Circuit-Elements - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

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Circuit-Elements, KCL & KVL, Ohm’s Law Docsity.com Circuit Elements • Passive Elements – R, C, L  Active, Independent Sources • (a) ≡ V-source • (b) ≡ Battery – Basically a V-Source with a small series resistance • (c) ≡ I-Source Docsity.com Power Example • Determine Power Absorbed Or Supplied By Each Element ( )( ) [ ] ( )( ) [ ] ( )( ) [ ] ( )( ) [ ]WAVP WAAP WAVP WAVP WAVP DS 144436 8241 56228 48224 ][48)4)(12( 36 3 2 1 −=−= −=×Ω−= == == ==  Note That • Power-Supplied = Power-Absorbed – In both Cases = 152W Docsity.com Charge Conservation • One Of The Fundamental Conservation Principles In Electrical Engineering: “CHARGE CANNOT BE CREATED NOR DESTROYED” Docsity.com Node, Loops, Branches • NODE: Point Where Two, Or More, Elements Are Joined (e.g., Big Node 1)  LOOP: A Closed Path That Never Goes Twice Over A Node (e.g., The Blue Line) • The red path is NOT a loop (2x on Node 1)  BRANCH: a Component Connected Between Two Nodes (e.g., R4 Branch) Docsity.com KCL Algebra • Two Equivalent KCL Statements – Algebraic Sum Of Currents (Flowing) OUT Of A Node Is ZERO – Algebraic Sum Of Currents Flowing INTO to A Node Is ZERO • Example: Use Sign Convention ( ) ( ) ( ) ( ) ( ) ( ) 0 0 54321 =+−−− =∑ − tititititi ti NodeINto Docsity.com KCL Problem Solving • KCL Can Be Used To Find A Missing Current – ∑(Currents INto Node-a) = 0 A 5 A 3 ? = X I a b c d ( ) AIAAI XX 2or035 −==−++ Which Way are Charges Flowing in Branch a-b? b a c d e 2A -3 A 4A I be = ? • Iab = 2A • Icb = −3A • Ibd = 4A • Ibe = ? • Nodes = a,b,c,e,d • Branches = a-b, c-b, d-b, e-b ( ) AIAAAI bebe 5or0432 −==−−++−  Notation Practice Docsity.com UnTangling • A node is a point of connection of two or more circuit elements, and Region of Constant Potential • It may be stretched-out or compressed or Twisted or Turned for visual purposes…But it is still a node Equivalent Circuits Docsity.com Example • Find Currents  – Use +OUT 1 2 3 4  The Presence of the Dependent Source Does NOT Affect KCL • KCL Depends Only On The Topology  Again, Node-4 Eqn is (Linearly) Dependent on the Other 3 Docsity.com KCL Convention: In = Out • An Equivalent Algebraic Statement of Charge Conservation ∑∑ = Node of OUT CurrentsNode INTO Currents 0501 =+ mAI mAmAmAIT 204010 ++= 1 Find I TI Find Docsity.com Examples: In = Out mAI mAmAI 6 410 1 1 = −= mAImA 412 1 +=mAII 321 += 1I Find 21 I and I Find mAImA 410 1 += mAmAmAmAII mAmAmAI 5383 8412 12 1 =−=−= =−= Docsity.com Energy Conservation • One Of The Fundamental Conservation Laws In Electrical Engineering is Kirchoff’s Voltage Law: THE CONSERVATION OF ENERGY PRINCIPLE: “ENERGY CANNOT BE CREATED NOR DESTROYED” Docsity.com Kirchoff’s Voltage Law (KVL) • KVL Is A Conservation Of Energy Principle – A Positive Charge Gains Energy As It Moves to a Point With Higher Voltage and Releases Energy If It Moves to a Point With Lower Voltage + AV B BV)( AB VVqW −=∆ q −+ abV a b + q abqVW =∆ LOSES +− cdV c d + q cdqVW =∆ GAINS Docsity.com "Gedanken" Experiment • If The Charge Comes Back To The Same Initial Point, Then The Net Energy Gain Must Be Zero (Conservative Network) – Otherwise, With Repeated Loops The Charge Could End Up With Infinite Energy, Or Supply An Infinite Amount Of Energy. Mathematically + AV B BV q CV − + ABV − + BC V −+ CAV 0)( =++ CABCAB VVVq Docsity.com KVL Problem Solving • KVL Is Useful To Determine Voltages – Find A Loop Including The Unknown Voltage • The Loop Does NOT Have To Be Physical – Example: VR1, VR3 Are Known – VR2, is UNknown – Determine Vbe − + beV 3131 ][300][30 RRbeRbeR VVVVVVVV −−=⇒=+++−  Using the “Virtual” Loop Shown Docsity.com Linear Dependence • Background: For KCL We Saw That Not All Possible KCL Equations Are Independent – The Same Situation Arises When Using KVL • The Third Equation Is The Sum Of The Other Two Docsity.com Dependence Quantified • For a Given Circuit Define – N ≡ Number of Nodes – B ≡ Number of Braches • Then – N−1 ≡ No. of Linearly Independent KCL Eqns – B−(N − 1) ≡ No. of Linearly Independent KVL Eqns • Example: For Previous Circuit We Have N = 6, B = 7. – Hence There Are Only Two Independent KVL Equations Docsity.com KVL Examples VV VVV ac ac 10 064 = =−− VV VVV bd bd 6 042 = =−− 1 2 1 2 VV VVVV ad ad 26 06812 = =−−− VV VVVV eb eb 10 01264 = =−+− 1 2 1 2 Docsity.com More Examples • Find Vbd – First Find VR1 by KVL ( ) VVVVV RRRR 1112101010112 1111 =⇒−=+⇒=+++− − + V +− V  Remember: ALWAYS Use The Passive Sign Convention With Resistors ( ) VVVVVVVVV RRbdbdRR 11111010100 2121 =+=+=⇒+−−= Docsity.com Virtual Loop Example • Find – Vx, Vab – Power Dissipated in the 2 kΩ Resistor • Need To Find A Closed Path With Only One Unknown Potential + - + - a b−+ V4 −+ xV − + 1V − + 2V Ω= kR 2 VVVV 4,1 2 21 == VV VVVV VVVV X X X 4 04124 0412 = =+−+ =+−+ VVVV VVV VVV ab Xab Xab 844 0 2 2 −=−−= −−= =++ 1 2 1 2 ( ) mW k V R VP kk 82 4 222 2 =Ω == ΩΩ Docsity.com Resistors cont. • Most Common Multiples of Resistance – kΩ = 103 Ω – MΩ = 106 Ω • Most Common Practical Occurrence kΩofunits⇒= mA VoltRtypical  Other V, I, R Combinations Potential Current Resistance Volts Amps Ohms (Ω) Volts mA kΩ mV Amps mΩ mV mA Ω Docsity.com Conductance = 1/Resistance • Recast Ohm’s Law to Find CURRENT as a Function of VOLTAGE: Gviv R i R vi =⇒=⇒= 1  G is Defined As CONDUCTANCE • Has Units of Amps/Volt Called Siemens Werner Siemens (1816-1892) German electrical engineer Docsity.com Physical Realization (4) SE — Ai_ — a ae (6) Circuit — z Symbol 8) m ee _. (9) (10) (12) Docsity.com Ohm’s Law Problem Solving • Recall Ohm’s Law: GviRiv == or − = + ][10 VV I = 2[A] R = 5[Ω] V = IR + V = 20[V] - I = 4[A] ][5 Ω=R R = V/I • Given Any 2 Values, Can Find the 3rd – Given Current & Resistance → Find Voltage – Given Current & Voltage → Find Resistance Docsity.com Problem Solving cont. • Given Voltage and Resistance → Find Current I = V/R R = 4[Ω] + V = 12[V] - ][3 AI =  Note The Use of the Passive Sign Convention • In a Resistor, Current ALWAYS Flows Vhi→Vlo – Use VOLTAGE POLARITY Reference Docsity.com Example  Conductance • Find i(t) Using Conductance – Ohm’s Law using G ( ) ( ) ( ) [ ] [ ] ( ) [ ] ( ) [ ]Ati V V Ati VSti tGvti 8 42 42 = ×   = ×= = Docsity.com Solving Power Relations • Given: P, i 2then i P i vR i Pv === R vviP R vi 2 then === 2then RiviPRiv === & then2 iRv R PiPRi ===  Given: v, R  Given: i, R  Given: P, R  Given: P, v v Pi RPvR P v === &then 2 Docsity.com Power & Passive-Convention • If Not Specified, The Ref. Direction For V ExOR I Can Be Chosen, Then the Other Qty Given By The Passive Sign Convention • Units: Beware the Multiples – V, A, W, Ω are SI units; mA, MΩ are NOT! • Prefixes Can Result in Exponent Errors – Strategy: Eliminate Prefixes for Complex Calculations » Example: R = 40 kΩ, I = 2 mA [ ] [ ]mWWARiP RiP 16010160)102(*)1040( Watts?1602*40 32332 22 =×=×Ω×== =≠= −− Docsity.com Example: i, v, P, R relations • Given HdLamp Ckt  Find • Lamp Current, I • Lamp Resistance, R • Charge Supplied to Lamp in 1 minute, Q1  Use P&v Pwr Rlns  Thus ( ) ( ) ( ) ( ) AI VVAI R VAVR 5 1260 4.2 6012 2 = = Ω= = + - − + V 12 HALOGEN LAMP W P 60 = R vPviP 2 and ==  And CoulQ SecAtIQ 300 605 = ⋅=⋅= Docsity.com All Done for Today Ohm’s “Law Book”  Die galvanische Kette : mathematisch bearbeitet (The Galvanic Circuit Investigated Mathematically), Berlin, Riemann, © 1827 Docsity.com WhiteBoard Work • Let’s Work This Problem 12 mA 3 mA 2 mA 4 mA Ix Iy Iz  Find zyx III Docsity.com WhiteBoard Work • Let’s Work Problem This Nice Little Problem 30 kΩ 12 V 2 Vx 10 kΩ + Vx _  Find Vx & P30kΩ Docsity.com Brice Mayer, PE PP & 2 & / é Zo Engineering instructor Chabot College 25555 Hesperian Bivd Hayward, CA 94545 eMail: bmayer@chabotcollege.' Docsity.com FIND Nox 3a -OVt o> A . -th ] ela A fe 2% GN _ r a + Vs B ” ev [| epev ¥ |6V 4 " 7 A "yea 4 FIND POWER BALARCE O544eG PASSIVE S/GNM CONVENTIONS "A SSUANE VOLTAGE REF —~SETS COKE ENT DIRETIOWNM BY PASSINGCE 8fGAL CONVYEMTIONG ReEecAte POWELL EOr- [ se & +V- P= YE P-Pesi7iVE => POR DISSIPATED P-NEG SD PUR en APB) PWR ACCOUNTIVG EXCEPT fer Vs Docsity.com ZT™M P= Vaex Tae PO) of 2Vx 2A +aw z gv x GA + 2eW 3 EN x £A az x GV *€A 24 GY PS GV x (-2A) —/2 2 -TOTAK Paw Now Pee Mos7 BALANCE PR, + 72Ww SOP R= -7eWw er Vs x (“¥A) +72VA =O Vs = —- 22ZVA — YA Vs SHE VY wo | OF DV2w Example • Use Power Balance To Compute I0 ( )[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] AV VAI IVWW IVWW PP abs 16 6 6176182 6176323010812 0 0 0 sup == ×+= ×+=+++ =∑∑  Thus a 1A Current Flows in the Direction Assumed in the Figure Docsity.com