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ENGR 202CIRCUIT ANALYSIS II

Dr. Mario Edgardo MagañaEECS

Oregon State University

• Acknowledgement

I would like to thank my colleague Dr. Annette Von Jouanne for providing the original MS Word class notes for this course.

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Quick Review of ENGR 201

st1→ Ohm’s Law IRV = (constant (dc), V & I upper case for dc, Ave, rms)

iRv = (time varying (ac), v & i lower case) nd2→Kirchoff’s Laws (What circuit analysis is based on) 1. Kirchoff’s Current Law (KCL) ∑∑ = outin '' sisi , cba iii += or 0entering' =∑ si , 0=−− cba iii or 0leaving' =∑ si , 0=++− cba iii (We use this for Node Voltage)

biai

ci

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2. Kirchoff’s Voltage Law (KVL) The sum of the voltages ( sv∑ ' ) around any closed loop equals zero. (Defn of Loop≡ Closed path - Begin at a given node and trace a path through the

circuit back to the original node without passing through any intermediate node more than once)

(resistors) (sources) e.g. ∑ ∑= risesdrops VV

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The systematic Approaches to Circuit Analysis (valid for dc and ac analysis – which we simplify with phasors)

1. Circuit Reduction

combining elements in series (share the same current) / parallel (share the

same voltage), Y−Δ transformations (3Δ=

ZZ y ), source transformations.

a

bRVV

R

R

b

a

1V 2V

sV sI

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Thev. and Norton Equivalent (R above is replaced by Req), for when only interested in “terminal behavior”, e.g. with a power-supply, we know there is a variety of circuitry in there, but we mainly want to know what voltage, current i.e. power it can supply

2. Circuit Recognition (of common configurations) voltage / current dividers, bridges

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1 IRV

=

)(,

11 out

i

out

outio RR

VI

RRRV

V+

=+

= 21

12

21

2111

21

211

)(,

)(,

RRRI

IRR

RIIRI

RRRIR

V+

=+

==+

=

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3. Node Voltage Method (need 1−en node voltage (KCL) equations) ≡en essential node where 3 or more circuit elements join.

We write KCL at the essential nodes ( en ) minus the reference node.

Typically end up with fewer equations than for Mesh Current Method (end up with 2 equations in the previous example)

1V 2V

sV sI

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4. Mesh Current Method ( need )1( −− ee nb mesh current (KVL) equations) ≡eb essential branch that connects essential nodes

without passing through an essential node. We write KVL for each mesh current loop

5. Superposition ---“kill”( short voltage source, open current source) all

independent sources but one, and sum the response for each.

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Alternating Current (AC) Circuits AC circuits are used for generation, transmission, distribution and utilization of electric energy, also the dominant form of signals in communications. So time varying, we deal mostly with sinusoidal excitation (input) and response (output). Favored by Edison favored by Westinghouse and Tesla After a bitter battle in electrical power generation and transmission, “ac” won over because: (in 1890≈ ) i) It is easier to generate; ii) It is easier to change the voltage (through a transformer) for efficient transmission of electric power, i.e. higher voltage, lower RI 2 losses, more suitable for higher power motors.

"dc"

IV or

t

iv or+

−

""Sinusoidal

ac

t

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Disadvantage: ac cannot be stored as easily as dc (no ac equivalent of batteries or capacitors). Why do we use sinusoidal time variation? - sin and cos are simple periodic, continuous functions, easy to deal with

(derivative and integral of a sinusoid is also a sinusoid, see below) - cosine is sine advanced by 90 tt

dtd ωωω cos)(sin =→ ; tdtt ω

ωω cos1sin −=∫ ;

- Any periodic waveform (no matter how complicated) can be represented

as a sum of sinusoids of different frequencies (Fourier Series in engr 203).

- From the superposition principle: we find the circuit response for each individual frequency and sum the response to get the full behavior

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Steady State Sinusoidal Analysis f = 1/T Hz, ω = 2πf (rad/sec) (angular freq.) Most often we are interested in the steady-state response of a circuit, which exists for ∞→t , rather than the transient, which “dies off” as ∞→t . The time depends on the time constant RL=τ (smaller dies away quicker).

v iR

L

mV

)(ti

T

)(tv

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What we find is that the sinusoidal excitation (input/source) i.e. )cos( φω += tVv m produces a steady-state (s.s.) sinusoidal response, i.e.

)cos( θφω −+= tIi m of the same frequency ( fπω 2= ). Theta (θ) is the phase shift due to capacitors or inductors, etc. ∴ Our ac circuit problems are a matter of determining:

1. magnitudes (i.e. mI )

2. phase angles (i.e. )(tan 1

RLωθ −= for RL circuit)

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Characteristics of a Sinusoidal Function

)cos( φω += tVv m ; )cos( θω += tIi m ; specified by 3 parameters: Vm = amplitude ω = angular freq. (rad/sec) φ,θ = phase angle

Period (T ) in ssec = the time between successive peaks of the same sign Frequency

Tf 1=⇒≡ , cycles per second ( HzHertz, )

ex: U.S. electric utility frequency fHz =60

msT 67.16601period ===⇒

v

tmV

mV

mV−

mV−

TPeriod

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The angular equivalent of the period is 360 or π2 radians ∴the angular frequency is ωπ

=(sec)

)(2T

rad , (T

f 1= )

which we refer to as omega: sec/22 radfT

ππω ==

- =φ phase shift (degree or radians), i.e. the shift of the cosine function with respect to t = 0, we have )cos( φω += tVv m (recall sin(A) = cos(A-90°) )

∴ Peak occurs when 0=+φωt , or ωφ

−=t ,

if no phase shift, 0=φ ,and peak occurs at 0=t .

Note: If φ is positive, sinusoid shifts to left If φ is negative, sinusoid shifts to right.

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Phasors are an effective way of dealing with ac circuit problems, and much simpler than dealing with sinusoidal quantities.

Defn

Phasor: a complex number that carries the magnitude and phase angle information of a sinusoidal function. ∴ V=∠≡+= φφω mm VtVv )cos( in phasor form (angular or polar) (bold in text) Vector in the Complex Plane We can put this in rectangular form as follows:

φcosmVa = , φsinmVb = , (from Euler’s identity) where ba j+=V in rectangular form

Re

Im

a

b

1φ

mV

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“ j ” is complex 90 operator that comes into play because of the phase angles introduced by sinusoidal sources, inductors and capacitors, etc. ( 12 −=j )

then 22 baVm += ,ab1tan−=φ

Do all in rectangular form on Calculator! →∴We now use complex arithmetic for all of the Systematic Approaches developed for dc analysis. → For example, we introduce the concept of impedance Z , and Ohm’s law becomes:

ZIV = ,where )(j)( ωω XRZ += R = real (resistive) component X = imaginary (reactive) component

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Element Impedance (Voltage and Current in) Complex Plane/Phasor

Diagram R R (i.e. no reactance)

(Ω )

Re

Im

RVRI

L 90∠= LLj ωω reactance )(Ω= LX L ω ELI the ICE man (current lags voltage)

Re

ImLV

LI

C 901901j

1−∠=∠−=

−=

CCCj

C ωωωω

reactance )(1Ω−=

CXC ω

ReIm

CICV

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It can be very useful to represent complex numbers in the complex plane.

Re

Im

x

y

θ

VV

θ∠= VV with angular frequency ω back in sinusoidal form

)cos( θω +=⇒ tVv

)cos( θω += tVv Polar: θ∠= VV Rectangular: YX j+=V ; θθ sin,cos VYVX == so related back to polar: 22 YXV +=

1tan ( )YX

θ −=

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Simple Circuit Example: ω ω Find )(0 ti if mVtvs 4000sin500= using cosine mVtvs )904000cos(500 −= 1st construct the frequency domain equivalent circuit ( want everything in 'Ω s)

Ω800 Ω400j

Ω− 1000j

mH100

Fμ25.0sv

)(tio

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Ω=×== j400)100j(4000)(10j -3LZL ω ;

Ω−=×

== j1000)1025j(4000)(0.

1j

16-C

ZC ω;

Ω−∠=

Ω=+=

87.361000j600-800j1000-j400800Z

mVs 90500 −∠=V ,use cosine.

A

Zs

μ13.53500

)87.3690(1050087.3610009010500

0

60

3

0

−∠=

+−∠×=−∠

−∠×==

−

−

I

I

VI

⇒ Atti μ)13.534000cos(500)(0 −=

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Example of Sinusoidal Steady State Analysis Node Voltage: Use the N-V method to find the s.s expression for )(tv , where ,cos10 Atis ω= Vtvs ωsin100= , 350 10 / 50 /secrad s kradω= × =

Ω20

Ω5jΩ− 22.2j Hμ100

Fμ9 sv)(tv

V

Ω5

+

−si

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1st Construct the frequency domain equivalent circuit (i.e. the phasor domain circuit) The phasor transform of the current source is 010∠=sI ; The phasor transform of the voltage source is 90100 −∠=sV ; (both in cos) The resistors transform directly as Ω5 and Ω20 ; The Hμ100 inductor in the frequency domain has an impedance of: 3(50 10 / )(100 ) 5j L j rad s H jω μ= × = Ω (note if had frequency in Hz , fπω 2=⇒ ) The Fμ9 capacitor has an impendence of :

j j j2.22250k 9Cω μ

− = − = − Ω⋅

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The node-voltage equation is: (∑ currents leaving node =0)

20020

901005j222.2j5

010 )( ×=−∠−

++−

++∠−VVVV 1j2 −= , j = √(-1)

(polar) 0)90100(4j9j4200 =−∠−+−++− VVVV

100j)90sin(j100)90cos(100-10

−⇒−+ (rectangular)100j200)55( −=+⇒ jV

°−∠=−=⇒ 57.7162.3130j10V ω

31.62cos(50,000 71.57 )v t V⇒ = −

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Example: Thev./Norton Equiv. Find the Thev. and Norton Equivalents with respect to terminals a, b Since Ω30j and Ω− 24j are in series, to find

abth VV = we can use voltage divider again

⇒ VVjabth 4.189.75247224j30j18)18(080

−∠=−=−+

∠== VV

To find thZ we kill independent voltage source (short it),

⇒

N

th

Z

Z

=

Ω+=−+−

= 4.5j8.124j30j18

)24j30j)(18(

(note, if had been negative, would be capacitor)

V080∠

Ω30j

Ω− 24j

Ω18

a

b

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Thevenin equivalent: Norton equivalent: Source Transformation I = V/Z

Vj )2472( −Vor 4.189.75 −∠

Ω8.1 Ω4.5ja

b

a

b

NZ

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Super Position Example Find 0V using superposition i) Kill voltage source (short) 00 =⇒ 'V in other words, current source has no effect on 0V

A302∠ V6012∠

Ω2

Ω2

Ω2

Ω− 2jΩ− 1jΩ1j+

−oV

A302∠Ω2

Ω2

Ω− 2j+

−

'oV

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ii) Kill current source (open)

2j26012

−∠

=i ,where 20 ⋅= i"V

This is principle of voltage divider!

V10548.82.8j2.22j2

)2)(6012(00 ∠=+−=

−∠

==⇒ VV"

V6012∠

Ω2

Ω2

Ω2

Ω− 2jΩ− 1jΩ1j+

−

''oV

i

parallel

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Example: Find vo

Vtvg )87.362000cos(201 −= Vg 87.36201 −∠=⇒ V

Vtvg )26.162000sin(502 −= , put it in cosine (standard),

V

Vg

26.10650

9026.16502

−∠=

−−∠=⇒ V

Ω10

Ω2j Ω− 5jmH1 Fμ100

1gv 2gvov

oV

+

−

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ω = 2000rad/sec, jωL = j2Ω , -j/ωC = -j5Ω Write KCL at 0V (∑ currents leaving node =0)

05j10

02j

20010 =−−

+−

+− gg VVVVV

5j2j

)5j1

101

2j1( 21

0 −+=−+ gg VV

V

0360j360 ∠=+=⇒ VV

0 36cos(2000 0)v t V⇒ = +