PHASOR ANALYSIS

KIRCHHOFF’S LAWS Kirchhoff’s Current Law

If sinusoidal excitation is applied to a circuit, then sinusoidal currents (4.1.1) flow through the elements. If cosinusoidal excitation is applied to a circuit, then cosinusoidal currents, say , flow through the elements. KCL holds for both excitations, then it also holds for the complex excitation (defined in Appendix B):

Dividing out the common factor , KCL for phasors is obtained.

The phasor algebraic sum of all currents at a node (cutset) is equal to zero

(4.2.1)

Kirchhoff’s Voltage Law

A similar development will also establish KVL.

The phasor algebraic sum of all voltages around a loop (closed path) is equal to zero (4.2.2)

ELEMENT LAWS Resistor

Setting sinusoids into the Ohm’s law, the following equation is obtained:

(4.2.3) Thus, the resistor rms voltage may be expressed by its rms current, the voltage phase by the current phase

(4.2.4a) (4.2.4b) Setting these equations in the voltage phasor (4.1.5b), the following equation is obtained

Finally, current-voltage law for phasors can be formulated.

)

(4.2.5a) (4.2.5b)

Circuit symbols for a resistor described in time and frequency domains are presented in Fig.

4.2.1.

Fig. 4.2.1 Circuit symbols for a resistor described in time and frequency domain

Inductor

Setting sinusoids into the inductor law (3.1.16a)

(4.2.6) Then,

(4.2.7a) (4.2.7b) Next,

Finally, current-voltage law for phasors can be formulated.

(4.2.8a)

Fig. 4.2.2 Circuit symbols for a coil described in time and frequency domain

Capacitor

Setting sinusoids into the capacitor law (3.1.8a), the following equation is obtained:

(4.2.9) Thus,

(4.2.10a) (4.2.10b) Next,

Finally, current-voltage law for phasors can be formulated.

(4.2.11a) (4.2.11b)

Circuit symbols for a capacitor described in time and frequency domains are shown in Fig.

4.2.3.

Fig. 4.2.3 Circuit symbols for a capacitor described in time and frequency domains )

90 sin(

2 )

sin(

2 t _i  CU t _u  ^o

I     

CU I 

90o



 _u

i 



) 90 exp(

) exp(

) exp(

)

(j I j CU j j ^o

I   _i  _u

) ( )

(j jCU j

I 

) 1 (

)

( 

  I j C j j

U 

i u

L

i u

C

GENERAL TWO-TERMINAL PHASOR CIRCUIT, PHASOR IMPEDANCE Element equations (4.2.5), (4.2.8) and (4.2.11) can be expressed in the general form

(4.2.12a) (4.2.12b) where

(4.2.13)

is called the element complex impedance, is called the complex admittance. As can be observed, these impedances (admittances) are the s-domain impedances (admittances), collected in Table 3.1.2, with .

Consider a general phasor subcircuit with two accessible terminals, as presented in Fig. 4.2.4.

Fig. 4.2.4 General phasor two-terminal subcircuit and its equivalents

The equivalent impedance of such subcircuit can be defined, as the ratio of the phasor voltage to the phasor current:

(4.2.14a) The reciprocal of impedance, the ratio of the phasor current to the phasor voltage is the equivalent admittance:

(4.2.14a) It is important to stress that impedance (admittance) is a complex number that scales one phasor to produce another, but it is not a phasor. The modulus (magnitude) of impedance is the ratio of effective values of the voltage and the current and the angle is the difference of the voltage and the current angles, as presented in equations (4.2.15) – for simplicity of description argument is omitted (fixed frequency is assumed).

)

(4.2.15) The impedance (admittance) can be expressed in a rectangular form, as presented in equations (4.2.14), where:

 is the resistive component of Z, or simply resistance,

 is the reactive component of Z, or simply reactance,

 is the conductive component of Y, or simply conductance,

 is the susceptive component of Y, or simply susceptance.

For the given components of impedance (admittance) its magnitude and angle can be determined, or vice-versa:

(4.2.16a) (4.2.16b) These relationships are graphically expressed in Fig. 4.2.5.

Fig. 4.2.5 Graphical representation of impedance

For the given components of impedance, components of admittance can be determined, or vice-versa:

(4.2.17a)

(4.2.17b) For the fixed frequency , taking into account rectangular form of the impedance, its series equivalent circuit can be found, taking into account rectangular form of the admittance, its parallel equivalent circuit can be found, as presented in Fig. 4.2.4.

The resistance is always nonnegative while reactance can be positive (inductive) or negative (capacitive), and correspondingly the impedance angel can be positive or negative

(4.2.18) All possible cases are considered next.

i



The voltage leads the current by 90 degrees, as presented in Fig. 4.2.6 ( is assumed as the reference phasor, ). The resistance is equal to zero and subcircuit has pure inductive character, its equivalent consists of one element, inductor, as presented in the same Fig. 4.2.6.

Fig. 4.2.6 Voltage and current phasors and circuit equivalent,



The voltage leads the current by the angle less than 90 degrees, as presented in Fig. 4.2.7 ( ). The resistance is greater than zero, reactance is positive, . Then, the circuit has inductive character. Its equivalent consists of two elements, resistor and inductor, as presented in the same Fig. 4.2.7.

Fig. 4.2.7 Voltage and current phasors and circuit equivalent, 0°< φ < 90°



There is no shift between the voltage and the current, as presented in Fig. 4.2.8 ( ).

The resistance is greater than zero, reactance is equal to zero. Then, the subcircuit has resistive character, its equivalent consists of one element, resistor, as presented in the same Fig. 4.2.8.

90o

 

) (j I 0o

i 



o

o 90

0   0o i 

 X L_s

0o



0o i 





The current leads the voltage by the angle less than 90 degrees, as presented in Fig. 4.2.9 ( ). The resistance is greater than zero, reactance is negative, . Then, the circuit has capacitive character. Its equivalent consists of two elements, resistor and capacitor, as presented in the same Fig. 4.2.9.

Fig. 4.2.9 Voltage and current phasors and circuit equivalent, −90°< φ < 0°



The current leads the voltage by the angle of 90 degrees, as presented in Fig. 4.2.10 ( ). The resistance is equal to zero, reactance is negative, . Then, the subcircuit has pure capacitive character, its equivalent consists of one element, capacitor, as presented in the same Fig. 4.2.10.

Fig. 4.2.10 Voltage and current phasors and circuit equivalent

Example 4.2.1

Find the series and parallel equivalents of the circuit presented in Fig. 4.2.11, rad/s.

Fig. 4.2.11 Circuit for Example 4.2.1



o

o 0

90  

 

0o i 

 X 1/C_s

90o



  0o i 

 X 1/C

1000 ,

10 /

1 , 15 ,

10 ,

5 ₂

1  R   L  C  

R

I(j)

I(j) U(j)

j C



 1

U(j)

L

C

The subcircuit impedance is

(4.2.19)

As can be observed, for the given frequency of 1000 rad/s, the subcircuit has inductive character.

The series equivalent consists of resistance and reactance. Then, the series inductance is mH.

From (4.2.17a), parameters of the parallel equivalent can be calculated: S, S. Then, the parallel resistance is equal to and the parallel inductance is equal to mH.

Both equivalent circuits are presented in Fig. 4.2.12.

Fig. 4.2.12 Equivalent circuits for Example 4.2.1

ALGORITHM OF AC STEADY-STATE ANALYSIS

The Kirchhoff’s laws (4.2.1) and (4.2.2), together with element laws (4.2.5), (4.2.8) and (4.2.11), can be used to formulate circuit equations in the phasor-domain. The analysis is therefore identical to the resistive circuit analysis, with impedances replacing resistances and phasors replacing dc currents and voltages, nodal analysis can be applied. Then, algorithm of ac steady-state analysis can be formulated.

Algorithm 4.2.1 Phasor method of ac steady-state analysis Step 1. Built a phasor circuit.

Step 2. Formulate phasor equations, nodal method can be applied.

Step 3. Solve the equations to find phasors describing currents and voltages.

Step 4. Express the solution graphically, by means of the phasor diagram.



Transformation of the solution to the time-domain is trivial. Once phasors are found, they can be converted immediately to the time-domain sinusoidal answers. Phasor diagram is helpful in checking correctness of the solution, also allows to read phase shifts between phasors.

Example 4.2.2

Find the mesh current, element voltages and voltage ; draw the phasor diagram:

V, , , .

Fig. 4.2.13 Phasor circuit of Example 4.2.2



The mesh current is:

A (4.2.20)

It lags the supply voltage, what means that, for rad/s, RLC series circuit has inductive character.

Next, element voltages can be calculated

V (4.2.21a)

V (4.2.21a)

V (4.2.21a)

These voltages satisfy KVL equation

(4.2.22) This solution can be expressed graphically. Fig. 4.2.14 presents three phasor diagrams:

a) all phasors are anchored in the origin of the complex plane, b) voltage phasors are shifted, following KVL equation,

c) voltage phasors are shifted, following KVL equation, such that the circuit topology is mapped.

In this latter case, location of the circuit nodes is uniquely defined and thus, all other voltages may be read directly from the diagram. The voltage between nodes C and A is

V V (4.2.23)

Fig. 4.2.14 Phasor diagrams for RLC series circuit (Example 4.2.2)



Drill problems 4.2

1. What reactance of: a) inductive character, b) capacitive character, should be connected in series with j100  coil such that at supply, coil voltage drops by 50%, i.e.

down to 100 V?

Fig. P.4.2.1

2. Sketch the phasor diagram and read the voltage : ,

V. Repeat calculations (drawing) for: a) , b) .

Fig. P.4.2.2

V

200 U

) (j

U_BA R₁ R₂  X_L  X_C 10

10 ) (j 

E R₁ 0 R₂ 0

c)

D

A

B C b)

a)

200

C

A B

D

3. Find the effective value of the mesh current and the coil voltage.

Fig. P.4.2.3

4. For the parallel RL circuit, find the total rms current if rms currents of elements are:

.

5. For the series RL circuit, find the total rms voltage if rms voltages of elements are:

.

6. Repeat Problems 4.2.4 and 4.2.5 with L replaced by C.

7. Find the effective value of the mesh current and the capacitor voltage.

Fig. P.4.2.7

8. The rms voltages of the RLC series circuit are: . Find the

supply rms voltage.

9. A two-terminal circuit is a series connection of resistor and energy storage element.

Identify character of this element and find both constants R and C or L if the circuit

voltage and current are: V, A.

10. Sketch the phasor diagram mapping the topology. Assume: , , a) , b) 2 , c) 0.5 .

Fig. P.4.2.10 A 3 , A

4 

 _L

R I

I

V 3 ,

V

4 

 _L

R U

U

V 3 ,

V 4 V,

10  

 _{C L}

R U U

U

t

u12 2sin100 _{i3sin(100t45}^o₎ E j

E( ) X_L R X_C 

XL X_L X_L

B D C R

A

V

V

V

V

11. At rad/s the rms currents are: . Find the rms

currents for: a) , b) and .

Fig. P.4.2.11

12. Find the coil ammeter indication, if the capacitor ammeter indication is 2 A and .

Fig. P.4.2.12

13. Find the capacitor ammeter indication, if the coil ammeter indication is 2 A and .

Fig. P.4.2.13

14. Find resistance and reactance of the series equivalent at rad/s, for .

Fig. P.4.2.14

15. The rms current taken from a voltage source V by a series combination of and is 4 A. Find the inductance.

16. Find the rms current taken from a voltage source V by a capacitor of 100 nF in series with a resistance of 10 k.

1 100

 I_L(₁)16A, I_C(₁)4A )

( ), ( ),

(jk₁ I jk₁ I jk₁

I _{L C} k 2 k 1/2 U()const







R 1/ C 10

L 









R 1/ C 10

L 



5000

 μF

20 ,

mH 1 ,

10  

 L C

R

t e20 2sin10



3

R L?

t e20 2sin10³

A A

AC

17. Find the Thevenin equivalent, V, , . Fig. P.4.2.17

18. If the current that flows through the RLC branch is V and H, F, find the branch voltage.

19. In the circuit of Problem 4.2.12 parameters are not known but it is known that the resistor rms current is 6 A and the capacitor rms current is 8 A. Find the coil rms current.

20. It is known that the resistor rms voltage is 8 V and the capacitor rms voltage is 4 V. Find the coil rms voltage.

Fig. P.4.2.20

21. A series combination of a resistance and a capacitance produces a 2 A rms current that leads the applied voltage by . If the amplitude of this voltage is 200 V (50 Hz), what are the resistance and capacitance ?

t

e10 2sin100 _R1k C0.1μF

t i10 2sin10 6

. 0 ,

2 

 L

R C0.05

45o

W dokumencie Circuit theory (Stron 186-199)