Dc Ckt Analysis Ppt2

7/28/2019 Dc Ckt Analysis Ppt2

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& 7 Choi ( ( & 6 6 S U L Q J Lecture

CIRCUIT ELEMENTS AND CIRCUIT ANALYSIS

Lecture 2 review:

• Terminology: Nodes and branches

• Introduce the implicit reference (ground) node – definesnode voltages• Introduce fundamental circuit laws: Kirchhoff’s Current

and Voltage Laws

Today:• Basic circuit elements: Ideal voltage sources and current

sources; linear resistors• Resistor formulas• Resistors in series and the parallel• Formal nodal analysis

Choi ( ( & 6 6 S U L Q J Lecture

ELECTRIC CIRCUITSCircuit: Interconnection of electrical elements; modeled with two (or more)terminals Terminology: Nodes and Branches

There are also 3-terminal elements:

% U D Q F K

1 R G H

9 (Branches)

1 R G H V

CIRCUIT TERMINOLOGY

Labels:Nodes: connection points of two or more circuit elements(together with “wires”)Branches: one two-terminal element and the nodes at either end

Idealizations:Wires are “perfect conductors” – voltage is the same at any pointon the wire (add capacitor circuit element to model thisphenomenon)Is this reasonable? Yes, in many cases.What are the limits?

“long” wires” ⇒ voltage propagates at nearly c = 3 × 10 8 m/s

Circuit Elements of DC Circuits

Ideal Voltmeter and Ideal Ammeter

• Ideal voltmeter is a device which measures thevoltage across its terminals while drawing nocurrent. The current thru an ideal voltmeter is zero.

• Power = Voltage X Current = ?

• Ideal ammeter is a device which measures thecurrent going thru it while maintains the voltageacross its terminal to be zero.

• Power = Voltage X Current = ?

CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE &CURRENT SOURCES

Describe a two-terminal circuit element by plotting current vs. voltage

U H O H D V L Q J S R Z H U

D E V R U E L Q J S R Z H U

U H O H D V L Q J S R Z H U

D E V R U E L Q J S R Z H U

F K D U J L Q J

: H D V V X P H

: 3 D V V R F L D W H G V L J Q V V R L

F R P H V R X W R I

W H U P L Q D O

Ideal voltage source

Ideal current source

: H D V V X P H

: 3 D V V R F L D W H G V L J Q V V R L

F R P H V R X W R I W H U P L Q D O

Ideal/Independent Voltage Sources

(a) + terminal is higher than - terminal by V volt.(b) v s(t) is time varying(c) The voltage vary sinosoidally (sine wave) with amplitude of

How (not) to jump start a car?

Good battery-

+Dead battery

Good battery-

+Dead battery

Example• Find the potential difference V AB.

Remember the “Chain-rule”?

VAB = VAC+VCD+VDB = (VA-VC) + (V C-VD) + (V D-VB)or = -6v + 4v + (-3v) = -5v.VAB = VAD+VDBIf CE is any circuit element, will it affect the result?

Ideal Current Sources• Ideal current source (similar to voltagesource)

27 milliampere is goingin the direction as shown

A sinusoidal current sourcewith a magnitude of 16milliampere is going in thedirection as shown

Current Source Example

• Find the current I.

ABy itself, the 10V voltagesource does not affectthe current.

Apply KCL at node A:Sum of all current into nodeA equal to zero.

I + 3A = 0I = -3A

Exercise

• Find the current I. A

By itself, the 10V voltage sourcedoes not affect the current.

At node A, apply KCL:Sum of all the current going intonode A:-3mA -5mA + I =0I = 8mA.

Ideal Resistance

• Ohm’s law stated that:IA→B = (VA-VB)/ R

Examples• Terminal A is at potential -3v with respect to ground,

and terminal B is at -2v. The resistance is 10 Ω. What

is the current?• Apply Ohm’s law:

IA→B = (VA-VB)/ R = (-3 - (-2))/(10) = -0.1A.

• Suppose the current IA→B is known to be positive, doesthe voltage drop from A to B positive or negative?

If IA→B > 0, then V A - VB >0 => V A > VB

Examples (cont.)

• Suppose the “actual direction” of current is from B to Awith a magnitude of 10A, the resistant is 3 Ω. What is V AB?

Remember the direction of potential drop is also the directionof the current flow.

By Ohm’s law: I B→A = (VB-VA)/ R => (VB-VA)= VBA = R I B→A => VAB = -VBA = - R I B→A = - 3 (10) = -30v.

Example• Reference for the voltage V 1 and current I 1 have beenchosen as indicated. Measurements show that the valueof I1 is -200A. If R 1=3Ω, what is V 1?

I1= (VA-VB)/R 1Notice the direction of I 1 andVAB(=V1) is consistent .V1=I1R1 = (-200)(3) = -600v

Example

• R 1=3000 Ω, R 2=4700 Ω, and R 3=3300 Ω. From measurement

VA=-2v, V B=3v, V C=-3v, V D=12.735V (w.r.t.). Find I 1, I2,and I 3.

I1= (VB-VA)/R 1=(3-(-2))/3000 = 1.666mA

I2= (VC-VB)/R 2=(-3-3)/4700= -1.28mA

I3= (VB-VD)/R 3

=(3-12.735)/3300=-2.95mA.

I-V Graph for Ideal Resistance

• For an ideal resistor,the slope (1/R) is constant.

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$ Q V Z H U 9 ⇒ ,

9 , 5 ⇒ , P $

Z K H Q 9 9

IV CHARACTERISTICSOF A RESISTOR

If we use associated current andvoltage (i.e., i is defined as into +terminal), then v = iR (Ohm’slaw)

Question: What is the I-V characteristic for a 10K Ω resistor? Draw on axesbelow.

V10 20 30

I (mA)

Resistors in Parallel and Series

In series:

v = i R1+i R2 = i (R1+R 2) = i (R), where R=R 1+R 2

Series Parallel

continue

• In Parallel

Apply KCL at node Di = i 1 +i 2 = (v/R 1-v/R 2)

= v (1/R 1-1/R 2)= v (R 2 + R 1)/(R 1R2)= v /[(R 1R2)/(R 2 + R 1)]= v /R

R = [(R 1R2)/(R 2 + R 1)], where R 1 is in parallel with R 2

Resistors example• The resistors shown has 2 terminals A and B. It is desirable to replace it

with a single resistor connected between terminals A and B. What

should the value of this resistor be so that the resistance between theterminals is unchanged?By inspection, R 2 and R 3 are inseries. And this series combinationis in parallel with R 1.

Resistant between A and B:

R = (R 2+R 3) || R 1

= R 1(R2+R 3)/(R 1+(R 2+R 3))

remember product over sum.=(10,000)(100,000+47,000)/(10,000+100,000+47,000)= 9360 Ω

Power Dissipation in Ideal ResistorResistor can convert electrical energy into thermal energy(heat).In Chapter 1, we learned that in any circuit element:Power = Voltage across the circuit element X Current

flow thru the circuit elementPower = V I

But in the case of resistor, Ohm’s law holdV = I R (voltage difference between resistor terminals

equals current flow thru resistor X resistance)Power = ( I R) I =I 2 R

Or Power = V (V/R) = V2

/RIn practice, manufacturers state the max. power dissipationof a resistor in watts.

Example• The power dissipation of a 47000 Ω resistor is stated by the

manufacturer to be 1/4 Watt. What is the maximum dc voltage thatmay be applied? What is the largest dc current that can be madeto flow through the resistor without damaging it?

From the formula: P = V 2 /RP max = Vmax

2 /Rget V max

2 = P max RVmax = √[(1/4)(47000)]

= 108.4V

From the formula: P = I 2RP max = I max

get I max 2 = P max /R= √[(1/4)/(47000)]= 2.3mA

CONDUCTIVITY, RESISTIVITY, AND RESISTANCE(optional-for those of you who are interested)

If we apply a voltage V across a block of conductingmaterial of length L and cross-sectional area A, thecurrent, , , is proportional to V and A, and inversely,proportional to L.

ResistorA t

The proportionality constant σ is called the conductivity and has units:

6 L H P H Q V 6 6 F P O H Q J W K R O W D J H & X U U H Q W Y

σ =Ω

More familiar, resistivity, ρ F P

F P O H Q J W K X U U H Q W Y R O W D J H F X Q L W V =Ω=×= σ ρ

Define resistance and Ohm’s Law:

W : $ L I

×===≡

, ~ 9 $ /

FORMAL CIRCUIT ANALYSIS USING KCL:NODAL ANALYSIS

2 Define unknown node voltages (those not fixed byvoltage sources )

3 Write KCL at each unknown node, expressing currentin terms of the node voltages (using the constitutiverelationships of branch elements*)

* with inductors or floating voltages we will use a modified Step 3

1 Choose a Reference Node

4 Solve the set of equations (N equations for N unknownnode voltages)

NODAL ANALYSIS USING KCL – The Voltage Divider –

1 Choose reference node

2 Define unknown node voltages

3 Write KCL at unknown nodes

' ' −=−

4 Solve:

EXAMPLE OF NODE ANALYSIS

Define the node voltages (except reference node and the one set bythe voltage source); write down set of equations for node voltages V aand V b

node voltage known Ì

V a V b

←reference nodeApply KCL:

, - , ,

=−+− $

R 4V 1 R 2

You can solve for V a ,Vb .

: K D W L I Z H X V H G G L I I H U H Q W U H I Q R G H "

RESISTORS IN SERIES

Circuit with several resistors in series – Can we find an equivalent resistance?

. & / W H O O V X V V D P H F X U U H Q W

I O R Z V W K U R X J K H Y H U \ U H V L V W R U

. 9 / W H O O V X V

9 5 , 5 , 5 , 5 , =⋅+⋅+⋅+⋅

& O H D U O \

5 5 5 5 9 ,

7 K X V H T X L Y D O H Q W U H V L V W D Q F H R I U H V L V W R U V L Q V H U L H V L V W K H V L P S O H V X P

(Here its more convenient to use KVL than node analysis)

GENERALIZED VOLTAGE DIVIDER

Circuit with several resistors in series

−+−

−+ V 1

• We know 5 5 5 5 9 ,

• Thus,

# # # #

' ⋅+++

and $ $

# # # #

' ⋅+++

0 9 . 0 9 .

WHEN IS VOLTAGE DIVIDER FORMULA CORRECT?

−+ −

5 5 5 5

Correct if nothing else

connected to nodes

−+ −

5 5 5 5

+++≠

because R 5 removes condition ofresistors in series – i.e . , L

What is V 2? Answer: $ $

5 5 5 5 5

Dc Ckt Analysis Ppt2

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