EnergySys244 - Week 1 - 2.1 to 2.7

Express the sinusoidal AC equations of voltage and current

Derive the formula for root mean square

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### Left Side (Derivation)

$$V(t) = V_m \cos(\omega t + \delta)$$

$$V_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) \, dt}$$

$$= \sqrt{\frac{1}{T} \int_{0}^{T} V_m^2 \cos^2 \theta \, dt}$$

$$= \sqrt{\frac{1}{T} \int_{0}^{T} V_m^2 \left( \frac{\cos 2\theta + 1}{2} \right) dt}$$

$$= \frac{V_m}{\sqrt{2}} \sqrt{\frac{1}{T} \int_{0}^{T} (\cos 2\theta + 1) \, dt}$$

$$= \frac{V_m}{\sqrt{2}}$$

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### Right Side (Identities & Notes)

Top (Trigonometric Identities):

$$\cos 2x = \cos^2 x - \sin^2 x$$

$$\cos 2x = 2\cos^2 x - 1$$

$$\frac{\cos 2x + 1}{2} = \cos^2 x$$

Derive the expression for v(t) as a phasor

Left Side (Mathematical Derivation)

1. Euler's Identity:
   ejθ=cos⁡θ+jsin⁡θ

2. Time-domain Voltage:
   v(t)=Vmcos⁡(ωt+δ)

3. Substitution (using Vm=2VrmsVm​=2​Vrms​):
   =2⋅Vrms⋅Re[ejθ]

4. Expanding the angle θθ:
   =2⋅Vrms⋅Re[ej(ωt+δ)]

5. Separating time and phase components:
   =2⋅Vrms⋅Re[ejωt⋅ejδ]

6. Phasor Definition (V‾V​):
   V‾=2⋅Vrms⋅ejδ

7. Polar Form:
   =2⋅Vrms∠δv​

8. Rectangular Form (using Euler's):
   =2⋅Vrms[cos⁡δv+jsin⁡δv]

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Right Side (Phasor Diagram)

A complex plane diagram is drawn with:

• Axes: Real axis (ReRe) pointing right, Imaginary axis (ImIm) pointing up.

• Vector V‾V​: A red arrow originating from (0,0)(0,0) at an angle δvδv​.

• Horizontal Component: Labeled as Vcos⁡δvVcosδv​ along the Real axis.

• Vertical Component: Labeled as jVsin⁡δvjVsinδv​ along the Imaginary axis.

Derive the expression for the reactance of an inductor

Here is the transcription of the handwritten notes in the image:

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### Derivation of Inductive Impedance

1. Basic Inductor Equation:

$$V_L = L \cdot \frac{di_L}{dt}$$

2. Assume Sinusoidal Current:

$$i_L = i_m \cdot \cos(\omega t + \theta)$$

3. Derivative of Current:

$$i_L' = -\omega \cdot i_m \cdot \sin(\omega t + \theta)$$

4. Substitute into Voltage Equation:

$$V_L = -\omega L \cdot i_m \cdot \sin(\omega t + \theta)$$

5. Convert Sine to Cosine (using $\sin(A) = \cos(A - 90^\circ)$):

$$= -\omega L \cdot i_m \cdot \cos(\omega t + \theta - 90^\circ)$$

6. Express in Complex Exponential Form:

$$= -\omega L \cdot i_m \cdot e^{j(\theta - 90^\circ)}$$

Alternative expansion shown:

$$= -\omega L \cdot i_m \cdot e^{j\theta} \cdot e^{-j90^\circ}$$

7. Expand using Euler's Formula:

$$= -\omega L \cdot i_m \cdot e^{j\theta} \cdot [\cos(-90^\circ) + j\sin(-90^\circ)]$$

8. Simplify ($\cos(-90^\circ)=0, \sin(-90^\circ)=-1$):

$$= -\omega L \cdot i_m \cdot e^{j\theta} \cdot [0 + j(-1)]$$

$$= -\omega L \cdot i_m \cdot e^{j\theta} \cdot (-j)$$

$$= j\omega L \cdot i_m \cdot e^{j\theta}$$

9. Final Phasor Relationship:

$$= j\omega L \cdot \underline{I}$$

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### Conclusion (Right Side)

$$\frac{\underline{V}_L}{\underline{I}} = j\omega L$$

$$X_L = j\omega L$$

Derive the expression for the reactance of a capacitor

Here is the transcription of the handwritten notes in the image:

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### Derivation of Capacitive Impedance

1. Basic Capacitor Equation:

$$i_C = C \cdot \frac{dv_c}{dt}$$

2. Assume Sinusoidal Voltage:

$$v(t) = V_m \cos(\omega t + \theta)$$

3. Derivative of Voltage:

$$v' = -\omega V_m \sin(\omega t + \theta)$$

4. Substitute into Current Equation:

$$i_C = -\omega C \cdot V_m \sin(\omega t + \theta)$$

5. Convert Sine to Cosine (using $\sin(A) = \cos(A - 90^\circ)$):

$$= -\omega C \cdot V_m \cos(\omega t + \theta - 90^\circ)$$

6. Express in Complex Exponential Form:

$$= -\omega C \cdot V_m e^{j(\omega t + \theta - 90^\circ)}$$

7. Separate Time and Phase Components:

$$= -\omega C \cdot V_m \cdot e^{j\omega t} \cdot e^{j\theta} \cdot e^{-j90^\circ}$$

8. Expand using Euler's Formula:

$$= -\omega C \cdot V_m \cdot e^{j\omega t} \cdot e^{j\theta} \cdot [\cos(-90^\circ) + j\sin(-90^\circ)]$$

9. Simplify ($\cos(-90^\circ)=0, \sin(-90^\circ)=-1$):

$$= -\omega C \cdot V \cdot (-j)$$

10. Final Phasor Relationship:

$$i_C = j\omega C \cdot V$$

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### Conclusion (Right Side)

$$\frac{\underline{V}}{\underline{i}_C} = \frac{1}{j\omega C} = \frac{-j}{\omega C}$$

$$X_C = \frac{-j}{\omega C}$$

Derive the formula for insant power in terms of avg power and reactant power

Derive the equation for the instantaneous power of a resistor

$$p_R(t) = P + P \cos(2\omega t + 2\theta)$$

$$= P \left[ 1 + \cos(2\omega t + 2\theta) \right]$$

Derive the equation for instant power of an inductor

$$p_L = Q \sin(2\omega t + 2\theta) \quad\quad \theta_v - \theta_i = \beta = 90^\circ$$

$$= \frac{V_m I_m}{2} \sin(\beta) \sin(2\omega t + 2\theta)$$

$$= \frac{V_m I_m}{2} \sin(2\omega t + 2\theta) \quad \text{or} \quad V_{\text{rms}} I_{\text{rms}} \sin(2\omega t + 2\theta)$$

Derive the equation for instant power of a capacitor

$$p_c = Q \sin(2\omega t + 2\sigma) \qquad \text{where } \beta = -90^\circ$$

$$= \frac{V_m I_m}{2} \sin(\beta) \sin(2\omega t + 2\sigma)$$

$$= -\frac{V_m I_m}{2} \sin(2\omega t + 2\sigma)$$

Break down the formula for instant power

1. Real (Average) Power

• In-phase Current Component:

  $$I_r = I_{\text{rms}} \cos(\delta - \beta) = \frac{I_m}{\sqrt{2}} \cos(\delta - \beta)$$

• Real Power ($P$):

  $$P = \frac{V_m I_m}{2} \cos(\delta - \beta)$$

• Unit:

  $$\text{Unit} = \text{Watt } [\text{W}]$$

2. Reactive Power

• Quadrature Current Component:

  $$I_x = I_{\text{rms}} \sin(\delta - \beta) = \frac{I_m}{\sqrt{2}} \sin(\delta - \beta)$$

• Reactive Power ($Q$):

  $$Q = \frac{V_m I_m}{2} \sin(\delta - \beta)$$

• Unit:

  $$\text{Unit} = \text{VAR } (\text{volt-ampere reactive})$$

Describe the power factor

1. Equations

Power Factor ($pf$):

$$pf = \cos(\delta - \beta)$$

Reactive Factor ($rf$):

$$rf = \sin(\delta - \beta)$$

2. Operating ConditionsLeading / Capacitive Load

$$\beta > \delta$$

Current leads voltage:

$$\theta_i > \theta_v$$

The phase difference is negative:

$$\delta - \beta < 0$$

Resulting in a negative reactive factor:

$$rf < 0$$

Lagging / Inductive Load

$$\delta > \beta$$

Current lags voltage:

$$\theta_v > \theta_i$$

The phase difference is positive:

$$\delta - \beta > 0$$

Resulting in a positive reactive factor:

$$rf > 0$$

Conclusion

$$\beta > \delta \quad \text{vs.} \quad \delta > \beta$$

Describe and break down complex power

1. Definition & Trigonometric Form

Complex power $S$ is defined as the sum of real power $P$ and reactive power $Q$:

$$S = P + jQ$$

Using maximum peak values $V_m$ and $I_m$, and phase difference $\theta = \theta_v - \theta_i$:

$$S = \frac{V_m I_m}{2}\cos(\theta) + j\frac{V_m I_m}{2}\sin(\theta)$$

2. Converting to RMS & Exponential Form

Using Euler's identity:

$$e^{j\theta} = \cos\theta + j\sin\theta$$

Converting peak values to RMS:

$$V = \frac{V_m}{\sqrt{2}}$$

$$I = \frac{I_m}{\sqrt{2}}$$

Rewriting the expression:

$$S = \frac{V_m I_m}{2} e^{j\theta}$$

$$S = \left(\frac{V_m}{\sqrt{2}}\right)\left(\frac{I_m}{\sqrt{2}}\right) e^{j(\theta_v - \theta_i)}$$

$$S = V I e^{j(\theta_v - \theta_i)}$$

3. Splitting into Phasors

Separate the exponential term into individual voltage and current components:

$$S = \left(V e^{j\theta_v}\right) \cdot \left(I e^{-j\theta_i}\right)$$

Since:

$$\underline{V} = V e^{j\theta_v}$$

$$\underline{I}^* = I e^{-j\theta_i}$$

We get:

$$S = \underline{V} \cdot \underline{I}^*$$

In polar notation:

$$S = S \angle (\theta_v - \theta_i)$$

4. Expressing in Terms of Impedance

Using Ohm's Law in the phasor domain:

$$\underline{V} = \underline{I} \cdot Z$$

$$\underline{I} = \frac{\underline{V}}{Z}$$

In terms of Current:

$$S = (\underline{I} \cdot Z) \cdot \underline{I}^*$$

Since:

$$\underline{I} \cdot \underline{I}^* = \vert{}\underline{I}\vert{}^2$$

The final expression is:

$$S = \vert{}\underline{I}\vert{}^2 Z$$

In terms of Voltage:

Taking the conjugate of both sides:

$$\underline{I}^* = \left(\frac{\underline{V}}{Z}\right)^* = \frac{\underline{V}^*}{Z^*}$$

Substituting back into complex power:

$$S = \underline{V} \left( \frac{\underline{V}^*}{Z^*} \right)$$

Since:

$$\underline{V} \cdot \underline{V}^* = \vert{}\underline{V}\vert{}^2$$

The final expression is:

$$S = \frac{\vert{}\underline{V}\vert{}^2}{Z^*}$$

Describe what is meant by load convention

• Current enters positive terminal of circuit element

• Power is absorbed if P/Q is positive

• Power is delivered if P/Q negative

Describe what is meant by generator convention

• Current leaves positive terminal of circuit element

• Power is delivered if P/Q is positive

• Power is absorbed if P/Q is negative

Describe the concept of power correction

• S = VI*

  • S / V = I*

• pf = cos( $\theta$ )

  • P = $\frac{VmIm}{2}$ cos ($\theta$)

    • increasing pf increases P which increases S

• if S increases and V is constant, I* is decreasing

• therefore power correction is used to decrease total current flowing (which reduces wastage in the form of heat, etc)

Describe the process of writing nodal equations in matrix form

1. Mark a reference node.

2. All essential nodes' voltages are with respect to the reference node.

3. For every voltage source in series with an impedance:

   • Transform it to a current source in parallel with the impedance.

4. Convert impedances into admittances.

5. Use the following format:

$$\underline{Y} \underline{V} = \underline{I}$$

For $\underline{Y}$ (Admittance Matrix):

• $Y_{kk} =$ sum of admittances connected to node $k$

• $Y_{kn} =$ negative sum of admittances directly between nodes $<span>$k$</span> and$$n$

For $\underline{V}$ (Node Voltage Vector):

• $V_{n0} =$ node voltage for node $n$ relative to reference node $0$

For $\underline{I}$ (Current Source Vector):

• $I_n =$ net sum of source currents connected to node n (entering node is usually $+$ , leaving node is $-$)

Draw the basic template of a balanced three phase circuit. State its characteristics

• All magnitudes are equal

• All voltages have a phase difference of 120$\degree$

• Line to neutral voltages - Ean, Ecn, Ebn represent the voltage between their respective nodes and the neutral node

• Ean has a phase difference of 0

  • Ebn has a phase difference of -120

  • Ecn has a phase difference of 120

• Line to line voltages - Eab = Ean - Ebn

  • Other similiar voltages are about the same

• 
  

Draw 3 phase voltages connected to a wye config

Draw three phase voltages connected to a delta config

Derive the equation for the conversion of line to neutral TO line-to-line voltages

$$E_{ab} = V_p \angle 0^\circ - V_p \angle -120^\circ$$

$$= V_p - V_p \cdot e^{-j120^\circ}$$

$$= V_p \left[ 1 - e^{-j120^\circ} \right]$$

$$= V_p \left[ 1 - (\cos(120^\circ) - j \sin(120^\circ)) \right]$$

$$= V_p \left[ 1 - \left( -\frac{1}{2} - j \frac{\sqrt{3}}{2} \right) \right]$$

$$= V_p \left[ 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right]$$

$$= V_p \left[ \frac{3}{2} + j \frac{\sqrt{3}}{2} \right]$$

$$= V_p \left[ \sqrt{3} \angle 30^\circ \right]$$

$$= \sqrt{3} V_p \angle 30^\circ$$

Derive the equation for the conversion of line to neutral current, Ia TO line-to-line currents

$$I_a = I_p - I_p \angle 120^\circ$$

$$= I_p \left[ 1 - e^{j120^\circ} \right]$$

$$= I_p \left[ 1 - (\cos(120^\circ) + j \sin(120^\circ)) \right]$$

$$= I_p \left[ 1 - \left( -\frac{1}{2} + j \frac{\sqrt{3}}{2} \right) \right]$$

$$= I_p \left[ 1 + \frac{1}{2} - j \frac{\sqrt{3}}{2} \right]$$

$$= I_p \left[ \frac{3}{2} - j \frac{\sqrt{3}}{2} \right]$$

$$= I_p \left[ \sqrt{3} \angle -30^\circ \right]$$

$$= \sqrt{3} I_p \angle -30^\circ$$

State the equation for line to neutral current of a node $a$

in wye config

• Ia = Ean / Zy

State the equation for line to neutral current of a node $a$

in delta config

• Ia = -Iac * Iab

State the equation for line to line current of a Iab

in delta config

• Iab = Eab / $\Delta$Z

  • Eab = Ean - Ebn

State the equation for the conversion of a wye impedance to a delta impedance

Zy = $\frac{Z\Delta}{3}$

Convert a wye 3-phase to its equivalent line-to-neutral circuits

Describe the power of a three phase circuit

$$S_a = \underline{V}_{an} \underline{I}_a^*$$

$$= V_{an} \angle \delta \cdot I_a \angle -\beta$$

$$= V_{an} I_a \angle (\delta - \beta)$$

$$S_{3\phi} = S_a + S_b + S_c$$

$$= 3 S_\phi$$

$$= 3 V_{an} I_a \angle (\delta - \beta)$$

$$= 3 \underline{V}_{an} \underline{I}_a^*$$

$$= 3 V_{an} I_a \left[ \cos(\delta - \beta) + j \sin(\delta - \beta) \right]$$

$$= 3 V_{an} I_a \cos(\delta - \beta) + j 3 V_{an} I_a \sin(\delta - \beta)$$

$$P = \operatorname{Re}\{S_{3\phi}\} = 3 V_{an} I_a \cos(\delta - \beta)$$

$$Q = \operatorname{Im}\{S_{3\phi}\} = 3 V_{an} I_a \sin(\delta - \beta)$$

$$\vert{}S\vert{} = 3 V_{an} I_a$$

Describee the advantages of a balanced 3-phase system over a single phase system