«Electrical Engineering.

УДК 621.3 Е50 Укладачі: Куземко Н.А., канд. техн. наук, доцент. Рецензенти: Лупенко А.М., докт. техн. наук, професор. Методичний посібник розглянуто й затверджено на засіданні кафедри світлотехніки та електротехніки Тернопільського національного технічного університету імені Івана Пулюя. Протокол № 4 від 22 листопада 2017 р. Схвалено та рекомендовано до друку на засіданні методичної комісії факультету прикладних інформаційних систем та електроінженерії Тернопільського національного технічного університету імені Івана Пулюя. Протокол № 4 від 24 листопада 2017 р. Е50

1. Electrical circuits calculations methods

Task 1. Draw the scheme (fig.1.1) according to your variant in table below. Write down the system of equations according to Kirchhoff’s laws.

Task 2. Write down the system of equations according to the loop currents method and calculate the circuit using this method.

Task 3. Verify the calculations using the equation of power balance.

Task 4. Write down the system of equations according to the nodal potential method and write down the expressions of branches’ currents.

Task 5. Define the external loop point potentials and draw the potential diagram for the external loop.

Var. E1 E2 E3 E4 E5 E6 R1 R2 R3 R4 R5 R6 № V V V V V V Ω Ω Ω Ω Ω Ω 00 12 15 18 – – – 3 4 5 6 7 8 01 – 15 18 21 – – 4 5 6 7 8 3 02 – – 18 21 24 – 5 6 7 8 3 4 03 – – – 21 24 27 6 7 8 3 4 5 04 15 18 21 – – – 7 8 3 4 5 6 05 – 18 21 24 – – 8 3 4 5 6 7 06 – – 21 24 27 – 3 4 5 6 7 8 07 – – – 24 27 12 4 5 6 7 8 3 08 18 21 24 – – – 5 6 7 8 3 4 09 – 21 24 27 – – 6 7 8 3 4 5 10 – – 24 27 12 – 7 8 3 4 5 6 11 – - - 27 12 15 8 3 4 5 6 7 12 21 24 27 – – – 3 4 5 6 7 8 13 – 24 27 12 – – 4 5 6 7 8 3 14 – – 27 12 15 – 5 6 7 8 3 4 15 – – – 12 15 18 6 7 8 3 4 5 16 24 27 12 – – – 7 8 3 4 5 6 17 – 27 12 15 – – 8 3 4 5 6 7 18 – – 12 15 18 - 3 4 5 6 7 8 19 – – – 15 18 21 4 5 6 7 8 3 20 27 12 15 – – – 5 6 7 8 3 4

Example for task 1. The graph for the circuit (fig.1.2) is shown at fig.1.3. There are six branches (unknown currents), p=6 and number of independent nodes is three at the circuit q=3. The currents’ (I1-I6) directions are chosen arbitrarily. The nodes are noted as

The equations according to the Kirchhoff’s first law for the nodes 1-3 (if the current flows into the node it is assumed with “+“, and out of the node with “-“) are:

for N1:

   

I

₁

I

₃

I

₄

0

, for N2:

   

I

₁

I

₂

I

₅

0

, for N3:

   

I

₃

I

₂

I

₆

0

.

The equations according to the Kirchhoff’s second law (p q3) for the loops,

1

L

,

L

₂,

L

₃ (the directions along the loops are chosen clockwise , if the directions of the bypass and the voltage or e.m.f. are the same, they are denominated with “+“, if opposite with “-“) are:

for L1:



R I

_{1 1}



R I

_{5 5}



R I

_{4 4}

  

E

₁

E

₅



E

₄,

for L2:



R I

_{2 2}



R I

_{5 5}



R I

_{6 6}

  

E

₂

E

₅



E

₆,

for L3:



R I

6 6



R I

3 3



R I

4 4

  

E

6

E

3



E

4.

So, the equations system according to the Kirchhoff’s laws is:

1 3 4

0

I

I

I

   

1 2 5

0

I

I

I

   

3 2 6

0

I

I

I

   

1 1 5 5 4 4 1 5 4

R I

R I

R I

E

E

E







  



2 2 5 5 6 6 2 5 6

R I

R I

R I

E

E

E







  



6 6 3 3 4 4 6 3 4

R I

R I

R I

E

E

E







  



Example for task 2 . For the circuit on fig.1.2

E

1



10

V

,

E

3



15

V

,

E

6



20

V

,

V

E

E

E

₂



₄



₅



0

,

R

₁

 10



,

R

₂

 6



,

R

₃

 5



,

R

₄

 7



,

R

₅

 8



,

R

₆

 5



. The system of equations according to the loop currents method for the circuit with

                 3 3 33 2 32 1 31 2 3 23 2 22 1 21 1 3 13 2 12 1 11 L L L L L L L L L L L L E I R I R I R E I R I R I R E I R I R I R , where

R

₁₁



R

₁



R

₅



R

₄ 108725,

R

₂₂



R

₂



R

₆



R

₅ 65819, 3 6 4 33

R

R

R

R







75517 – are the individual resistances of the loops, which are equal to the sum of all the resistances of the loop;









_{21 5}

8

12

R

R

R

,

R

₁₃



R

₃₁



R

₄



7



,

R

₂₃



R

₃₂



R

₆



5



– are mutual resistances of the loops, i.e. the resistances of the branches which are mutual for the respective loops;

4 5 1 1 E E E E_L    100010V, E_L2 E₂ E₅ E₆ 002020V, 4 3 6 3 E E E

E_L    2015035V – are loops’ e.m.f. These are equal to the algebraic sum of the electromotive forces alongside the loops.

When substituted in the equations system above, these values make:

                  35 17 5 7 20 5 19 8 10 7 8 25 3 2 1 3 2 1 3 2 1 L L L L L L L L L I I I I I I I I I .

When solved the values make:

I

_L1



1

.

153

A

,

I

_L2



0

.

108

A

,

I

_L3



2

.

565

A

.

Branches’ currents can be defined from the loops’ currents as:

I

₁



I

_L1



1

.

153

A

,

A

I

I

₂



_L2



0

.

108

,

I

₃





I

_L3





2

.

565

A

,

I

₄



I

_L3



I

_L1



2

.

565



1

.

153



1

.

412

A

,

A

I

I

I

₅



_L1



_L2



1

.

153



0

.

108



1

.

045

,

I

₆



I

_L3



I

_L2





2

.

565



0

.

108



2

.

457

A

.

Example for task 3. The power balance equation is used to verify the calculations. For the circuit on fig.1.2, according to the calculations at example 2 the total power of the circuit’s sources should be equal to the total power of the circuit’s consumers



P

_R





P

_E.

Thus, total power of the sources is:

6 6 3 3 1 1

I

E

I

E

I

E

P

_E











10



1

.

153



15



(



2

.

565

)



20



2

.

457



99

.

145

W

.

When the e.m.f. and the current have the same directions the source power

P

_E



EI

is considered with “+“, if opposite with “-“.

The total power of the consumers makes:

        2 6 6 2 5 5 2 4 4 2 3 3 2 2 2 2 1 1I R I R I R I R I R I R P_R

W

13

.

99

)

457

.

2

(

5

)

045

.

1

(

8

)

412

.

1

(

7

)

565

.

2

(

5

)

108

.

0

(

6

)

153

.

1

(

10



2





2







2





2





2





2





So,



P

_E





P

_R.

                3 3 33 2 32 1 31 2 3 23 2 22 1 21 1 3 13 2 12 1 11 J G G G J G G G J G G G          .

Where

G

₁₁



G

₁



G

₃



G

₄,

G

₂₂



G

₁



G

₂



G

₅,

G

₃₃



G

₂



G

₃



G

₆ – are the individual conductivities of the nodes. They are equal to the sum of the branch conductivities that are coming into the node;

1 21

12

G

G

G





,

G

₂₃



G

₃₂



G

₂,

G

₁₃



G

₃₁



G

₃ – are the mutual conductivities of the nodes. They are equal to the conductivities of the branches that connect the respective nodes. Branches’ conductivities are:

G 

₁

1 R

/

₁,

G 

₂

1 R

/

₂,

G 

₃

1 R

/

₃,

G 

₄

1 R

/

₄,

5

5

1 R

/

G 

,

G 

₆

1 R

/

₆.

Nodal potentials are



₁,



₂,



₃ related to the node 0 with zero potential.

3 3 4 4 1 1 1

G

E

G

E

G

E

J









,

J

₂



G

₁

E

₁



G

₅

E

₅



G

₂

E

₂, 3 3 2 2 6 6 3

G

E

G

E

G

E

J







– the algebraic sum of the currents of currents’ sources, that are flowing into the respective nodes. If the current J of the source flows into the node, it is marked with the sign “+“, when it flows out – with the sign “-“.

Branches’ currents are defined this way:



₁



R

₄

I

₄



E

₄, ( ) ( _{1 4}) ₄,

4 4 1 4 E G R E I       5 5 5 2



R

I





E



, ( ) ( _{2 5}) ₅, 5 5 2 5 E G R E I       6 6 6 3



R

I



E



, ( ) ( _{6 3}) ₆, 6 3 6 6 E G R E I     1 1 1 2 1







R

I





E



, _{1 2 1 1} 1 1 2 1 1 ( ) ) ( G E R E I         , 2 2 2 3 2







R

I



E



, 2 2 3 2 2 2 3 2 2 ( ) ) ( G E R E I         , 3 3 3 3 1







R

I



E



, 3 3 3 1 3 3 3 1 3 ( ) ) ( G E R E I         .

Example for task 5. The external loop of the circuit (fig. 1.2) is shown at fig. 1.4. The parameters of the circuit are:

E

1



10

V

,

E

3



15

V

,

E

6



20

V

,

R

1

 10



,

R

2

 6





 5

3

R

,

R

₄

 7



_,

R

₅

 8



_,

R

₆

 5



_{and the branch currents are:}

I

₁



1

.

153

A

,

A

I

₂



0

.

108

,

I

3





2

.

565

A

(example for task 2).

The loop points potentials are:



₁



0

,

 

r

₁

0

,

V

I

R

b 2 2

1

.

53

6

0

.

108

2

.

178

3





















,



r

₃



R

₁



R

₂



10



6



16



,

V

E

c





3



3





2

.

178



15



12

.

822



,



r

₃



R

₁



R

₂



10



6



16



,

V

I

R

c 3 3

12

.

822

5

(

2

.

565

)

0

1



















,



r

1



R

1



R

2



R

3



10



6



5



21



.

The potential diagram for the external loop is shown at fig.1.5.

Task 6. Draw the scheme (fig. 1.6). Define the branches’ currents with the two nodes method and the operating mode of every source according to your variant in table below. Verify the calculations using the first Kirchhoff’s law.

09 200 300 150 20 30 40 10 10 200 400 120 30 50 80 10 11 100 120 130 50 30 10 80 12 100 125 140 30 50 80 10 13 100 130 140 50 40 30 20 14 100 135 145 20 30 40 50 15 120 130 155 50 20 30 10 16 120 145 130 40 30 50 20 17 130 160 145 30 50 20 40 18 130 155 140 50 10 40 30 19 140 150 160 30 20 10 50 20 200 300 100 40 50 20 10

Example for task 6. The parameters of the circuit are:

E 

₁

10

V

,

E 

₂

20

V

,

E 

₃

30

V

,

R

₁



10



,

,

5

2





R

R

₃



10



,

R

₄



5



.

Choose arbitrarily the directions of branches’ currents

I

₁

,

I

₂

,

I

₃

,

I

₄ and the direction of the inter-node voltage

V

ab (fig.1.7).

The branches’ conductivities are:

Sm

R

G

₁



1

/

₁



1

/

10



0

.

1

,

G

2



1

/

R

2



1

/

5



0

.

2

Sm

,

Sm

R

G

₃



1

/

₃



1

/

10



0

.

1

,

G

₄



1

/

R

₄



1

/

5



0

.

2

Sm

. Calculate the inter-node voltage as:

       4 3 2 1 3 3 2 2 1 1 G G G G E G E G E G V_ab 13.33 2 . 0 1 . 0 2 . 0 1 . 0 30 1 . 0 20 2 . 0 10 1 . 0 _         V.

For chosen positive currents’ directions, their values are defined according to the second Kirchhoff’s law.

1 1

1

I

E

R

V

_ab





, so

I

₁



(

E

₁



V

_ab

)

/

R

₁



(

E

₁



V

_ab

)

G

₁



(

10



13

.

33

)



0

.

1





0

.

33

A

. Sign «-» means, that actual direction of the current is opposite to the chosen one.

2 2

2

I

E

R

V

_ab





, so

I

₂



(

E

₂



V

_ab

)

G

₂



(

20



13

.

33

)



0

.

2



1

.

33

A

. Sign «+» means that actual direction of the current is the same as chosen one.

3 3 3

I

E

R

V

_ab





, so

I

₃



(

E

₃



V

_ab

)

G

₃



(

30



13

.

33

)



0

.

1



1

.

67

A

. 4 4

I

R

V

_ab



, so

I

₄



V

_ab

/

R

₄



V

_ab



G

₄



13

.

33



0

.

2



2

.

67

A

.

Verify the calculations by the first Kirchhoff’s law for node

a

:

I

1



I

2



I

3



I

4



0

so 0.331.331.672.67 0.

Define electrical sources operating modes. If the directions of the real branch current and e.m.f. are the same P  EI 0, the source works as a generator, if not

0   EI

P the source works as a consumer. Thus, the sources

E

₂

, E

₃ at fig. 1.7 work as generators and

E

₁ as a consumer.

Task 7. Draw the scheme (fig. 1.8) with values according to your variant in table below. Define the branches’ currents by the superposition method.

Example for task 7. For the circuit at fig.1.9



 5

1

R

,

R

₂

 15



,

R

₃

 10



,

E

₁



20

V

,

V

E

₂



25

.

The superposition principle means that every e.m.f. acts in the circuit independently. Thus, the calculation of one circuit (fig. 1.9) with two sources, for example, can be reduced to the calculation of two circuits each with a single source (fig. 1.10, 1.11). Two partial circuits with partial currents should be calculated according to this principle. There is a single e.m.f. E₁ in the first partial circuit (fig. 1.10):

        6 10 15 15 10 3 2 3 2 23 R R R R R .

The total resistance of this scheme:

R





R

₁



R



₂₃



5



6



11



.

The partial current makes

I

1





E

1

/

R





20

/

11



1

.

818

А

, and voltage makes

V

R

I

V

_ab





₁



₂₃





1

.

818



6



10

.

91

.

The partial branches’ currents are:

I



₂



V

_ab



/

R

₂



10

.

91

/

15



0

.

727

А

,

А

R

V

I

₃





_ab



/

₃



10

.

91

/

10



0

.

091

.

There is a single e.m.f.

E

₂ in the second partial circuit (fig. 1.11).         3.33 10 5 10 5 3 1 3 1 13 R R R R R .

The total resistance of this scheme is:

R





R

2



R

13





15



3

.

33



18

.

33



.

The partial current makes

I

₂





E

₂

/

R





25

/

18

.

33



1

.

364

А

, and voltage

V

R

I

V

_ab





₂



₁₃





1

.

364



3

.

33



4

.

54

.

The partial branches’ currents are:

I

₁





V

_ab



/

R

₁



4

.

54

/

5



0

.

909

А

,

А

R

V

I

₃





_ab



/

₃



4

.

54

/

10



0

.

454

.

Therefore, the real branches’ currents are defined as an algebraic sum of the respective partial currents (fig.1.9):

3 3

3

I

I

I











0

.

091



0

.

454



0

.

545

А

.

Task 8. Draw the scheme (fig. 1.12) according to your variant in table below. Define the second branch current using the equivalent generator method.

Var. E1 E2 E3 E4 E5 R1 R2 R3 R4 R5 № V V V V V Ω Ω Ω Ω Ω 00 12 - 15 – – 3 4 5 6 7 01 – - 18 - 27 4 5 6 7 8 02 – – 18 21 - 5 6 7 8 3 03 – – – 21 24 6 7 8 3 4 04 15 - 21 – – 7 8 3 4 5 05 – - 15 24 – 8 3 4 5 6 06 – – 21 - 27 3 4 5 6 7 07 15 – – 24 - 4 5 6 7 8 08 18 - - – 21 5 6 7 8 3 09 25 - - - 27 6 7 8 3 4 10 – – 24 - 12 7 8 3 4 5 11 – - - 27 12 8 3 4 5 6 12 21 - - 18 – 3 4 5 6 7 13 – - 27 12 – 4 5 6 7 8 14 – – 27 12 - 5 6 7 8 3 15 – – – 12 15 6 7 8 3 4 16 24 - 12 – – 7 8 3 4 5 17 – - - 15 21 8 3 4 5 6 18 – – 12 - 18 3 4 5 6 7 19 18 – – 15 - 4 5 6 7 8 20 27 - - – 15 5 6 7 8 3

Example for task 8. For the circuit at fig.1.13

R

₁

 5



,

R

₂

15



,

R

₃

10



,

The branch with unknown current (

I

₂) is selected from the circuit on fig. 1.13. An equivalent generator (fig. 1.14) replaces the rest of the circuit. Its parameters are Eeqv –

equivalent e.m.f., which is equal to the open circuit voltage on the clamps ab of an open second branch and Reqv– equivalent resistance, which is equal to the

input resistance of the circuit in respect to the open second branch clamps ab (the e.m.f. is shortened). The equivalent generator parameters (Eeqv and Reqv)

are to be calculated. For the circuit at fig.1.13 Reqv is:

3 1 3 1 R R R R R_eqv        1.333 10 5 10 5 . eqv

E (fig. 1.15) is defined the following way: V_abOC  E_eqv  E₁ R₁I

V 65 . 21 ) 33 . 0 ( 5 20     , where 1 2 1 3 E E I R R    _{5 10} 0.33A 25 20 _  

 . According to the fig.1.14

unknown current makes: I2  Eеqv/(Reqv R2) 1.33 A

15 33 , 1 65 . 21    .

Example 9. Define the non-linear element static and dynamic resistance at work point for I₀ 0.006A, V₀ 4V .

Static resistance (fig. 1.16) is

 V₀/ I₀

R_ST 4/0.006 666.67,

Dynamic resistance (fig. 1.16) is

  

 V I

R_D / 4/0.0041000.

Example 10. For scheme at fig.1.17 R₁  5 , R₂  15 , E₁ 20 V, E₂ 25 V . Define the non-linear element R₃ current, if its volt-ampere relationship V(I) is represented by the table.

I, А 1,0 2,0 3,0 4,0 5,0 6,0 V, В 13 17 18 18.5 19.5 20.0

The equivalent generator method is used to solve this task. Firstly, the branch with non-linear resistance is selected. The rest of the scheme is represented as an active one-port network with the parameters EEQV VOCab, REQV (fig. 1.18).

The branch with non-linear resistance R₃ is

disconnected and the input resistance R_EQV is defined as the one between the open clamps

a and b (the sources are shortened):

2 1 2 1 R R R R R_EQV        3.75 15 5 15 5 .

The open circuit voltage is defined by two nodes method (G_N 1/R_N) as

1 1 1 R/ G  1/50.2Sm, G ₂ 1 R/ ₂ 1/150.067Sm. 2 1 2 2 1 1 G G E G E G V E_{EQV OCab}     21.25V 067 . 0 2 . 0 25 067 . 0 20 2 . 0 _      .

For scheme at fig. 1.18 write down the equation according to the second Kirchhoff’s law E_EQV  R_EQVI V_ab, so V(I)V_ab E_EQV R_EQVI. Solve

this equation by graphic method (fig. 1.19). The left part of this equation is a non-linear element VAC (curve)

) (I

V_AB . The right part of the equation is a line drawn across the points (E_EQV,0), (0,I_SC). By substituting the values EEQV, ISC  EEQV /REQV , the points (21.25,0) and

(0 .,567) are obtained. The line is drawn across these points. It crosses the curve at the point with

, 20 . 1 , 15V I А

V_A  _A  that defines a non-linear element working regime.

Tasks for individual work.

1. Current I_R2 2 А and resistances

  4

1

R , R2  2, R3  6  are given for the

circuit on the figure. Define the branches’ currents, sub-circuits’ voltages, input voltage.

2. Voltage V_R2  2V and resistances

  4

1

R , R₂  2, R₃  6  are given for the circuit on the figure. Define the branches’ currents, voltages across the elements and current of the unforked sub-circuit.

3. Current I_R2  2 А and resistances R₁  2 ,

  3

2

R , R₃  6  are given for the circuit on the figure. Define the branches’ currents, sub-circuits’ voltages and input voltage.

4. Voltage V_R2  4V and resistances R₁  2,

  4

2

R , R₃  6  are given for the circuit on the figure. Define the branches’ currents, voltages across the elements and the current of the unforked sub-circuit.

5. Define source internal resistance and electromotive force , source’s and consumer’s powers if voltage is V=190V, load resistance is R=139 Ω and efficiency factor is ɳ=0.99. Calculate the efficiency factor at the load equal to R/10.

6. Calculalate the current in the circle if the resistances are

  165

1

R , R₂ 197, R₃  198, R₄  193 and the maximum power makes

W

P₃ 7.4 when allocated to the resistor R ₃ in the unforked circuit. Define the input voltage and sub-circuits’ voltages, branches’ currents, power

of the circuit and the powers of the sub-circuits.

7. The consumer with resistance R_C  10  is connected to the clamps of the source with e.m.f. E40 W and internal resistance R₀ 1. Define the voltage across the source clamps and source efficiency factor.

8. Define the voltage source parameters, if the parameters of current source are A

J 5 , G₀ 0.5Sm. Draw the voltage source scheme and write down its equation. 9. Define the current source parameters, if the parameters of voltage source are

V

E 50 , R₀  5. Draw the current source scheme and write down its equation.

11. In short circuit mode source current is I_SC 48A. When the source is connected to the resistive element R19.5, the circuit current is 1.2A. Define the source e.m.f. and its internal resistance.

12. The battery consists of three connected in series sources with the following parameters: E1.5V , R₀  50. . The battery power is P2.25W .

Define the load resistance, power and the voltage across the source clamps.

13. Define the resistance R₁ at the circuit with connected in series elements: R₂ 16Ом, R₃  24Ом,V₂ 8V, I 2 A and

V

V 60 . Define the circuit total resistance.

14. Define the conductivity of an element R₃ in the circuit with parallel connection of elements: G₁ 0.05Sm, G₂ 0.1Sm,

А

I₁ 2 and I 14 А. Define the circuit total conductivity.

15. Define the branches’ currents and verify your calculations, if R₁  R₃  R₅ 9, R₂  3, R₄  9 and V 21V .

16. The elements R₁ 16, R₂  1  and R₃  3  are connected to the source with the parameters E40 V and R₀  80, . Define the source voltage, branches’

currents and voltages across the elements.

16. Define the load R power, if E₁200V, Е₂ 150V , R₁  20, R₂  30. What should be the R value to reach the maximum power on it?

2. Alternating current circuits calculation.

Task 1. The circuit of two connected in series coils and capacitor (fig. 2.1) is powered by AC voltage source of frequency f 50 Hz. The parameters of the circuit are given in table below. Define the circuit current and the voltages across the coils. Define the resonance frequency of the circuit. Write down the current and voltages instantaneous values. Draw the vector diagram.

Example for task 1. The circuit (fig. 2.3) has the following parameters: V 20V, f 50Hz

  5

1

R , R₂  10 , L₁  L₂ 127mH , F

C  318  . This circuit consists of two coils with parameters

R

₁

, L

₁,

R

₂

, L

₂ and capacitor C.

Angular frequency makes:

s rad f 3 3.14 50 314 / 2        .

Reactances elements are:

X

L1





L

1



314



0

.

0127



4



,











₂

314

0

.

0127

4

2

L

X

_L



,











1

/(

С

)

1

/(

314

0

.

000318

)

10

X

_С



.

Circuit resistance is:

R



R

1



R

2



5



10



15



.

Circuit reactance is:

X



X

_L1



X

_L2



X

_C



4



4



10





2



. Circuit impedance is: Z  R2  X2  152  (2)2 15.13Ω. Coils impedances are: 21 52 (4)2 6.4

2 1 1C  R  XL    Z Ω, 77 . 10 ) 4 ( 102 2 2 2 2 2 2C  R XL    Z Ω.

Circuit current makes: I V /Z 20/15.131.32A_.

Phase shift angle between input voltage and circuit current makes:



6

)

15

/

2

(

)

/

(











arctg

X

R

arctg



.

Input voltage initial phase is



V



0

.

Circuit current I initial phase is defined from the expression







_V





_I:



6













_{I V} .

First coil voltage makes:

V

_1C



Z

_1C

I

6.41.32 8.45V. Phase shift angle between this voltage and current I makes:



39

)

5

/

4

(

)

/

(

_{1 1} 1



arctg

X

L

R



arctg





.

Voltage

V

_1C initial phase is defined from the expression



₁





_V1





_I:

  

33

6

39

1 1





I







V







.

Second coil voltage is:

V

_2C



Z

_2C

I

10.771.32 14.22V. Phase shift angle between this voltage and current I is:



22

)

10

/

4

(

)

/

(

_{2 2} 2



arctg

X

L

R



arctg





.

Voltage

V

_1C initial phase is defined from the expression



₂





_V2





_I:

  

16

6

22

2 2





I







V







.

Capacitor voltage is:

V

_C



X

_C

I

101.32 13.2V.

Phase shift angle between this voltage and current I is:



C





90

. Voltage

V

_C initial phase is found from the expression



_C





_VC





_I:

  

96

6

90















_{C I} VC







.

The resonance condition for this circuit is

X 

_L

X

_C, it means

)

/(

1

)

(



₀

L

₁





₀

L

₂





₀

C

, so resonance frequency makes

) ( / 1 _{1 2} 0  C L L  1/ 0.000318(0.0127 0.0127) 352rad /s.

Instantaneous values of voltages and current are:

)

sin(

)

(

t

V

_m

t

_V

v









,

i

(

t

)



I

_m

sin(



t





_I

)

: ) 314 sin( 2 20 ) (t t v  В, i(t)1.32 2sin(314t6) V, ) 33 314 sin( 2 45 . 8 ) ( 1    t t

v_COIL В, v_2COIL(t)14.22 2sin(314t 16)V,

) 96 314 sin( 2 2 . 13 ) (t  t   vС V.

Task 2. The circuit, which consists of connected in parallel coil and capacitor (fig. 2.2), is powered by AC voltage source with frequency f 50Hz. The parameters of the circuit are given in table above. Define the circuit and branches currents, the phase shift angles. Define the resonance frequency of the circuit. Find the power balance for the circuit.

Example for task 2. Parameters for the circuit at fig. 2.4 are: V  20V , R₁  5,

  10 2 R , L2 12.7mH , С 318F . Angular frequency is f 50Hz :

f







2



2



3

.

14



50



314

rad /

s

. Elements’ reactances are: XL L3140.0127 4,

    1/( С) 1/(314 0.000318) 10 X_С  .

Branches’ impedances are:

4 . 6 ) 4 ( 52 2 2 2 1 1  R  XL    Z Ω. 14 . 14 ) 10 ( 102 2 2 2 2 2  R  XC     Z Ω.

Branches’ currents are: I₁ V / Z₁  A 13 . 3 4 . 6 / 20  , I₂ V / Z₂  20/14.14 1.39A_.

Phase shift angles between the input voltage and the branches’ currents are:



39

)

5

/

4

(

)

/

(

1



arctg

X

L

R



arctg





,



2



arctg

(

X

C

/

R

)



arctg

(



10

/

10

)





45

.

Input voltage initial phase is



V



0

.

Currents

I

₁,

I

₂ initial phases are found from the expressions



₁





_V





_I1,

The conductances and susceptances method can be used to define the total current. Serial connection is to be converted into parallel one. So, branches’ conductances and susceptances are accordingly (fig.2.5):

2 1 1 1 Z R G  0.122 4 . 6 5 2   Sm, 2 1 1 Z X B  L 0.0976 4 . 6 4 2   Sm. 2 2 2 2 Z R G  0.05 14 . 14 10 2   Sm, 2 2 2 Z X B  C 0.05 14 . 14 10 2   Sm.

The circuit conductance and susceptance (fig. 2.6) make accordingly:

172

.

0

05

.

0

122

.

0

2 1











G

G

G

Sm, BB₂B₁0.050.09760.0476 Sm. 0 

B . Thus, susceptance has an inductive character.

Circuit admittance is: Y  G2 B2  0.1722 (0.0476)2 0.178_Sm.

Total current makes: I VY  200.178  3.56A.

Phase shift angle between input voltage and total current is:





arctg

(

B

/

G

)



 15 ) 172 . 0 / 0476 . 0 ( 

arctg , if _V  0, then total current I initial phase can be found from the expression  V I. Therefore,

 15      I V .

The resonance condition for this circuit is B L BC, or B 1 B2. It means that









2 0 2 2 0 2 0 2 1 0 / 1 ) /( 1 C R C L R L        .

From this expression  can be found. ₀

Power balance equations are used to verify the obtained results. Thus, consumers active and reactive powers are:







2 2 2 2 1 1

I

R

I

R

P

_cons 53.132 101.392 68.77W,







2 2 2 1

X

I

I

X

Q

_{cons L C} 43.132 101.392 18.43VAr.

Source active and reactive powers are:

W

VI

P

_sour



cos





20



3

.

56

cos

15

•



68

.

77

,

VAr

VI

Q

_sour



sin





20



3

.

56

sin

15

•



18

.

43

.

Thus, if

P

_cons



P

_sour,

Q

_cons



Q

_sour, power balance is true. Therefore, the circuit parameters have been defined correctly.

G1 B1 G2 B2

Fig. 2.5

V _{G B}

Task 3. Draw the scheme (fig. 2.7) according to your variant at table below. Find out the circuit and branches’ impedances for frequency f 50Hz . Define the branches’ currents and the voltages. Verify the calculations (balance of power equation). Draw the

vector diagram of currents and voltages.

Example for task 3. For the circuit on fig. 2.7: V 70V_,

R

₁

 5



_,

R

₂

 0



_,



 4

3

R

,

L

₁



9

mH

,

L

₂



7

mH

,

L

₃



20

mH

,

C

₁



C

₃



0



F

,

C

₂



281



F

. The circuit according to this variant is shown at fig. 2.8

Attention! Calculation is for frequency f 60Hz.

Angular frequency, thus, is 2f 23.1460377 rad /s.

Reactances of inductive elements are X_Ln L_n: 39 . 3 10 9 377 3 1 1       L X_L  Ω, X_L2 L₂ 3777103  2.64Ω, 54 . 7 10 20 377 3 3 3       L X_L  Ω.

Reactances of capacitive elements are X_Cn 1/C_n:

44 . 9 10 281 377 / 1 / 1 ₂ 6 2       C X_C  Ω.

Branches’ impedances make :

 34 1 1 1 5 3.39 6,04 j L j e jX R Z      Ω,  90 2 2 2 (2.64 9.44) 6.8 6.8 j C L jX j j e jX Z         Ω,  62 3 3 3 4 7.54 8.54 j L j e jX R Z      Ω.

Unforked part of the circuit impedance is:

 34 1 1 1 12 5 3.39 6,04 j L j e jX R Z Z       Ω.

Parallel connection impedance makes:

            90 62  28 ₁₀28 3 2 3 2 23 07 . 4 05 . 58 74 . 0 4 05 . 58 54 . 7 4 8 . 6 54 . 8 8 . 6 j j j j j e e j e j j e e Z Z Z Z Z ) 87 . 8 18 . 11 ( 27 . 14 38 j e j     Ω.

Circuit impedance is: 

19 23 12 5 3.39 11.18 8.87 16.18 5.47 17.08 j e j j j Z Z Z           Ω.

Circuit current is: I V /Z 70/17.08ej19 4.1ej19 (3.88 j1.31)A. Voltages across the parts of the circuit make:

V j e e e I Z V12  12 6.04 j34 4.10 j19 24.76 j53 (14.95 19.74)    . V j e e e I Z V j j j ) 74 . 19 05 . 55 ( 48 . 58 10 . 4 27 . 14 38 19 20 23 23            . According to the second Kirchhoff’s law:

V j j V V V  12  23 (14.95 19.74)(55.05 19.74)70 .

Branches’ currents are:

A j e e e Z V I j j j ) 09 . 8 9 . 2 ( 6 . 8 8 . 6 / 48 . 58 / 20 90 70 2 23 2           , A j e e e Z V I3  23/ 3 58.48 j20 /8.54 j62 6.85 j82 (0.98 6.78)      . According to the first Kirchhoff’s law:

A j j j I I I  2  3 (2.9 8.1)(0.98 6.78)3.88 1.31

Complex total power makes:

Consumers’ active and reactive powers make: 2 3 3 2 1 1I R I R P   54.12 46.852  272W .      2 3 3 2 2 2 2 2 1 1I (X X )I X I X Q _{L L C L} VAr 92 85 . 6 54 . 7 6 . 8 8 . 6 1 . 4 39 . 3  2   2   2    . Vector diagram.

Assumed voltages’ vectors scale is M_V 15 V /cm, the voltage vectors’ lengths are:

cm M V V V 7 . 4 1570    , cm M V V V 65 . 1 15 76 . 24 12 12    , cm M V V V 9 . 3 15 48 . 58 23 23    .

Voltage initial phases (the angles between axis X and the vector) are: _V  0, _V12 53, _V23  20.

Assumed currents’ vectors scale is M_I 1.5 A/cm, the currents’ vectors lengths are:

cm M I I I 7 . 2 5 . 1 1 . 4 1 1    , cm M I I I 7 . 5 5 . 1 6 . 8 2 2    , cm M I I I 6 . 4 5 . 1 85 . 6 3 3    .

Currents’ initial phases (the angles between axis X and the vector) are: 

19 1  I  ,  70 2  I  , _I3  82.

Example 4. The voltage v12.56sin(314t/3)V is applied to inductivity mH

L0.02 . Write down the current instantaneous value. Draw the vector diagram for current and voltage effective values.

Inductivity reactance is: X_L L 3140.026.28. Current amplitude is: I_m V_m/X_L 12.56/6.282A. Phase shift angle for the element makes: _V  _I/2. Current initial phase is: _I _V  /3 /2 /6.

Current instantaneous value is: i I_msin(t_I )=2sin(314t/6)A. The vector diagram is shown at fig. 2.10.

Example 5. The voltage v141sin(314t /6) Vis applied to capacitance F

C 320 . Write down the current instantaneous value. Draw the vector diagram for current and voltage effective values.

Capacitance reactance is: X_C 1/(C)1/(314320106)10

. Current amplitude is: I_m V_m/X_C 141/1014.1A.

Phase shift angle for the element is:  _V  _I /2. Current initial phase is: _I _V   /6 /2 /3.

Current instantaneous value is: i I sin(t )14.1sin( 314t /3) А

The vector diagram is shown at fig. 2.11.

Example 6. The capacitor C400 F and coil with parameters R 50, H

L0.3 are connected in series to the AC source. The voltage effective value is V

V 200 , f 50Hz . Define the circuit current, the voltages across the capacitor and coil, the circuit’s active and reactive powers and the power factor.

Coil and capacitor reactances are: X_L L3140.3100,

       8 ) 10 400 314 /( 1 ) /( 1 6 C X_С  .

Circuit impedance is: 2 2

)

(X_L X_C

R

Z     502 (1008)2 105. Circuit current is: I V/Z 200/1051.9A.

Coil impedance is: 2 2

L COIL R X

Z    502 1002 112

. Coil voltage is: V_COIL  IZ_COIL 1.9112 213V.

Capacitor voltage is V_C  IX_C 1.9815.2V.

Elements reactive powers are: Q_L  X_LI2 1001,92 361VAr,

VAr I

X

Q_C  _C 2 81,92 29

.

Circuit reactive power is: QQ_L Q_C 36129 332VAr. Circuit active power is: P RI2 501,92 180.5W

. Power factor makes: cos P/S R/Z 50/1050.48.

Example 7. The input voltage for the circuit with a parallel connection of the coil with parameters R 6, X_L  12 and capacitor with reactanceX_C  6 is V 20V . Define the circuit current, the branches’ currents and the phase shift angles.

The coil serial connection R,X_L is transformed into parallel one of G,B_L. Conductance and susceptances then are:

   ) ( 2 2 L X R R G 0.033 Sm 12 6 6 2 2   ,    ) ( 2 _L2 L L X R X B _{0.067 Sm} 12 6 12 2 2   , Sm X B_C 1/ _C 1/60.167 . The circuit susceptance and admittance are:

Sm B B B  _C  _L 0.1670.067  0.1 , 2 2 B G Y    0.0332 (0.1)2 0.105Sm. The coil admittance makes: 2 2

L

COIL G B

Y    0.0332 (0.067)2 0.075Sm

. The circuit current is: I YV 0.105202.1A.

The branches’ currents are: I_COIL Y_COILV 0.075201,5A,

A V

В

I_С  _С 0.16720 3,34 .

o 72 ) 033 . 0 / 1 . 0 ( ) / (      arctg B G arctg  .

Phase shift angle between coil current and input voltage is:

o 64 ) 033 . 0 / 067 . 0 ( ) / (    arctg B_L G arctg L  .

The circuit active power makes: P  RI_COIL2 61.52 13.5W .

Example 8. The voltage V  220Vis applied to the coil. The current is I 4 A, active power makes P500W and f 50Hz . Define the cos before and after the capacitors’ battery C 400 F is connected in parallel to the coil.

The circuit resistance, impedance and reactance without capacitors make accordingly: R P/ I2 500/42  31.3 , Z V /I  220/4 55, 2 2 R Z X_L    552 31.32 45.2.

Power factor is: cos R/ Z 31.3/550.57.

When the capacitors’ battery is connected, the circuit will have parallel connection of the coil and capacitors. Serial connection of X_L,R elements is to be transformed into parallel connection of B_L,G elements.

Sm Z X B L L 0,015 55 2 . 45 2 2    , Sm Z R G 0.01 55 3 . 31 2 2    .

The capacitors’ susceptance makes: B_C C 314500106  0.0157 Sm

. The circuit susceptance is: В В_С  B_L0.01570.0150.0007Sm. The circuit admittance makes: Y  G2 B2  0.012 0.00072 0.01Sm_.

The power factor is: cos G/Y 0.01/0.011.

Example 9. Define the instantaneous values of currents i1, i2, i (fig. 2.12), when the

input voltage is v 35sin 314tV . Elements parameters are accordingly: R₁  7,

  2

2

R , Х_L  3 and Х_C  5.

This task is solved by using the vector diagrams method. Branches’ impedances are: 2 2

1 1 R XL Z    72 32  7.62 , 2 2 2 2 R XC Z    22 52 5.39 . Branches’ currents amplitudes are:

A 4.59 /7.62 35 / ₁ 1 V Z   I_{m m} , A 6.49 /5.39 35 / ₂ 2 V Z   I_{m m} .

Phase shift angles between V and branches’ currents are: ₁  arctg( X_L / R₁ ) arctg(3/ 7)32,

 68 2 5 2 2  arctg(XC / R )arctg( / )   ,

Currents initial phases makes I V :

  32 32 0 1     I  , _I2 68.

The basic vector is a voltage vector V . It is put along the axis X (fig. 2.13), the branches currents vectors are drawn at the angles o

Branches’ currents instantaneous values are: A ) t sin( . i₁  4 59  32 , i₂ 6.49sin(t68 )A.

The total current vector is the geometrical sum of branches’ currents vectors:

m m m I I I  1  2 (fig. 2.13):      I I I I cos( ) I_{m m m}2 _{m1 m2 1 2} 2 2 1 2   A . ) cos( . . . .59 649 2 459 649 32 68 727 4 2  2       .

Total current effective value makes: I  Im / 2 8.61/1.415.2A.

Phase shift angle between input voltage and total current is:

    2 2 1 1 2 2 1 1      cos I cos I sin I sin I arctg m m m m  30 68 49 6 32 59 4 68 49 6 32 59 4 _     ) cos( . cos . ) sin( . sin . arctg .

Circuit total current is: i 7.27sin(t30 )A.

Example 10. Define the circuit current (fig. 2.12) at the input voltage effective value of V 25V , f 50Hz and elements’ corresponding parameters: R₁  7,

  2

2

R , Х_L  3, Х_C  5.

This task is solved by using the active and reactive constituents method. Branches’ impedances are: 2 2

1 1 R XL Z    72 32 7.62, 2 2 2 2 R XC Z    22 52  5.39 . Branches’ currents effective values are:

A 3.28 /7.62 25 / ₁ 1 V Z   I , I₂ V /Z₂ 25/5.39 4.64A.

Phase shift angles between the voltage and branches currents are:

 32 7 3 1 1  arctg( XL / R ) arctg( / )  ,  68 2 5 2 2  arctg(XC / R )arctg( / )   .

The vector diagram for the circuit is shown at fig. 2.13. The branches currents vectors are shown as I₁, I₂ together with their active and reactive constituents.

Active and reactive currents constituents are defined accordingly: A I I_a1  ₁cos₁ 3.28cos(32)2.78 A . ) cos( . cos I I_a2  ₂ ₂ 464 68 174 , 1 1 1 I sin I_R  3.28sin321.74A, A I I_R2  ₂sin₂ 4.64sin(68) 4.3 . Active and reactive total current constituents make accordingly: A I I I_a  _a1  _a2  2.781.74  4.52 , A I I I_R  _R1  _R2 1.744.3 2.526 .