Power Source 1

220∠0∘

Power Source 2

Voltage is at an angle of ∠8∘

may be owing to the effect of transformers

from the power source

Incoming

Power

Owing to transformers in power source or other reason,

come in voltage is already  8∘above X axis.

Same

Circuit

With

Different

Power

Source

Angle

Circuit

Total

Impedance

ZTot

ZTot = R+XL+XC   =   (30+j0)+(0+j20)+(0-j10)

                         =   30+j10Ω or 31.623∠18.435∘Ω

Impedance Triangle

Since the imaginary part of the circuit total impedance is positive, so the whole locoal circuit is an inductive circuit.

Circuit

Total

Current

ATot

    = 6.6 -j2.2  

=6.957∠-18.435∘A

Total included angle = I angle = 18.435∘as V angle = 0∘

  = 6.842 -j1.26 

=6.957∠-10.435∘A

Total included angle = I angle + V angle = 18.435∘

The total included angle 18.435∘between V & I does not change since the total reactance of the circuit have not change. Only the position of the whole power source change.

As the angle of voltage does not affect the magnitude of current, so the angle of voltage can be neglected during the calculation of current except you want to know the angle of current to X axis line in Cartesian coordinate graph.

The included angle within the impedance triangle of the circuit is 18.435∘

So the circuit total reactance will add their effects to the incoming original included angle between V & I and results an angle of 18.435∘even voltage change its angle.

Obviously, the included angle between voltage and current will be affected by locale reactive elements.

If the circuit has a total reactance X > 0, than it is a inductive circuit, and it will cause the current towards lagging.

If the circuit has a total reactance X < 0, than it is a capacitive circuit, and it will cause the current towards leading.

Proved that the included angle between voltage and current is affected by the included angle of the total impedance triangle of the circuit.

Voltage

of each

element

V_R  =  

= 208.71∠-18.435∘V

A pure resistive element does not affect angle between V & I, as the whole circuit already affected by some reactive elements, the V & I come into R already changed as the circuits V & I angle, so the result bare and angle equal to circuit V & I angle of 18.435∘

= 139.14∠ 71.565∘V

X_L is a reactive element has a +90∘effect.

Angle of come in power 0∘

The impedance effect angle on whole circuit -18.435∘

90 + 0 + (-18.435) =71.565∘will exist after X_L

= 69.57∠-108.435∘V 

X_C is a reactive element has a -90∘effect.

Angle of come in power 0∘

The impedance effect angle on whole circuit -18.435∘

(-90) + 0 + (-18.435) =-108.435∘will exist after X_C

Counter check :

V_R  =  

= 208.71∠-10.435∘V 

R is a non-reactive element

Angle of come in power +8∘

The impedance effect angle on whole circuit -18.435∘

8 + (-18.435) = -10.435∘will exist after R.

= 139.14∠ 79.565∘V 

X_L is a reactive element has a +90∘effect.

Angle of come in power +8∘

The impedance effect angle on whole circuit -18.435∘

90 + 8 + (-18.435) =79.565∘will exist after X_L

= 69.57∠-100.435∘V 

X_C is a reactive element has a -90∘effect.

Angle of come in power +8∘

The impedance effect angle on whole circuit -18.435∘

(-90) + 8 + (-18.435) =-100.435∘will exist after X_C

Counter check :

Please note that the angle value of the above result shows the angle of voltage or current away from reference X axis depend on whether the angle value is positive or negative, even the answer is in unit of voltage.

i.e.

If the angle value is +ve, it represents the voltage angel away from reference X axis line.

If the angle value is -ve, it represents the current angel away from reference X axis line.

When current go through each element, the included angle between V & I will then affected by that element and a new included angle between V & I is formed. After current pass through the last element back to power source, the final circuit included angle between V & I will be detected.

In terms of

current

Active

Power (P)

P_Tot = |I|² R_Tot    (where R_Tot is real part of Z_Tot )

= (6.957 )² x 30             = 1452 W

P_Tot = I² R_Tot   (Calculate with complex number of I )

= (6.957∠-18.435∘)² x 30 = 1452∠-36.87∘==> 1452W

Note : ∠-36.87∘is a double of lagging angle -18.435∘of I

P_Tot = |I|² R_Tot  (where R_Tot is real part of Z_Tot )

= (6.957 )² x 30             = 1452 W

P_Tot = I² R_Tot   (Calculate with complex number of I )

= (6.957∠-10.435∘)² x 30 = 1452∠-20.87∘==> 1452W

Note : ∠-20.87∘is a double of lagging angle -10.435∘of I

It shows that complex number part of I is not needed as it does not affect the magnitude of real power calculation.

Reactive

Power (Q)

Calculate with different value

Q_XL = |I|² X_L = (6.957)² (+j20) = j968        ==> 968 VAr

    = I² |X_L| = (6.957∠-18.435∘)² (20)= 968∠-36.87∘

             ==> 968 VAr

    = I² X_L = (6.957∠-18.435∘)² (+j20)= 968∠53.13∘

           ==> 968 VAr

    = |I|² |X_L| = (6.957)² (20)  = 968VAr(Ind.)

Q_XC = |I|²  |X_C| = (6.957)² (10) = 484VAr(Cap.)

Q_Tot = Q_XL - Q_XC  = (968) - (484) = 484VAr(Ind.)

Calculate with total X_Tot  ( X_Tot is imaginary part of Z_Tot )

Q_Tot

= |I|² X_Tot = (6.957)² x +j10          = 484∠90∘    VAr

= I² |X_Tot| = (6.957∠-18.435∘)² (10) = 484∠-36.87∘VAr

= I² X_Tot = (6.957∠-18.435∘)² (+j10)= 484∠53.13∘ VAr

= |I|² |X_Tot| = (6.957)² (10) = 484 VAr(Ind.)

Calculate with different value

Q_XL = |I|² X_L = (6.957)² (+j20) = j968        ==> 968 VAr

    = I² |X_L| = (6.957∠-10.435∘)² (20) = 968∠-20.87∘ 

==> 968 VAr

    = I² X_L = (6.957∠-10.435∘)² (+j20) = 968∠69.13∘ 

==> 968 VAr

    = |I|² |X_L| = (6.957)² (20) = 968VAr(Ind.)

Q_XC = |I|² |X_C| = (6.957)² (10) = 484VAr(Cap.)

Q_Tot = Q_XL - Q_XC  = (968) - (484) = 484VAr(Ind.)

Calculate with total X_Tot ( X_Tot is imaginary part of Z_Tot )

Q_Tot

= |I|² X_Tot = (6.957)² x +j10           = 484∠90∘   VAr

= I² |X_Tot| = (6.957∠-10.435∘)² (10) = 484∠-20.87∘VAr

= I² X_Tot = (6.957∠-10.435∘)² (+j10) = 484∠69.13∘ VAr

= |I|² |X_Tot| = (6.957)² (10) = 484 VAr(Ind.)

It shows that, no matter you are calculating the reactive power Q for individual or whole circuit,

complex number part is not needed.

In terms of

Voltage

Active

Power (P)

P_Tot = |V |² / R_Tot  (R_Tot is the real part of Z_Tot)

= (208.71)² / 30              = 1452 W

P_Tot = V ² / R_Tot   (Calculate with complex number of V )

= ( 208.71∠-18.435∘)² / 30 = 1452∠-36.87∘==> 1452W

Note : ∠-36.87∘is a double of lagging angle -18.435∘of I

P_Tot = |V |² / R_Tot  (R_Tot is the real part of Z_Tot)

= (208.71)² / 30              = 1452 W

P_Tot = V ² / R_Tot   (Calculate with complex number of V )

= ( 208.71∠-10.435∘)² / 30 = 1452∠-20.87∘==> 1452W

Note : ∠-20.87∘is a double of lagging angle -10.435∘of I

It shows that complex number part of V is not needed as it does not affect the magnitude of real power calculation.

Reactive

Power (Q)

Q_Tot = Q_XL - Q_XC = 968 - 484 = 484VAr

Q_Tot = Q_XL - Q_XC = 968 - 484 = 484VAr

It shows that complex number part is not needed as it does not affect the magnitude of reactive power calculation.

All Type of

Power

In terms of

I & Z

All Type of Power = |I_Tot | ²x Z_Tot = ( 6.957)² x (30+j10)

in rectangular form è =1452 - j484

P is the real part in W,  Q is the imaginary part in VAr.

in polar/phasor form è =1530.54∠18.435∘

The magnitude represents Apparent Power S in VA

Same

May use current (magnitude only) and complex form of Z_Tot to calculate all type of power (apparent, active, reactive) at the same time of a circuit.

In terms of

V & Z

in rectangular formè

P is the real part in W,  Q is the imaginary part in VAr.

in polar/phasor form è ∘

The magnitude represents Apparent Power S in VA

Same

Since the included angle between V & I will affect the result of real part(P) and imaginary part(Q),

don’t include the angle value if you are not sure, just use the voltage magnitude only

( i.e. ∠0∘==> take voltage as reference ).

In terms of

V & I

S = V I

= 220∠0∘( 6.957∠-18.435∘)

in rectangular form è = 1452 - j484

= 220∠-18.435∘( 6.957∠0∘)

in rectangular form è = 1452 - j484

P is the real part in W,  Q is the imaginary part in VAr.

in polar/phasor form è = 1530.54∠-18.435∘

The magnitude represents Apparent Power S in VA

* Note *

When using this method, either voltage or current should take as reference.  i.e. V∠0∘or  I∠0∘and the angle used should be the total included angle between V & I rather than individual V angle alone nor I angle alone. You may put this total included angle between V & I in voltage or current.

You should find the total included angle between V & I , if they both have an angle theirself, you may then add up the absolute value of the two angles and then put the total angle on V or I only.

|S| = 

P = S cos(θ)   = 1530.54VA cos(-18.435∘)   = 1452W

Q = S sin(θ)   = 1530.54VA sin(-18.435∘)    = 484VAr

S = V I

S = 220∠8∘x 6.957∠-10.435∘

in rectangular form è = 1529.16 – j65 ✘

in polar/phasor form è =1530.54∠-2.435∘

S = 220∠8∘x 6.957∠0∘

in rectangular form è = 1515.64 – j213 ✘

in polar/phasor form è =1530.54∠8∘

S = 220∠0∘x 6.957∠8∘

in rectangular form è = 1515.64 – j213 ✘

in polar/phasor form è =1530.54∠-8∘

S = 220∠0∘x 6.957∠-18.435∘= 1452 - j484

S = 220∠-18.435∘x 6.957∠0∘= 1452 - j484

S = 220∠0∘x 6.957∠18.435∘= 1452 + j484

S = 220∠18.435∘x 6.957∠0∘= 1452 + j484

P is the real part in W,  Q is the imaginary part in VAr.

in polar/phasor form è = 1530.54∠18.435∘

The magnitude represents Apparent Power S in VA

The above examples prove that you should take V or I as reference and put the total angle value in V or I only.

* You may be already aware of that the included angle of power triangle is equal to the total included angle between V & I , as well as the included angle of the impedance triangle.

i.e. Reactance element affects the total included angle between voltage and current (leading or lagging), and in turn affects the included angle of power triangle.

Conclude

Given : E = 220∠0∘ and  Z_Tot = 30+j10

then

I_Tot = |E| / Z_Tot    =220/(30+j10)   =6.6 - j2.2

=6.957∠-18.435∘A

Power of each element in terms of | I | :

P_R  = | I_Tot |² x R  = ( 6.957 )² x 30 = 1452W

Q_XL = | I_Tot |² x X_L  = ( 6.957 )² x 20 = 968VAr

Q_XC = | I_Tot |² x X_C  = ( 6.957 )² x 10 = 484VAr

Power of each element in terms of individual voltage |V| :

P_R  = |V_R|² x R  = (208.71 )² / 30 = 1452W

Q_XL = |V_XL|² x X_L  = (139.14)² / 20 = 968VAr

Q_XC = |V_XC|² x X_C  = (69.57 )² / 10 = 484VAr

Calculate all the power of P, Q, S at one time by using Z_Tot

( Since Z_Tot already have the information about the circuit total included angle, so you no need to include angle in V or I any more.

i.e. absolute value of V or I is enough.

All type of Power = I² x Z_Tot

 = ( 6.957 )² x (30+j10)

 = 1452 +j484     ç P is real part, Q is imaginary part

 = 1530.54∠18.435∘ ç S is the magnitude value

All type of Power = V² / Z_Tot

= 220² / (30+j10)

= 1452 –j484

= 1530.54∠-18.435∘

Given : E = 220∠8∘ and  Z_Tot = 30+j10

then

I_Tot = |E| / Z_Tot   =220/(30+j10)       =6.6 - j2.2

=6.957∠-18.435∘A

or

I_Tot = E / Z_Tot   =220∠8∘/ (30+j10)  =6.842–j1.26

=6.957∠-10.435∘A

or try E = 220∠+12.435∘

I _Tot = 220∠+12.435∘/ (30+j10)      =6.913–j0.78

=6.957∠-6.435∘A

Found that no matter how you change the individual angle of V or I, the circuit total included angle between V & I still fixed the same, as only the reactive element within the circuit may affect it. The individual angle of V or I only affect the starting angle of plotting position on graph.

* The real thing may affect the amount (magnitude) of current and on top of the given total included angle between V & I is the impedance triangle, the given individual angle of voltage or current affect only the plotting coordinate position of V & I on graph.