begin notes on ac
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itags: [ mmme2051 ]
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itags: [ mmme2051, electronics, electromechanical_devices ]
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								uni/mmme/2051_electromechanical_devices/ac_intro.md
									
									
									
									
									
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---
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author: Akbar Rahman
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date: \today
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title: MMME2051 // Introduction to Alternating Current (AC)
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tags: [ alternating_current, ac ]
 | 
			
		||||
uuid: 0c90c691-cbf8-43e9-bfa5-7b277c853151
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lecture_slides: [ ./lecture_slides/MMME2051EMD_Lecture2B.pdf, ./lecture_slides/MMME2051EMD_Lecture3A.pdf, ./lecture_slides/MMME2051EMD_Lecture3B.pdf ]
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---
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This section builds on [complex numbers](/permalink?uuid=b9be8780-1ab7-402f-9c67-8cc74a74f7a9).
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# Sinusoidal Waves
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This module will be using the following format:
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$$y(t) = A\cos(\omega t + \Phi)$$
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where $A$ is amplitude, $\omega$ is frequency, $\omega t + \Phi$ is phase angle, and $\Phi$ is phase
 | 
			
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offset.
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[Explore the effects of changing the variables in Desmos](https://www.desmos.com/calculator/dmzytwau2y)
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# Phasor
 | 
			
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- a phasor is a complex number that represents the initial position of a rotating vector
 | 
			
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- use the amplitude ($|V|$) and phase offset ($\Phi$) of a cosine function
 | 
			
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- for all AC steady state analysis ($\omega$ is constant), these two variables are the only two needed
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#### Example
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For voltage $v$ given by
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$$v = 150 \cos (50t + 25)$$
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it may be represented in the phasor form
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$$150 \angle 25$$
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#### Example
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For current $i$ given by
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$$i = 10 \cos \left(50t -\frac{pi}{6}\right)$$
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it may be represented in the phasor form
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$$10 \angle \frac{pi}{6}$$
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## Phasors in Resistive Circuits
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Convert all variables to phasors or to complex form
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Apply KCL, KVL, Ohm's Law
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\begin{align*}
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v &= iR \\
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V\angle \Phi = IR \angle\theta \\
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I \angle \theta = \frac VR \angle \Phi
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\end{align*}
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## Phasors in Inductive Circuit
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Ohm's law generalised to incorporate complex resistance, reactance, $X$:
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\begin{align*}
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v &= iX \\
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V\angle\Phi_v &= i\angle\Phi_iX \\
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&= i\angle\Phi_ij\omega L \\
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\frac{V}{j\omega L}\angle\Phi_v &= I\angle\Phi_i
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# Power
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## Resistive Circuits
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$$P_\text{avg} = V_\text{rms}I_\text{rms}$$
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## Inductive Circuits
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$$P = \frac{V^2}{2\omega L}\sin{2\omega t}$$
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## Capacitive Circuits
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$$P = \frac{\omega CV^2}{2}\sin{2\omega t}$$
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## Real Circuit (Resistive + Reactive)
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$$P = V_\text{RMS}I_\text{RMS}(\cos \gamma + \cos{(2\omega t + \gamma)}$$
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$$P_\text{avg} = V_\text{RMS}I_\text{RMS}\cos \gamma$$
 | 
			
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where $\cos \gamma$ is the power factor (PF) and $\gamma$ is phase deviation between voltage and current.
 | 
			
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The PF tells us what fraction of the current does useful work.
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## Apparent, Active, and Reactive Power
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Apparent Power:
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$$S = V_\text{RMS}I_\text{RMS}$$
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- as power still flows losses still occur
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- AC equipment is rated for apparent power as it handles both used and unused power
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Active Power:
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$$P = S\cos\gamma$$
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- this is the real power transferred to the load
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Reactive Power:
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$$P = S\sin\gamma$$
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A
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# Resonance
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The inductive load of on a circuit is $Z_C = \frac{1}{j\omega L}$.
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If the frequency of the power supply matches $\omega$, you get resonance and the circuit becomes
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purely resistive so there is a sharp drop in impedance.
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---
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author: Akbar Rahman
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date: \today
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title: MMME2051 // AC Power
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tags: [ ac, alternating_current, power ]
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uuid: c269b4b7-7835-4b50-8d4f-ff5bc63a8a3d
 | 
			
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lecture_slides: [ ./lecture_slides/MMME2051EMD_Lecture3B.pdf ]
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exercise_sheets: [ ./exercise_sheets/Exercise Sheet 4 - Power factor and three phase.pdf ]
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---
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# Definitions
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- Phase voltage - voltage across any phase
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- Line voltage - voltage between two live lines
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- Phase current - current through any phase
 | 
			
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- Line current - current through any live line
 | 
			
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 | 
			
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# Three-Phase Load
 | 
			
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$$P = \sqrt{3} V_lI_l\cos\gamma$$
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 | 
			
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- 3-phase devices (source and load) are usually balanced, meaning that the impedance in each
 | 
			
		||||
  phase is equal ($Z_1 = Z_2 = Z_3$).
 | 
			
		||||
- For loads, this means that the voltage across them are the same, apart from the phase angles:
 | 
			
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 | 
			
		||||
  \begin{align*}
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  v_{1N} = V\cos{2\pi ft} \\
 | 
			
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  v_{2N} = V\cos{2\pi ft - \frac{2\pi}{3} \\
 | 
			
		||||
  v_{3N} = V\cos{2\pi ft + \frac{2\pi}{3}
 | 
			
		||||
  \end{align*}
 | 
			
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 | 
			
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- Balanced loads and sources ensure that line/phase currents have equal magnitudes and that the
 | 
			
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  neutral current is 0
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## Star Load
 | 
			
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$$|V_\text{line}| = \sqrt 3 |V_\text{phase}|$$
 | 
			
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 | 
			
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$$I_\text{line} = I_\text{phase}$$
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 | 
			
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## Delta Load
 | 
			
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 | 
			
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 | 
			
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$$|V_\text{line}| = |V_\text{phase}|$$
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 | 
			
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$$I_\text{line} = \sqrt 3 I_\text{phase}$$
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								uni/mmme/2051_electromechanical_devices/basic_circuitry.md
									
									
									
									
									
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---
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author: Akbar Rahman
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date: \today
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title: MMME2051 // Basic Circuitry
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tags: []
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uuid: 6767b7f0-705a-43a1-9e02-aeee6b454538
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---
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# Symbols & Notations Used in Circuit Diagrams
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# Series Circuits
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Summary of this section in tabular form found in slides p30.
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## Resistors
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\begin{align*}
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I &= I_1 = I_2 = I_3 \\
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V &= V_1 + V_2 + V_3 \\
 | 
			
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R &= R_1 + R_2 + R_3
 | 
			
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\end{align*}
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## Inductors
 | 
			
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 | 
			
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More inductors in series makes is harder for current to change rapidly.
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 | 
			
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 | 
			
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\begin{align*}
 | 
			
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I &= I_1 = I_2 = I_3 \\
 | 
			
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V &= V_1 + V_2 + V_3 \\
 | 
			
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L &= L_1 + L_2 + L_3
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\end{align*}
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## Capacitors
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More capacitors in series makes is easier for voltage to change rapidly.
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 | 
			
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\begin{align*}
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I &= I_1 = I_2 = I_3 \\
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V &= V_1 + V_2 + V_3 \\
 | 
			
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\frac 1C &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
 | 
			
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\end{align*}
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 | 
			
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# Parallel Circuits
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## Resistors
 | 
			
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\begin{align*}
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I &= I_1 + I_2 + I_3 \\
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V &= V_1 = V_2 = V_3 \\
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\frac 1R &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
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\end{align*}
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## Inductors
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More inductors in parallel makes is easier for current to change rapidly.
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\begin{align*}
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I &= I_1 + I_2 + I_3 \\
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V &= V_1 = V_2 = V_3 \\
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\frac 1L &= \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}
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\end{align*}
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## Capacitors
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More capacitors in parallel makes is harder for voltage to change rapidly.
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\begin{align*}
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I &= I_1 + I_2 + I_3 \\
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V &= V_1 = V_2 = V_3 \\
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C &= C_1 + C_2 + C_3
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\end{align*}
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@@ -32,6 +32,10 @@ where $V$ is voltage across a component, $I$ is current through it, and $R$ is r
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- Impedance is used when there are energy storage elements to a component.
 | 
			
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- Resistance, a special case of impedance, can be used when there is no storage element
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 | 
			
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## Admittance
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 | 
			
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$$Y \frac1Z$$
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# Kirchhoff's Laws
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## Current
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@@ -84,3 +88,11 @@ Capacitors try to minimize changes in voltage.
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If a capacitor is shorted, the current through the connecting wires will be extremely high, causing
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the wires to heat up.
 | 
			
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# Root Mean Square (RMS)
 | 
			
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$$x_{\text{RMS}} = \sqrt{\frac{x_1^2 + \dots + x_n^2}{n}}$$
 | 
			
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For a sinusoidal wave:
 | 
			
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$$x_\text{RMS} = \frac{A}{\sqrt2}$$
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								uni/mmme/2051_electromechanical_devices/kirchhoff.md
									
									
									
									
									
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---
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author: Akbar Rahman
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date: \today
 | 
			
		||||
title: MMME2051 // Kirchhoff's Current Law, Voltage Law
 | 
			
		||||
tags: [ kirchhoff, kcl, kvl ]
 | 
			
		||||
uuid: 88e2eb6a-7f6a-4ea0-9850-81305028e7b5
 | 
			
		||||
lecture_slides: ./lecture_slides/MMME2051EMD_Lecture2A.pdf
 | 
			
		||||
---
 | 
			
		||||
 | 
			
		||||
# Application of Kirchhoff's Current/Voltage Laws (KCL, KVL)
 | 
			
		||||
 | 
			
		||||
(lecture slides 14-21)
 | 
			
		||||
 | 
			
		||||

 | 
			
		||||
 | 
			
		||||
1. Identify all the loops in the circuit and assign each loop a "loop current" variable:
 | 
			
		||||
 | 
			
		||||
 | 
			
		||||
   
 | 
			
		||||
 | 
			
		||||
1. Identify "branch current" values (apply KCL)
 | 
			
		||||
 | 
			
		||||
 | 
			
		||||
   
 | 
			
		||||
 | 
			
		||||
1. Apply KVL to each loop:
 | 
			
		||||
 | 
			
		||||
   Loop 1: $10 - 2 - V_1 - V_2 = 0$  
 | 
			
		||||
   Loop 2: $V_2 - V_4 = 0$  
 | 
			
		||||
   Loop 3: $V_4 - V_3 - V_5 = 0$
 | 
			
		||||
 | 
			
		||||
   
 | 
			
		||||
 | 
			
		||||
1. Apply Ohm's Law to KVL
 | 
			
		||||
 | 
			
		||||
   Loop 1 (origin at node A):
 | 
			
		||||
 | 
			
		||||
   \begin{align*}
 | 
			
		||||
   0 &= 10 - 2 - V_1 - V_2 \\
 | 
			
		||||
     &= 8 - I_1R_1 - (I_1-I_2)R_2 = 0 \\
 | 
			
		||||
   8 &= I_1(R_1+R+2) - I_2R_2 \\
 | 
			
		||||
     &= 6I_1 - I_2
 | 
			
		||||
   \end{align*}
 | 
			
		||||
 | 
			
		||||
   Loop 2 (origin at node B):
 | 
			
		||||
 | 
			
		||||
   \begin{align*}
 | 
			
		||||
   0 &= V_2 - v_4 \\
 | 
			
		||||
     &= (I_1-I_2)R_2 - (I_2-I_3)R_4 \\
 | 
			
		||||
     &= I_1 - 3I_2 + 2I_3
 | 
			
		||||
   \end{align*}
 | 
			
		||||
 | 
			
		||||
   Loop 3 (origin at node C):
 | 
			
		||||
 | 
			
		||||
   \begin{align*}
 | 
			
		||||
   0 &= V_4 - V_3 - V_5 \\
 | 
			
		||||
     &= (I_2-I_3)R_4 - I_3R_3 - I_3R_5 \\
 | 
			
		||||
     &= 2I_2 - 5I_3
 | 
			
		||||
   \end{align*}
 | 
			
		||||