Network Theorems - Alternating Current examples.pdf

Network Theorems- a c examples – Professor J R Lucas 1 November 2001

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

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∠ -30o = – j 20 I2 – j 160 (I1 – I2) ⇒ 7∠ -30o = – 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

i.e. I1 = o

o

81.4863.10

48.5265.17

∠∠

= 1.660∠ 3.67o A

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

i.e. I2 = o

o

908

27.7938.13

∠∠

= 1.673∠ -10.73o A

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.

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1

I

I2


Example 2

Let us solve the same problem as earlier, but using Superposition theorem.

This circuit can be broken into its two constituent components as shown.

Using series parallel addition of impedances, we can obtain the supply currents as follows.

Equivalent Zs1 = 20 + (-j160)//j20, Zs2 = j20 + 20//(-j160)

= 20 + 140

20160

j

jj

−×−

, = j20 +16020

)160(20

j

j

−−×

= 20 + j 22.857, = j20 + 19.692 – j 2.462 = 19.692 + j17.538

= 30.372∠ 48.81o Ω, = 26.370∠ 41.69o Ω

source current I1A = o

o

81.48372.30

0100

∠∠

, −I2B = o

o

69.41370.26

3070

∠−∠

= 3.293∠ -48.81o, = 2.655∠ -71.69o,

Using the current division rule (note directions of currents and signs),

IA = 140

2081.48293.3

j

jo

−×−∠ , IB =

16020

2069.71655.2

jo

−×−∠

= 0.470∠ 131.19o, = o

o

87.8225.161

69.7110.53

−∠−∠

= 0.329∠ 11.18o

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.

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1

I

I2

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

I1A

IA

I2A 20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1B

IB

I2B

IB

+


Example 3

Let us again consider the same example to illustrate Thevenin’s Theorem.

Consider the capacitor disconnected at P and Q.

Current flowing in the circuit under this condition = 2020

30700100

j

oo

+−∠−∠

o

o

o

j

j37.3863.1

4528.28

63.4169.52

2020

3562.60100 −∠=∠

∠=+

+−=

∴ Thevenin’s voltage source = 100∠ 0o – 20 × 1.863∠ -3.37o = 62.80 + j 2.19 = 62.84∠ 2.00o

Also, Thevenin’s impedance across Q = 20//j20 = 2020

2020

j

j

+×

= 14.142∠ 45o = 10 + j 10

∴ Thevenin’s equivalent circuit is

From this circuit, it follows that

I = 1601010

00.284.62

jj

o

−+∠

= o

o

19.8633.150

00.284.62

−∠∠

= 0.418∠ -88.2o A

which is again the same result.

Example 4

Let us again consider the same example to illustrate Norton’s Theorem.

Consider the capacitor short-circuited at P and Q.

The Norton’s current source is given as 20

3070

20

0100

j

oo −∠+∠= 5 –1.75 –j3.031

= 3.25 – j 3.031 = 4.444∠ -43.00o

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1 I2 P

I

Q

P

I

Q

10 + j 10 Ω

19.89 µF −j160 Ω

Eth = 62.84∠ 2.00o V

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1 I2 P

I

Q


Norton’s admittance = 05.005.020

1

20

1j

j−=+ S

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.

I = 4.444∠− 43.0o × 1601010

1010

jj

j

−++

= 4.444∠− 43.0o ×o

o

19.8633.150

45142.14

−∠∠

= 0.418∠ -88.2o A

Example 5

Using Millmann’s theorem find the current in the capacitor.

VSN = ∑

∑Y

VY . =

20

1

160

1

20

1

307020

10

160

10100

20

1

jj

jjoo

+−

+

−∠⋅+⋅−

+∠⋅

= 19.4106643.0

00.43444.4

04375.005.0

031.325.3

05.000625.005.0

031.375.105

−∠−∠=

−−=

−+−−+ o

j

j

jj

j

= 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.

I

−j160 4.444∠ -43.0o

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1

I

I2 S

N

20 Ω j20 Ω

−j160 Ω

S

C

A B

C

A B YAB

YBC YCA ≡


YAB =

20

1

160

1

20

120

1

20

1

jj

j

+−

+

×=

205.17

1

205.220

1

jj +=

+−, ∴ ZAB = 17.5 + j 20 Ω

YBC =

20

1

160

1

20

120

1

160

1

jj

jj

+−

+

×−

= 140160

1

16020160

1

jjj −=

−+, ∴ ZBC = 160 – j 140 Ω

YCA =

20

1

160

1

20

1160

1

20

1

jj

j

+−

+

−×

=160140

1

16020160

1

jj −−=

−+−, ∴ ZCA = –140 – j 160 Ω

Example 7

Determine the star equivalent of the delta connected network shown.

ZA = 140160160140205.17

)160140)(205.17(

jjj

jj

−+−−+−−+

= 2805.37

19.131603.21281.48575.26

j

oo

−−∠×∠

= 2001.000.2037.825.282

38.825650 =−∠=−∠−∠ o

o

o

Ω (same as original value in Ex 6).

ZB = 140160160140205.17

)140160)(205.17(

jjj

jj

−+−−+−+

= 2805.37

19.41603.21281.48575.26

j

oo

−−∠×∠

= 2099.8900.2037.825.282

62.75650jo

o

o

=∠=−∠

∠ Ω (same as original value in Ex 6).

ZC =140160160140205.17

)160140)(140160(

jjj

jj

−+−−+−−−

=2805.37

19.131603.21219.41603.212

j

oo

−−∠×−∠

= 16001.9000.16037.825.282

38.17245200jo

o

o

−=−∠=−∠−∠ Ω (same as original value in Ex 6).

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.

C

A B 17.5+j20 Ω

160−j140 Ω − 140 − j160Ω ≡

ZA ZB

ZC

S

C

A B


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.

Solution

20 µF corresponds to 5021020

16 ×××× − πj

= -j159.15 Ω

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

Note that the direction of ∆I is marked in the same direction as the original I, so that the source would in fact send a current in the opposite direction.

i.e. – ∆I = 20//2015.159

8.1355.0

jj

o

+−∠

101015.159

8.1355.0

2020

202015.159

8.1355.0

jjj

jj

oo

++−∠=

+×+−

∠=

= o

o

16.8648.149

8.1355.0

−∠∠

= 0.00237∠ 88.0o, giving ∆I as –0.00237∠ 88.0o, or 0.00237∠ 268.0o A

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

From this circuit, it follows that

I = 15.1591010

00.284.62

jj

o

−+∠

= o

o

16.8648.149

00.284.62

−∠∠

= 0.420∠ -88.2o A which is the same result.

E1 = 100∠ 0o V

20 Ω j20 Ω

63.66 mΗ

19.89 µF −j160 Ω

E2 = 70∠ -30o V

I1

I

I2

20 Ω j20 Ω

63.66 mΗ

20 µF −j159.15 Ω

∆I1

∆I

∆I2

0.355∠ 1.8o

P

I

Q

10 + j 10 Ω

20 µF −j159.15 Ω

62.84∠ 2.00o V

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Network Theorems - Alternating Current examples.pdf