ee40 lec 3 basic circuit analysis prof. nathan cheungee40/fa09/lectures/lec_03.pdf · ee40 lec 3...

EE40

Lec 3

Basic Circuit Analysis

Prof. Nathan Cheung09/03/200909/03/2009

Reading: Hambley Chapter 2

Slide 1EE40 Fall 2009 Prof. Cheung

Chapter 2

• Outline– Resistors in Series – Voltage Divider– Conductances in Parallel – Current Divider– Node-Voltage Analysisg y– Mesh-Current Analysis– SuperpositionSuperposition– Thévenin equivalent circuits– Norton equivalent circuitsNorton equivalent circuits– Maximum Power Transfer


Voltage Divider


Resistors in Series

• Resistors in seriesResistors in series can be combined into one equivalent resistance:


Current Divider


Resistors in Parallel

• Resistors in parallelResistors in parallel can be combined into one equivalent resistance:

Conductance G = 1/( Resistance R)


[ unit = Siemens = 1/ohm]

Example Calculation with Series and Parallel Resistance

D t i V I I d I• Determine V2, I1, I2, and I3:

• From (c):( )

and A2210

24I1 =+

=

• Returning to (b):V422I2V 12 =×==

4V A33.134

3VI 2

2 ===

A67.04VI 23 ===


A67.066

I3

Series and Parallel Combinations in general

Some circuits must be analyzed (not amenable to simple inspection)R1

R1R2

-+ R2

V

IR3

1

−

+

R R

V R3

R4

R4 R5

R5

Special cases:R3 = 0 OR R3 = ∞


Some Interesting Series and Parallel Combinations

All R=1Ω, what is Rab ?

See Hambley P.2.15 What is Rab ?Hambley P2.9[Hint: see bottom figure]


[Hint: see bottom figure]Ans. Rab = 5/6 Ω

Measuring VoltageTo measure the voltage drop across an element in a real circuit, insert a voltmeter (digital multimeter in voltage mode) in parallel with the elementvoltage mode) in parallel with the element.

Voltmeters are characterized by their “voltmeter input resistance” (Rin). Ideally, this should be very high ( in) y, y g(typical value 10 MΩ)

Ideal Voltmeter

Rin


Effect of Voltmeterundisturbed circuit circuit with voltmeter inserted

VSS

R1

R2_+

+V2 VSS

R1

R2 R_++V2′R2

–V2 SS R2 Rin

–V2

Compare to R2

+

=21

2SS2 RR

RVV

+

=′1in2

in2SS2 RR||R

R||RVV

Compare to R2

21 1in2 ||

Example: V1VK900R,K100R,V10V 212SS =⇒===


210 , ?inR M V ′= =

Measuring CurrentTo measure the current flowing through an element in a real circuit, insert an ammeter (digital multimeter in current mode) in series with the elementcurrent mode) in series with the element.

Ammeters are characterized by their “ammeter input resistance” (Rin). Ideally, this should be very low (typicalresistance (Rin). Ideally, this should be very low (typical value 1Ω).

Ideal Ammeter

Rin


Effect of Ammeter

Icircuit with ammeter insertedundisturbed circuit

Measurement error due to non-zero input resistance:

R

Imeas

R1

ammeterI

R1Rin

V1R2

_+V1

R2

_+

1VI = 1meas

VI =

Example: V1 = 1 V, R1= R2 = 500 Ω, Rin = 1Ω

21 RRI

+ in21meas RRR ++

1VCompare to


1 1 , ?500 500 meas

VI mA I= = =Ω+ Ω

R2 + R2

Source Combinations

• Voltage sources in • Current sources inVoltage sources in series can be replaced by an

Current sources in parallel can be replaced by an

equivalent voltage source:

equivalent current source:

–+

v1≡

+

+ i i ≡ i +i

–+

v2≡ –

+

v1+v2 i1 i2 ≡ i1+i2


Example


Node-Voltage Circuit Analysis1 Id tif ll t di d l t f1. Identify all extraordinary nodes, select one of

them as a reference node (ground), and then assign node voltages to the remaining (nex – 1) g g g ( ex )extraordinary nodes.

(Look for the one with the most connections)

2. At each of the (nex – 1) extraordinary nodes, apply the form of KCL requiring the sum of allapply the form of KCL requiring the sum of all currents leaving a node to be zero.

3. Solve the (nex – 1) independent simultaneous ( ex ) pequations to determine the unknown node voltages.


Nodal Analysis Example

Find the power dissipated in R5:



Step 1:



Step 2:• Node 1:

I1 + I2 + I3 = 0

Rewrite currents with node voltages:

−− VVVVV 04

21

32

01

1

1 =−

++−

+RVV

RRVV

RV



• Node 2:I4 + I5 + I6 = 04 5 6

0320

12 =−

+−−

RVVI

RVV

64 RR

• Node 3:I7 + I8 + I9 = 07 8 9

00233 =+

−+ IVVV


065 RR


Step 3: Solve three simultaneous equations (Appendix B):

1111 V

32

02

41

4321

1111RR

VVR

VRRRR +

=

−

+

++

111 V

06

32

641

4

111 IRVV

RRV

R=−

++

−

0365

26

111 IVRR

VR

−=

++

−


Solve for I7 = V3/R5; P = I72/R5

Node-Voltage Analysis: Dependent Sources

• Treat as independent source in organizing and writing node equations, but include another equation that expresses the relationship of the dependent source.


Nodal Analysis with Dependent Sources

35 VVVVNode 1:

0634

3.5 2111 =−

++− VVVV

0126

212 =−+−

XIVVV

Node 2:

126

Dependant Source:( )( )

3622 2121 VVVVII X

−=

−==


Nodal Analysis with Dependent Sources

35 VVVVNode 1:

0634

3.5 2111 =−

++− VVVV

0126

212 =−+−

XIVVV

Node 2:

Sub in Ix for Node equations: 126

Dependant Source:( )

9 V1 – 2 V2 = 15.9

6 V + 7 V 0 ( )36

22 2121 VVVVII X−

=−

==– 6 V1 + 7 V2 = 0

V1 = 2.18 V and V2 = 1.87 V


Hence, Ix = 0.1 A

Nodal Analysis: “Supernodes”

• To deal with “floating” voltage source (neither side is connected to the reference node) we use supernodes:

Definition: A supernode is the combination of two extraordinary nodes


Definition: A supernode is the combination of two extraordinary nodes (excluding the reference node) between which a voltage source exists.

Nodal Analysis: “Supernodes”

• To deal with “floating”To deal with floating voltage source (neither side is connected to the reference node) we

duse supernodes:

T i• Two equations:– KCL for supernode

A ili ti f


– Auxiliary equation for voltages (KVL)

Supernode Example

• KCL @ supernode: • KVL @ supernode:0+++ IIII 18VV04321 =+++ IIII 1812 =−VV

02842

4 211 =−++− VVV

326 21 =+VV

842


Solution: V1 = 2 V; V2 = 20 V

Example 1: Use of both KCL and KVL

• Find I1 and I2:


Suggested Exercise: Solve the node voltages first

Example 2: Use of both KCL and KVL

• Determine Ix:

Solving these equations leads to I = 0.84 A.


Solving these equations leads to Ix 0.84 A.

ee40 lec 3 basic circuit analysis prof. nathan cheungee40/fa09/lectures/lec_03.pdf · ee40 lec 3...

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