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Basic Circuit Analysis Methods (KVL and KCL method, Node method)

6.002x CIRCUITS AND ELECTRONICS

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Remember, our EECS playground

Review

Observe the lumped matter discipline LMD

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Lumped circuit element +

-‐

power consumed by element =

Review

i

v

vi

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4

LMD allows us to create the lumped circuit abstraction

Review

+!–! R1

R3 R2

V

5

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KVL: For all loops

KCL: For all nodes

0=∑ j jν

0=∑ j ji

Review Review

Maxwell’s equations simplify to algebraic KVL and KCL under LMD!

tdlE B

∂∂

−=⋅∫φ

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DEMO

+!–!

Review

R1 R4

R3

R2 R5

d

c

b

a

V0

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Let’s Begin by Building aToolchest of Analysis Techniques

+!–!

R1 R4

R3

R2 R5

Analyzing a circuit means:

Find all the element v’s and i’s

V0

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Method 1: Basic KVL, KCL method of Circuit analysis

Goal: Find all element v’s and i’s

1. write element v-i relationships (from lumped circuit abstraction) 2. write KCL for all nodes 3. write KVL for all loops

lots of unknowns lots of equations lots of fun solve

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i +

-

Element e

Current is taken to be positive going into the positive voltage terminal


Then power consumed by element e

is positive = vi

This convention is called: Associated variables discipline

Goal: Find all element v’s and i’s Labeling element v’s and i’s

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For R For voltage source For current source

You will need this for step 1: Element Relationships

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Let’s Apply KVL, KCL Method to this Example

The Demo Circuit

+!–!

R1 R4

R3

R2 R5

V0

Goal: Find all element v’s and i’s

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KVL, KCL Example

a

d

c

+!–!

Note the use of associated variables…

R4

R5 R2

R1

R3 b V0

Label all v’s and i’s Goal: Find all element v’s and i’s

5050 , ii ……νν12 unknowns

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Step 1 of KVL, KCL Method 12 unknowns

1. Element relationships ( )iv, 5id

c

1ν+ –

5ν+ –

+ – 2ν+ –

4ν+ – 0i

4i

+!–!

R4

R5 R2

R1

R3 b

v3 V0

i1

i3

i2 0ν+ –

a

5050 , ii ……ννL2

L3

L1

L4

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2. KCL at the nodes

(use convention, e.g., sum currents leaving the node)

5050 , ii ……νν5id

c

1ν+ –

5ν+ –

+ – 2ν+ –

4ν+ – 0i

4i

+!–!

R4

R5 R2

R1

R3 b

v3 V0

i1

i3

i2 0ν+ –

a

L2

L3

L1

L4


3. KVL for loops

(use convention, e.g., as you go around loop, assign first encountered sign to each voltage)

5050 , ii ……νν5id

c

1ν+ –

5ν+ –

+ – 2ν+ –

4ν+ – 0i

4i

+!–!

R4

R5 R2

R1

R3 b

v3 V0

i1

i3

i2 0ν+ –

a

L2

L3

L1

L4

KVL, KCL Method

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ugh @#!

1. Element v, i relationships v0 = V0

v1 = i1R1

v2 = i2R2

v3 = i3R3

v4 = i4R4

v5 = i5R5

2. KCL at the nodes

redundant

0410 =++ iii0132 =−+ iii0435 =−− iii0520 =−−− iii

a: b: d:

c:

3. KVL for loops

0431 =−+ vvv0210 =++− vvv

0253 =−+ vvv0540 =++− vvv redundant

L1:

L2:

L3: L4:

Method 3 – the node method will be much better!

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Other Analysis Methods Method 2— Apply element combination rules

A …R1 R2 R3 RN

B G2 G1 GN

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Method 2 — Apply element combination rules

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Method 2— Apply element combination rules

+!–!

Example

R1

R3 R2

V

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1. 2. 3. 4. 5.

Select reference node ( ground) from which voltages are measured. Label voltages of remaining nodes with respect to ground. These are the primary unknowns. Write KCL for all but the ground node, substituting device laws and KVL. Solve for node voltages. Back solve for branch voltages and currents (i.e., the secondary unknowns).

Particular application of KVL, KCL method

Method 3 — Node analysis

6.002x workhorse!

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Example: Old Faithful, plus current source

+!–!

R1 R4 R3

R2 R5

V0

Method 3 — Node analysis

1. Select reference ground node

2. Label node voltages with respect to ground.

g

I1

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Step 3 of Node Method

For convenience, write i

i RG 1

=

To avoid mistakes, use convention – E.g., always sum the currents leaving a node

g

+!–!

I1

e1

R4 R3 R1

R2 R5

e2 V0

V0

3. Write KCL for nodes, substituting device laws and KVL.

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Move constant terms to RHS & collect unknowns

2 equations, 2 unknowns Solve for e’s (compare units)

0)()()( 21321101 =+−+− GeGeeGVe

0)()()( 152402312 =−+−+− IGeGVeGee

KCL at e1

KCL at e2

4. Solve for node voltages

g

+!–!

I1

e1

R4 R3 R1

R2 R5

e2 V0

V0

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5. Back solve for branch voltages and currents

g

+!–!

I1

e1

R4 R3 R1

R2 R5

e2 V0

V0

e1 e2 Once you have solved for and , easy to find branch v’s and i’s

For example:

i1 + – v1

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In matrix form:

⎥⎦

⎤⎢⎣

⎡+

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡++−

−++

104

01

2

1

5433

3321

IVGVG

ee

GGGGGGGG

conductivity matrix

unknown node

voltages sources

Revisit Step 4 of Node Method for Cultural Interest )()()( 10323211 GVGeGGGe =−+++

140543231 )()()( IGVGGGeGe +=+++−

4. Solve for node voltages

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⎥⎦

⎤⎢⎣

⎡+

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡++−

−++

104

01

2

1

5433

3321

IVGVG

ee

GGGGGGGG

( )( ) 23543321

104

01

3213

3543

2

1

GGGGGGGIVG

VGGGGG

GGGG

ee

−++++

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡++

++

=⎥⎦

⎤⎢⎣

⎡

Solve

( )( ) ( )( )5G3G4G3G

23G5G2G4G2G3G2G5G1G4G1G3G1G1I0V4G3G0V1G5G4G3G

1e ++++++++

++++=

( )( ) ( )( )5343

23524232514131

1043210132 GGGGGGGGGGGGGGGGG

IVGGGGVGGe++++++++

++++=

(same denominator)

Notice: linear in , , no negatives in denominator – we will use this later

V0 I1

Step 4 of Node Method 4. Solve for node voltages

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E.g., solve for , given

K2.81

GG

5

1 =⎭⎬⎫

K9.31

GG

4

2 =⎭⎬⎫

K5.11G3 = 01 =I

02 6.0 Ve =

If , then VV 30 = 02 8.1 Ve =


g

+!–!

I1

e1

R4 R3 R1

R2 R5

e2 V0

V0

e2

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