3-Phase System

The illustration below is an example of a 3-phase system. It is a common method of AC power generation, transmission, and distribution. Its component values involve Van, Vbn, and Vcn, which are the phase voltages, Vl which is the line voltage, Zl which is the line impedance, Z which is the load impedance, and Ia, Ib, and Ic which is the phase currents. The 3 phases carry voltages and currents which are 120 degrees out of phase with each other. This illustration represents Y-Y connection, other samples are delta-Y, Y-delta, and delta-delta.

From a 3-phase system, we could create a single phase system. But a 3-phase system is more economical than a single-phase or two-phase system because it uses less conductor material (wire) to transmit electric power, at the same voltage. Written below are the common formulas for obtaining the unknown value in a 3-phase system, you could also use the other formulas introduced on the previous posts:

The Power Triangle and its Components

From the previous discussions, AC impedance (Z) is defined as a complex quantity that made up of real resistance (R) and imaginary reactance (X). But for now, we will discuss about Apparent power (S) which is also a complex quantity that is made up of real active power (P) and imaginary reactive power (Q).

The object illustrated above is a power triangle, it consists a real power (P), apparent power (S), reactive power (Q) and phase angle (Ɵ). The relationship between these three can be expressed using vectors, Pythagorean Theorem, simple sohcahtoa and etc. For vectors, Real power is horizontal vector, Reactive power is vertical vector, and Apparent power is the hypotenuse of the right angled triangle. 

For the Pythagorean theorem:

S=√Q^2 + P^2

For the sohcahtoa:

sin Ɵ = Q/S

cos Ɵ = P/S

tan Ɵ = Q/P

To get the phase angle (Ɵ):

Ɵ = arccos* Power factor (P.F.)

Other formulas that 

include S, P, and Q are:

P = I^2*(R)

Q = I^2*(X)

S = I^2*(Z)

Formulas also discussed at ac power analysis can be used.


The Power Factor (P.F.)

- Power factor is defined as the ratio of the actual electrical power dissipated by an AC circuit to the product of the r.m.s. values of current and voltage. It is also the ratio of the real power flowing to the load, to the apparent power in the circuit: 

P.F. = P/S

This formula from phase 

angle could also be used:

P.F. = cos Ɵ

AC Power Analysis

Power is defined as the time rate of doing work. In an AC circuit, power quantities are continuously varying. Capacitive, resistive, and inductive loads help us identify the amount of power in the circuit and in its elements. Power can be absorbed or supplied by circuit elements. There are formulas for finding the Instantaneous, Average, and Maximum power in an AC circuit.

For Instantaneous Power:

P(t)= ½ Vm Im cos(Ѳv – Ѳi) + ½ Vm Im cos(2ῳt + Ѳv + Ѳi)

-To solve the instantaneous power, you just simply substitute the solved/given voltage, current and their phase angles and also the angular frequency. P is also not time dependent.

For Average Power:

P(t)= ½ Vm Im cos(Ѳv – Ѳi)

-This formula is just similar to instantaneous power, but the ½ Vm Im cos(2ῳt + Ѳv + Ѳi) was removed.

For Maximum Power:

P(t)= ½ (I^2) Rl

But if you substitute other formulas, expand, and use derivation, the final formula would be:

1/8 (Vth^2 / Rth)

-Vth and Rth are only the real part of the obtained voltage and Impedance.

Thevenin's theorem/ Norton's Theorem

Thevenin's theorem/ Norton's theorem is another method used for circuit analysis and simplification. It involves converting circuit sources and impedances to a thevenin equivalent. The picture below shows the Thevenin's and Norton's equivalent circuit of the original circuit. No current flows in ab because it represents an open circuit.
 

Thevenin is commonly used for voltage analysis while Norton is used in current analysis.
You could use other circuit solving methods to identify Vth, Rth, and In.

Source Transformation

Is a method used to simplify a circuit by transforming a series voltage source and impedance to a parallel current source and impedance, and vice versa. This process is commonly used to lessen the number of loops in a circuit for easy analysis.

The direction of the arrow of the current source follows the positive sign of the voltage source.

In this circuit, we should identify the voltage at the 4Ω resistor. We will start transforming at the right side of the circuit using simple ohms law.

V= 2A (2Ω)

V= 4 V

 

The resulting circuit would be:

Combine the series impedance's.

Then, transform it again.

I=V/R

I= 4/ 2-j1

I= 1.788∟26.56 A

Combine the parallel impedances.

Z= j3(2-j1)/ j3+(2-j1)

Z= 2.37 ∟18.43 Ω

 

Then transform again so that we can subtract the voltage source and finally obtain V.

V= (1.788∟26.56) (2.25+j0.75)

V= 2.998 + j2.998

V= V1-V2

V= (2.998 +j2.998)-(10)

V= 7.6168 ∟156.82 V

We can now use voltage division to obtain V.

V=(7.616∟156.82)(4)/4+ (2.25+j0.75)

V= 4.84 ∟149.97 V