Sinusoidal function

Fundation

$$\begin{aligned} & e\left( t \right ) = E_m sin \left(\omega t+\Psi_e \right )\\ & u\left( t \right ) = U_m sin \left(\omega t+\Psi_u \right )\\ & i\left( t \right ) = I_m sin \left(\omega t+\Psi_i \right) \end{aligned}$$ \(e_1,e_2\)-instantaneous values

\(E_m\) –amplitude value

\(\Psi_e\)-phase at \(t = 0 \)

\(T=\frac{1}{f}=\frac{2\pi}{\omega}\)-period [s]

\(f\)-frequency [Hz]

\(\omega=2\pi f\)–angular frequency [rad/s]

Root mean square

AC current with an amplitude \(I_m\) would produce the same Joules of heat in a resistive load as DC current I, if the value of DC current equals to the root mean square (RMS) value of AC current:

$$I=\frac{I_m}{\sqrt{2}}$$ Also E and U is similar to I: $$E=\frac{E_m}{\sqrt{2}}\qquad U=\frac{U_m}{\sqrt{2}}$$

Steady State Sinusoidal Analysis Using Phasors

Complex number

For a standard formula: $$y=ax^2+bx+c$$ The solve of this formula is : $$x_{1,2}=\frac{-b}{2a}\pm \frac{\sqrt{-\Delta}}{2a}=\frac{-b}{2a}\pm \frac{j \sqrt{D}}{2a}$$ In the algebraic form it is: $$\underline{x}_{1\text{,}2}=\Re\left(x_1 \right )\pm \Im \left(x1 \right )$$ Now we can see that complex number consists of two parts: $$\begin{aligned} & \Re e\left(x_1 \right ) & - & \text{real part}\\ & \Im m\left(x_1 \right ) & - & \text{imaginary part}\\ & j=\sqrt{-1} & - & \text{imaginary unit} \end{aligned}$$

The parts of a complex number may be interpreted as coordinates of a complex plane:

$$\underline{X}=X_1+jX_2$$ $$\left | \underline{X} \right |=\sqrt{X_1^2+X_2^2}$$ $$tan\left(\alpha \right )=\frac{\Im m\left(\underline{X}\right)}{\Re e\left(\underline{X}\right)}=\frac{X_2}{X_1}$$ And vary IMPORTANT the exponential form is : $$\underline{X}=\left|\underline{X}\right|e^{j\alpha}$$

Euler's formula states that, for any real number \(\alpha\): $$e^{j\alpha}=cos\;\alpha+jsin\;\alpha$$ And multiply by a complex number modulus the \(\left|\underline{X}\right|e^{j\alpha}\) will going to change : $$\left|\underline{X}\right|e^{j\alpha}=\left|\underline{X}\right|cos\;\alpha+j\left|\underline{X}\right|sin\;\alpha$$

So the algebraic form of a complex number is dervied: $$X_1=\left|\underline{X}\right|cos\;\alpha\qquad X_2=\left|\underline{X}\right|sin\;\alpha$$ Conclusion: $$\underline{X}=\left|\underline{X}\right|e^{j\alpha}\Rightarrow\left|\underline{X}\right|=X_m\text{,}\alpha=\omega t+\psi$$ $$X_me^{j\left(\omega t+\psi \right )}=X_mcos\left(\omega t+\psi \right )+jX_msin\left(\omega t+\psi \right )$$

Complex function

It's a example of complex function: $$e\left(t \right )=E_msin\left(\omega t+\psi_e \right )$$

And we can present the excitation in the form of a complex function: $$E_me^{j\left(\omega t+\psi_e \right )}=E_mcos\left(\omega t+\psi_e \right )+jE_msin\left(\omega t+\psi_e \right )$$

And another example: $$I_me^{j\left(\omega t+\psi_i \right )}=X_1+jX_2=I_mcos\left(\omega t+\psi_i \right )+jI_msin\left(\omega t+\psi_i \right )$$

The obvious choice of complex functions when all energy sources of a circuit generate signals of the same frequency is t=0: $$\underline{X}=X_me^{j\psi}=X_mcos\psi+jX_msin\psi$$

Transformation of circuit function

We have an EMF function: $$e\left(t \right )=E_msin\left(\omega t+\psi_e \right )=311\cdot sin\left(314\cdot t+\frac{\pi}{4} \right )$$ And we will express it as a complex number: $$\underline{E}=311e^{j\frac{\pi}{4}}=311cos\frac{\pi}{4}+j311sin\frac{\pi}{4}=220+j220$$ We can varify them: $$E_m=\sqrt{220^2+220^2}=311\left[B \right ]$$ $$\psi_e=arctan\left(\frac{220}{220} \right )=0.785=\frac{\pi}{4}$$

Algorithm of phasor analysis

  1. Transform the input function into the complex function at t=0.
  2. Transform the circuit into equivalent circuit in which all currents, voltages and impedances are considered as complex numbers.
  3. Calculate the currents and voltages by using KVL, or any other circuit analysis method.
  4. Transform the complex functions of currents and voltages into functions in time domain.

Behavior of circuit elements under alternating current conditions

Resistance

resistance

$$i_R\left(t \right )=I_{R\text{.}m}sin\left(\omega t+\psi_i \right )$$ $$u_R\left(t \right )=U_{R\text{.}m}sin\left(\omega t+\psi_u \right )$$

AC current flowing through the resistance varies in proportion to the applied voltage across it following the same sinusoidal pattern. Phasor of current is “in-phase” with phasor of voltage \(\psi_u=\psi_i\)

$$\underline{Z}_R=\frac{\underline{U}_R}{\underline{I}_R}=\frac{U_R}{I_R}e^{j\left(\psi_u-\psi_i \right )}$$

power

The instantaneous power consumed in the resistor is given by: $$p\left(t \right )=u_R\left(t \right )\cdot i_R\left(t \right )=U_{R.m}sin\left(\omega t \right )\cdot I_{R\text{.}m}sin\left(\omega t \right )=U_{R\text{.}m}I_{R.m}sin^2\left(\omega t \right )$$ $$sin^2\left(\omega t \right )=\frac{1-cos\left(2\omega t \right )}{2}$$ $$p\left(t \right )=\frac{U_{R\text{.}m}I_{R\text{.}m}}{2}\left(1-cos\left(2\omega t \right ) \right )=U_RI_R\left(1-cos\left(2\omega t \right ) \right )$$ Active power [W] is the average power consumed by a resistance over one cycle: $$P=\frac{1}{T}\int_{0}^{T}p\left(t \right )dt=U_RI_R=RI_R^2=\frac{U_R^2}{R}\Leftrightarrow P=\frac{P_m}{2}=\frac{U_{R\text{.}m}\cdot I_m}{\sqrt{2}\cdot\sqrt{2}}=U_RI_R$$

Inductor

inductive reactance

$$i_L\left(t \right )=I_{L\text{.}m}sin\left(\omega t+\psi_i \right )\rightarrow \underline{I_L}=\frac{I_{L\text{.}m}}{\sqrt{2}}e^{j\psi_i}=I_Le^{j\psi_i}$$

$$\begin{aligned}&u_L\left(t \right )=L\frac{di_L}{dt}=\omega LI_{L\text{.}m}cos\left(\omega t+\psi_i \right )\\&=U_{L\text{.}m}sin\left(\omega t+\psi_i+\frac{\pi}{2} \right )=U_{L\text{.}m}sin\left(\omega t+\psi_u \right )\rightarrow \underline{U_L}=U_Le^{j\psi_u}\end{aligned}$$

The amplitude value and phase of voltage across the inductor:

$$U_{L\text{.}m}=\omega LI_{L\text{.}m}\qquad \psi_u=\psi_i+\frac{\pi}{2}$$

$$\underline{Z_L}=\frac{\underline{U_L}}{\underline{I_L}}=\frac{U_L}{I_L}e^{j\left(\psi_u-\psi_i \right )}=\omega Le^{j\frac{\pi}{2}}=jX_L$$

\(X_L=\omega L\)-inductive reactance\(\left[\omega\right]\) $$\varphi =\psi_u-\psi_i=\frac{\pi}{2}$$

energy

The instantaneous power: $$\begin{aligned}p\left(t \right )&=i_L\left(t \right )\cdot u_L\left(t \right )=I_{L\text{.}m}sin\left(\omega t \right )\cdot U_{L\text{.}m}sin\left(\omega t+\frac{\pi}{2} \right )\\&=U_{L\text{.}m}I_{L\text{.}m}cos\left(\omega t \right)sin\left(\omega t \right )=\frac{U_{L\text{.}m}I_{L\text{.}m}}{2}sin\left(2\omega t \right ) \end{aligned}$$ Reactive power is the portion of electricity that helps establish and sustain the magnetic field of an inductor (volt-amperes reactive (VAr): $$Q_L=\frac{U_{L\text{.}m}I_{L\text{.}m}}{2}=\frac{U_{L\text{.}m}}{\sqrt{2}}\cdot\frac{I_{L\text{.}m}}{\sqrt{2}}=U_LI_L=I_L^2X_L=\frac{U_L^2}{X_L}$$ And especially the average reactive power of an inductor over one cycle is zero.

Capacitor

capacitive reactance

$$u_c\left(t \right )=u_{C\text{.}m}sin\left(\omega t+\psi_u \right )\rightarrow \underline{U_C}=\frac{U_{C\text{.}m}}{\sqrt{2}}e^{j\psi_u}=U_Ce^{j\psi_u}$$ $$\begin{aligned}i_C\left(t \right )&=C\frac{du_C}{dt}=\omega CU_{C.m}cos\left(\omega t+\psi_u \right )\\&=I_{C\text{.}m}sin\left(\omega t+\psi_u+\frac{\pi}{2}\right )=I_{C\text{.}m}sin\left(\omega t+\psi_i \right )\rightarrow \underline{I_C}=I_Ce^{j\psi_i} \end{aligned}$$ The amplitude value and phase of voltage across the capacitive: $$I_{C.m}=\omega CU_{C\text{.}m}\rightarrow I_C=\omega CU_C$$ $$\underline{Z_C}=\frac{\underline{U_C}}{\underline{I_C}}=\frac{U_C}{I_C}e^{j\left(\psi_u-\psi_i \right )}=\frac{e^{-j\frac{\pi}{2}}}{\omega C}=-jX_C$$ \(X_C=\frac{1}{\omega C}\)-capacitive reactance \(\left[\omega\right]\) $$\varphi=\psi_u-\psi_i=-\frac{\pi}{2}$$

energy

The instantaneous power: $$\begin{aligned}p\left(t \right )&=u_C\left(t \right )\cdot i_C\left(t \right )=U_{C\text{.}m}sin\left(\omega t \right )\cdot I_{C\text{.}m}sin\left(\omega t+\frac{\pi}{2} \right )\\&=U_{C\text{.}m}I_{C\text{.}m}cos\left(\omega t \right )sin\left(\omega t \right )=\frac{U_{C\text{.}m}I_{C\text{.}m}}{2}sin\left(2\omega t \right )\end{aligned}$$

Reactive power is the portion of electricity that helps establish and sustain the electric field of a capacitor (volt-amperes reactive (VAr): $$Q_C=\frac{U_{C\text{.}m}I_{C\text{.}m}}{2}=\frac{U_{C\text{.}m}}{\sqrt{2}}\cdot\frac{I_{C\text{.}m}}{\sqrt{2}}=U_CI_C=I_C^2X_C=\frac{U_C^2}{X_C}$$ And also the average reactive power of a capacitor over one cycle is zero.

Power in AC circuit

Power triangle

Complex power (volt-ampere (VA)): $$\underline{S}=P+jQ=Se^{j\varphi}$$ $$P=\sum_{n}P_{R_n}\qquad Q=\sum_{n}Q_{L_n}-\sum_{n}Q_{C_n}$$ $$\underline{S}=U_{R\text{.}m}I_{R\text{.}m}cos\varphi+jU_{R\text{.}m}I_{R\text{.}m}sin\;\varphi$$ Apparent power(VA): $$s=\left|\underline{S}\right|=\sqrt{P^2+Q^2}$$ Complex power (in phasor form): $$\underline{S}=Se^{j\left(\psi_u-\psi_i \right)}=U_{R\text{.}m}e^{j\psi_u}\cdot I_{R\text{.}m}e^{-j\psi_i}$$

Power factor

Power factor: $$cos \varphi=\frac{P}{S}=\frac{P}{\sqrt{P^2+Q^2}}$$ $$-1\leq cos \varphi \leq 1$$

Reactive power compensation

Reactive power compensation is one of the most effective ways to reduce consumed electric energy and improve power quality. Reactive power compensation provides:

reducing costs; reducing network losses; avoiding penalty charges from utilities for excessive consumption of reactive power; increasing system capacity and save costs on new installations; improving system power factor; increasing power availability; improving voltage regulation in the network.

Electric power consumers

Active power consuming loads: light bulbs, electric motors, electric heaters, and so on.

Reactive power consuming loads: induction motors, induction generators, heavily loaded transmission line, solid state converters, transformers, under-excited synchronous machines, arc furnaces, discharged lighting, constant loads such as induction heating, space heating, water heating, and air conditioning.


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