6.002 Fall 2000 Lecture 1 2 6.002 CIRCUITS AND ELECTRONICS Basic Circuit Analysis Method (KVL and KCL method) 6.002 Fall 2000 Lecture 2 2 0= ∂ ∂ t B φ 0= ∂ ∂ t q Outside elements Inside elements Allows us to create the lumped circuit abstraction wires resistors sources Review Lumped Matter Discipline LMD: Constraints we impose on ourselves to simplify our analysis 6.002 Fall 2000 Lecture 3 2 LMD allows us to create the lumped circuit abstraction Lumped circuit element + - v i power consumed by element = vi Review 6.002 Fall 2000 Lecture 4 2 KVL: loop KCL: node 0= ∑ j j ν 0= ∑ j j i Review Review Maxwell’s equations simplify to algebraic KVL and KCL under LMD! 6.002 Fall 2000 Lecture 5 2 KVL 0=++ bcabca vvv 0=++ badaca iii KCL DEMO 1 R 2 R 4 R 5 R 3 R a b d c + – Review 6.002 Fall 2000 Lecture 6 2 Method 1: Basic KVL, KCL method of Circuit analysis Goal: Find all element v’s and i’s write element v-i relationships (from lumped circuit abstraction) write KCL for all nodes write KVL for all loops 1. 2. 3. lots of unknowns lots of equations lots of fun solve 6.002 Fall 2000 Lecture 7 2 Method 1: Basic KVL, KCL method of Circuit analysis For R, For voltage source, For current source, Element Relationships IRV = 0 VV = 0 II = 3 lumped circuit elements R 0 V o I + – J 6.002 Fall 2000 Lecture 8 2 KVL, KCL Example The Demo Circuit + – 1 R 2 R 4 R 5 R 3 R a b d c 00 V= ν + – 1 ν + – 5 ν + – 3 ν + – 2 ν + – 4 ν + – 6.002 Fall 2000 Lecture 9 2 Associated variables discipline ν i + - Element e Then power consumed by element e i ν = is positive Current is taken to be positive going into the positive voltage terminal 6.002 Fall 2000 Lecture 10 2 KVL, KCL Example The Demo Circuit + – 1 R 2 R 4 R 5 R 3 R a b d c 00 V= ν + – 1 ν + – 5 ν + – 3 ν + – 1 L 2 L 4 L 3L 2 ν + – 4 ν + – 2 i 1 i 0 i 5 i 3 i 4 i [...]... plus current source V0 R4 e2 R2 + V e1 – 0 R1 R 3 R5 I1 Gi = KCL at e1 (e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0 1 Ri KCL at l2 (e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0 move constant terms to RHS & collect unknowns e1 (G1 + G2 + G3 ) + e2 (−G3 ) = V0 (G1 ) e1 (−G3 ) + e2 (G3 + G4 + G5 ) = V0 (G4 ) + I1 2 equations, 2 unknowns (compare units) 6.002 Fall 2000 Lecture 2 Solve for e’s Step 4 17 In . 6.002 Fall 2000 Lecture 1 2 6.002 CIRCUITS AND ELECTRONICS Basic Circuit Analysis Method (KVL and KCL method) 6.002. e 0)()()( 152402312 =−+−+− IGeGVeGee KCL at 2 l move constant terms to RHS & collect unknowns )()()( 10323211 GVGeGGGe =−+++ 140543231 )()()( IGVGGGeGe