 − j160 + 160 − j140             37.5 − j 280

                       5650∠7.62 o
                 =                    = 20.00∠89.99 o = j 20 Ω (same as original value in Ex 6).
                     282.5∠ − 82.37 o


                      (160 − j140)(−140 − j160)          212.603∠− 41.19 o × 212.603∠− 131.19 o
         ZC =                                          =
                 17.5 + j 20 − 140 − j160 + 160 − j140               37.5 − j 280

                 45200∠− 172.38 o
             =                    = 160.00∠− 90.01o = − j160 Ω (same as original value in Ex 6).
                  282.5∠− 82.37 o


In order to show that the working is correct, I have selected the reverse problem for this
example and used the results of the previous example to find the original quantities. You can
see that the answers differ only due to the cumulative calculation errors.




Network Theorems- a c examples – Professor J R Lucas   5                                November 2001
Example 8
Determine using compensation theorem, the current I, if the available capacitor is 20 µF,
instead of the 19.89 µF already assumed in the earlier problems.
                                                                  j20 Ω
                                         I1    20 Ω             63.66 mΗ I2

                                                         I
                                              19.89 µF
             E1 = 100∠0o V                                                        E2 = 70∠-30o V
                                              −j160 Ω



Solution
                                                 1
         20 µF corresponds to                  −6
                                                              = -j159.15 Ω
                                       j 20 × 10 × 2 × π × 50
         change of impedance ∆Z = –j159.15 – (–j160) = j0.85 Ω.
from earlier calculations
         I = 0.418∠-88.2o A
∴ using compensation theorem, I . ∆Z = 0.418∠-88.2o× j0.85 = 0.355∠1.8o V
∴ changes in current in the network can be obtained from
                                                                                              j20 Ω
Note that the direction of ∆I is marked in the same                     ∆I1    20 Ω         63.66 mΗ ∆I2
direction as the original I, so that the source would in
fact send a current in the opposite direction.
                                                                          0.355∠1.8o
                    0.355∠1.8 o
i.e.     – ∆I =
                − j159.15 + 20 // j 20                                                ∆I
                                                                              20 µF
           0.355∠1.8 o           0.355∠1.8 o
   =                         =                                             −j159.15 Ω
                   20 × j 20 − j159.15 + 10 + j10
       − j159.15 +
                   20 + j 20

         0.355∠1.8 o
   =                    = 0.00237∠88.0o, giving ∆I as –0.00237∠88.0o, or 0.00237∠268.0o A
       149.48∠− 86.16 o



i.e. correct current I = 0.418∠-88.2o + 0.00237∠268.0o = 0.013 – j 0.4177 –0.00008–j0.00237
                         = 0.013 – j 0.420 = 0.420∠-88.2o A
Comparing result using Thevenin’s equivalent circuit derived in example 3
                10 + j 10 Ω                            From this circuit, it follows that
                                   P
62.84∠2.00o V                                                   62.84∠2.00 o      62.84∠2.00 o
                               I                       I =                     =
                                       20 µF                 10 + j10 − j159.15 149.48∠ − 86.16 o
                                       −j159.15 Ω
                                                         = 0.420∠-88.2o A which is the same result.

                                   Q




Network Theorems- a c examples – Professor J R Lucas     6                                  November 2001

								
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