dc theory.doc

8/14/2019 DC THEORY.doc

1/18

DC THEORY

Parameter Measuring Unit Symbol Description

Voltage Volt V or EUnit of Electrical Potential

V = I R

Current Ampere I or iUnit of Electrical Current

I = V R

Resistance Ohm R or Unit of DC Resistance

R = V I

Con!uctance Siemen " or Reciprocal of Resistance

G = 1 R

Capacitance #ara! CUnit of Capacitance

C = Q V

Charge Coulomb $Unit of Electrical Charge

Q = C V

In!uctance %enry & or %Unit of In!uctance

VL= -L(di/dt)

Po'er (atts (Unit of Po'er

P = V Ior I2 R

Impe!ance Ohm )Unit of AC Resistance

Z2= R2+ X2

#re*uency %ert+ %+Unit of #re*uency

= 1 T

, (h - The Watt-H!". /he amount of electrical energy consume! in the circuit by a loa! of one 'att

!ra'ing po'er for one hour. eg a &ight 0ulb1 It is commonly use! in the form of #Wh23ilo'att4

hour5 'hich is 6.777 'att4hours or $Wh2Mega'att4hour5 'hich is 6.777.777 'att4hours1

, !0 - The %e&i'e. /he !ecibel is a one tenth unit of the 0el 2symbol 05 an! is use! to represent

gain either in 8oltage. current or po'er1 It is a logarithmic unit e9presse! in dan! is

commonly use! to represent the ratio of input to output in amplifier. au!io circuits or

lou!spea:er systems1

#or e9ample. the !0 ratio of an input 8oltage 2Vin5 to an output 8oltage 2Vout5 is e9presse! as;7log672Vout


2/18

67 !egrees or ra!1

, - Ti3e C,*ta,t. /he /ime Constant of an impe!ance circuit or linear first4or!er system is the

time it ta:es for the output to reach B1F of its ma9imum or minimum output 8alue 'hen

subGecte! to a Step Response input1 It is a measure of reaction time1

4i"&h55* Ci"&!it La6

(e sa' in the Re*i*t"*tutorial that a single e*ui8alent resistance. 2 R/5 can be foun! 'hen t'o or more

resistors are connecte! together in either series. parallel or combinations of both. an! that these circuits obey

7h38* La61 %o'e8er. sometimes in comple9 circuits such as bri!ge or / net'or:s. 'e can not simply use

OhmHs &a' alone to fin! the 8oltages or currents circulating 'ithin the circuit1 #or these types of calculations 'e

nee! certain rules 'hich allo' us to obtain the circuit e*uations an! for this 'e can use 4i"&h55* Ci"&!it La61

In 6J. a "erman physicist. G!*ta9 4i"&h55!e8elope! a pair or set of rules or la's 'hich !eal 'ith the

conser8ation of current an! energy 'ithin electrical circuits1 /hese t'o rules are commonly :no'n asK Kirchoffs

Circuit Laws'ith one of 3irchoffs la's !ealing 'ith the current flo'ing aroun! a close! circuit.4i"&h55* C!""e,t

La6: (4CL)'hile the other la' !eals 'ith the 8oltage sources present in a close! circuit.4i"&h55* Vtae La6:

(4VL)1

4i"&h55* .i"*t La6 - The C!""e,t La6: (4CL)

4i"&h55* C!""e,t La6or 3C&. states that the ?total current or charge entering a junction or node is exactly

equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the

node?1 In other 'or!s the algebraic sum of A&& the currents entering an! lea8ing a no!e must be e*ual to +ero.

I2e9iting5L I2entering5= 71 /his i!ea by 3irchoff is commonly :no'n as the C,*e"9ati, 5 Cha"e1

4i"&h55* C!""e,t La6

%ere. the currents entering the no!e. I6. I;. Iare all positi8e in 8alue an! the ; currents lea8ing the no!e. I

an! IJare negati8e in 8alue1 /hen this means 'e can also re'rite the e*uation as

I6L I;L I4 I4 IJ= 7
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3/18

/he term ;dein an electrical circuit generally refers to a connection or Gunction of t'o or more current carrying

paths or elements such as cables an! components1 Also for current to flo' either in or out of a no!e a close!

circuit path must e9ist1 (e can use 3irchoffHs current la' 'hen analysing parallel circuits1

4i"&h55*


4/18

Components are connecte! in seriesif they carry the same current1

Components are connecte! in parallelif the same 8oltage is across them1

>a3?e ;1

#in! the current flo'ing in the 7Resistor. R

/he circuit has branches. ; no!es 2Aan! 05 an! ; in!epen!ent loops1

Using 4i"&h55* C!""e,t La6. 4CLthe e*uations are gi8en as

At no!eAK I6L I;= I

At no!e 0K I= I6L I;

Using 4i"&h55* Vtae La6. 4VLthe e*uations are gi8en as

&oop 6 is gi8en as K 67 = R69 I6L R9 I= 67I6L 7I

&oop ; is gi8en as K ;7 = R;9 I;L R9 I= ;7I;L 7I

&oop is gi8en as K 67 4 ;7 = 67I64 ;7I;

As Iis the sum of I6L I;'e can re'rite the e*uations as

E*1 No 6 K 67 = 67I6L 72I6L I;5 = J7I6L 7I;

E*1 No ; K ;7 = ;7I;L 72I6L I;5 = 7I6L B7I;

(e no' ha8e t'o ?


5/18

Substitution of I6in terms of I;gi8es us the 8alue of I6as 4716 Amps

Substitution of I;in terms of I6gi8es us the 8alue of I;as L71; Amps

As K I= I6L I;

/he current flo'ing in resistor Ris gi8en as K 4716 L 71; = 71;B Amps

an! the 8oltage across the resistor Ris gi8en as K 71;B 9 7 = 661 8olts

/he negati8e sign for I6means that the !irection of current flo' initially chosen 'as 'rong. but ne8er the less still8ali!1 In fact. the ;78 battery is charging the 678 battery1

??i&ati, 5 4i"&h55* Ci"&!it La6*

/hese t'o la's enable the Currentsan! Voltagesin a circuit to be foun!. ie. the circuit is sai! to be?Analyse!?. an! the basic proce!ure for using 4i"&h558* Ci"&!it La6*is as follo'sK

1@Assume all 8oltages an! resistances are gi8en1 2 If not label them V6. V;.111 R6. R;.etc1 5

2@&abel each branch 'ith a branch current1 2 I6. I;. Ietc1 5

A@#in! 3irchoffHs first la' e*uations for each no!e1

B@#in! 3irchoffHs secon! la' e*uations for each of the in!epen!ent loops of the circuit1

@Use &inear simultaneous e*uations as re*uire! to fin! the un:no'n currents1

As 'ell as using 4i"&h55* Ci"&!it La6to calculate the 8arious 8oltages an! currents circulating aroun! a linear

circuit. 'e can also use loop analysis to calculate the currents in each in!epen!ent loop 'hich helps to re!uce

the amount of mathematics re*uire! by using Gust 3irchoffHs la's

Ci"&!it ,a0*i*

In the pre8ious tutorial 'e sa' that comple9 circuits such as bri!ge or /4net'or:s can be sol8e! using4i"&h558*

Ci"&!it La6*1 (hile 3irchoffs &a's gi8e us the basic metho! for analysing any comple9 electrical circuit. there

are !ifferent 'ays of impro8ing upon this metho! by using $e*h C!""e,t ,a0*i*or;da Vtae ,a0*i*

that results in a lessening of the mathHs in8ol8e! an! 'hen large net'or:s are in8ol8e! this re!uction in maths

can be a big a!8antage1

#or e9ample. consi!er the circuit from the pre8ious section1

$e*h ,a0*i* Ci"&!it


6/18

One simple metho! of re!ucing the amount of mathHs in8ol8e! is to analyse the circuit using 3irchoffHs Current

&a' e*uations to !etermine the currents. I6an!I;flo'ing in the t'o resistors1 /hen there is no nee! to calculate

the current Ias its Gust the sum of I6an!I;1 So 3irchoffHs secon! 8oltage la' simply becomesK

E*uation No 6 K 67 = J7I6L 7I;

E*uation No ; K ;7 = 7I6L B7I;

therefore. one line of mathHs calculation ha8e been sa8e!1

$e*h C!""e,t ,a0*i*

A more easier metho! of sol8ing the abo8e circuit is by using $e*h C!""e,t ,a0*i*or L? ,a0*i*'hich

is also sometimes calle! $a>6eD* Ci"&!ati, C!""e,t*metho!1 Instea! of labelling the branch currents 'e

nee! to label each ?close! loop? 'ith a circulating current1 As a general rule of thumb. only label insi!e loops in a

cloc:'ise !irection 'ith circulating currents as the aim is to co8er all the elements of the circuit at least once1 Any

re*uire! branch current may be foun! from the appropriate loop or mesh currents as before using 3irchoffs

metho!1

#or e9ampleK K i6= I6. i;= 4I;an! I= I64 I;

(e no' 'rite 3irchoffHs 8oltage la' e*uation in the same 'ay as before to sol8e them but the a!8antage of this

metho! is that it ensures that the information obtaine! from the circuit e*uations is the minimum re*uire! to sol8e

the circuit as the information is more general an! can easily be put into a matri9 form1


/hese e*uations can be sol8e! *uite *uic:ly by using a single mesh impe!ance matri9 )1 Each element ON the

principal !iagonal 'ill be ?positi8e? an! is the total impe!ance of each mesh1 (here as. each element O## the

principal !iagonal 'ill either be ?+ero? or ?negati8e? an! represents the circuit element connecting all the

appropriate meshes1 /his then gi8es us a matri9 ofK


7/18

(hereK

Q V gi8es the total battery 8oltage for loop 6 an! then loop ;1

Q I states the names of the loop currents 'hich 'e are trying to fin!1

Q R is calle! the resistance matri91

an! this gi8es I6as 4716 Ampsan!I;as 471; Amps

As K I= I64 I;

/he current Iis therefore gi8en as K 4716 4 2471;5 = 71;B Amps

'hich is the same 8alue of 71;B amps. 'e foun! using4i"&h55D*circuit la' in the pre8ious tutorial1

$e*h C!""e,t ,a0*i*


8/18

;da Vtae ,a0*i*

As 'ell as using $e*h ,a0*i*to sol8e the currents flo'ing aroun! comple9 circuits it is also possible to use

no!al analysis metho!s too1;da Vtae ,a0*i*complements the pre8ious mesh analysis in that it is

e*ually po'erful an! base! on the same concepts of matri9 analysis1 As its name implies. ;da Vtae

,a0*i*uses the ?No!al? e*uations of 3irchoffHs first la' to fin! the 8oltage potentials aroun! the circuit1

So by a!!ing together all these no!al 8oltages the net result 'ill be e*ual to +ero1 /hen. if there are ?n? no!es in

the circuit there 'ill be ?n46? in!epen!ent no!al e*uations an! these alone are sufficient to !escribe an! hence

sol8e the circuit1

At each no!e point 'rite !o'n 3irchoffHs first la' e*uation. that isK ?the currents entering a node are exactly equal

in value to the currents leaving the node? then e9press each current in terms of the 8oltage across the branch1

#or ?n? no!es. one no!e 'ill be use! as the reference no!e an! all the other 8oltages 'ill be reference! ormeasure! 'ith respect to this common no!e1


;da Vtae ,a0*i* Ci"&!it

In the abo8e circuit. no!e Dis chosen as the reference no!e an! the other three no!es are assume! to ha8e

8oltages. Va. Vban! Vc'ith respect to no!e D1 #or e9ample

Va = 678an!Vc = ;78. Vbcan be easily foun! byK

again is the same 8alue of 71;B amps. 'e foun! using4i"&h558* Ci"&!it La6in the pre8ious tutorial1
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9/18

#rom both Mesh an! No!al Analysis metho!s 'e ha8e loo:e! at so far. this is the simplest metho! of sol8ing this

particular circuit1 "enerally. no!al 8oltage analysis is more appropriate 'hen there are a larger number of current

sources aroun!1 /he net'or: is then !efine! asK QI = QT QV 'here QI are the !ri8ing current sources. QV

are the no!al 8oltages to be foun! an! QT is the a!mittance matri9 of the net'or: 'hich operates on QV to

gi8e QI 1

;da Vtae ,a0*i*

In the pre8ious tutorials 'e ha8e loo:e! at sol8ing comple9 electrical circuits using4i"&h558* Ci"&!it

La6*.$e*h ,a0*i*an! finally;da ,a0*i*but there are many more ?Circuit Analysis /heorems?

a8ailable to calculate the currents an! 8oltages at any point in a circuit1 In this tutorial 'e 'ill loo: at one of the

more common circuit analysis theorems 2ne9t to 3irchoffs5 that has been !e8elope!. The9e,i,* The"e31

The9e,i,* The"e3states that ?Any linear circuit containing several voltages and resistances can e replaced

y just a !ingle "oltage in series with a !ingle #esistor?1 In other 'or!s. it is possible to simplify any ?&inear?

circuit. no matter ho' comple9. to an e*ui8alent circuit 'ith Gust a single 8oltage source in series 'ith a

resistance connecte! to a loa! as sho'n belo'1 The9e,i,* The"e3is especially useful in analy+ing po'er or

battery systems an! other interconnecte! circuits 'here it 'ill ha8e an effect on the a!Goining part of the circuit1
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10/18

The9e,i,* e!i9ae,t &i"&!it@

As far as the loa! resistor R&is concerne!. any ?one4port? net'or: consisting of resisti8e circuit elements an!

energy sources can be replace! by one single e*ui8alent resistance Rsan! e*ui8alent 8oltage Vs. 'here Rsis

the source resistance 8alue loo:ing bac: into the circuit an! Vsis the open circuit 8oltage at the terminals1


#irstly. 'e ha8e to remo8e the centre 7resistor an! short out 2not physically as this 'oul! be !angerous5 all

the emfs connecte! to the circuit. or open circuit any current sources1 /he 8alue of resistor Rsis foun! by

calculating the total resistance at the terminalsAan! 0'ith all the emfs remo8e!. an! the 8alue of the 8oltage

re*uire! Vsis the total 8oltage across terminalsAan! 0'ith an open circuit an! no loa! resistor Rsconnecte!1

/hen. 'e get the follo'ing circuit1

.i,d the !i9ae,t Re*i*ta,&e (R*)

.i,d the !i9ae,t Vtae (V*)


11/18

(e no' nee! to reconnect the t'o 8oltages bac: into the circuit. an! as VS= VA0the current flo'ing aroun! the

loop is calculate! asK

so the 8oltage !rop across the ;7resistor can be calculate! asK

VA0= ;7 4 2;7 9 71amps5 = 61 8olts1

/hen the /he8enins E*ui8alent circuit is sho'n belo' 'ith the 7resistor connecte!1

an! from this the current flo'ing in the circuit is gi8en asK

'hich again. is the same 8alue of 71;B amps. 'e foun! using4i"&h55D*circuit la' in the pre8ious tutorial1

The9e,i,* the"e3can be use! as a circuit analysis metho! an! is particularly useful if the loa! is to ta:e a

series of !ifferent 8alues1 It is not as po'erful as$e*hor;daanalysis in larger net'or:s because the use of

Mesh or No!al analysis is usually necessary in any /he8enin e9ercise. so it might as 'ell be use! from the start1

%o'e8er. /he8enins e*ui8alent circuits of T"a,*i*t"*. Vtae
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12/18

;"t,* The"e3

In some 'ays ;"t,8* The"e3can be thought of as the opposite to ?/he8enins /heorem?. in that /he8enin

re!uces his circuit !o'n to a single resistance in series 'ith a single 8oltage1 Norton on the other han! re!uces

his circuit !o'n to a single resistance in parallel 'ith a constant current source1 ;"t,* The"e3states that

?Any linear circuit containing several energy sources and resistances can e replaced y a single Constant

Current generator in parallel with a !ingle #esistor?1

As far as the loa! resistance. R&is concerne! this single resistance.RSis the 8alue of the resistance loo:ing

bac: into the net'or: 'ith all the current sources open circuite! an! ISis the short circuit current at the output

terminals as sho'n belo'1

;"t,* e!i9ae,t &i"&!it@

/he 8alue of this ?constant current? is one 'hich 'oul! flo' if the t'o output terminals 'here shorte! together

'hile the source resistance 'oul! be measure! loo:ing bac: into the terminals. 2the same as /he8enin51

#or e9ample. consi!er our no' familiar circuit from the pre8ious section1

/o fin! the Nortons e*ui8alent of the abo8e circuit 'e firstly ha8e to remo8e the centre7loa! resistor an!

short out the terminalsAan!0to gi8e us the follo'ing circuit1


13/18

(hen the terminalsAan! 0are shorte! together the t'o resistors are connecte! in parallel across their t'o

respecti8e 8oltage sources an! the currents flo'ing through each resistor as 'ell as the total short circuit current

can no' be calculate! asK

6ith -


14/18

Again. the t'o resistors are connecte! in parallel across the terminalsAan! 0'hich gi8es us a total resistance

ofK

/he 8oltage across the terminalsAan! 0'ith the loa! resistor connecte! is gi8en asK

/hen the current flo'ing in the 7loa! resistor can be foun! asK

'hich again. is the same 8alue of 71;B amps. 'e foun! using4i"&h55D*circuit la' in the pre8ious tutorials1

;"t,* The"e3


15/18

$a>i3!3 P6e" T"a,*5e"

(e ha8e seen in the pre8ious tutorials that any comple9 circuit or net'or: can be replace! by a single energy

source in series 'ith a single internal source resistance. RS1 "enerally. this source resistance or e8en impe!ance

if in!uctors or capacitors are in8ol8e! is of a fi9e! 8alue in Ohms1 %o'e8er. 'hen 'e connect a loa! resistance.

R&across the output terminals of the po'er source. the impe!ance of the loa! 'ill 8ary from an open4circuit state

to a short4circuit state resulting in the po'er being absorbe! by the loa! becoming !epen!ent on the impe!ance

of the actual po'er source1 /hen for the loa! resistance to absorb the ma9imum po'er possible it has to be

?Matche!? to the impe!ance of the po'er source an! this forms the basis of$a>i3!3 P6e" T"a,*5e"1

/he $a>i3!3 P6e" T"a,*5e" The"e3is another useful analysis metho! to ensure that the ma9imum amount

of po'er 'ill be !issipate! in the loa! resistance 'hen the 8alue of the loa! resistance is e9actly e*ual to the

resistance of the po'er source1 /he relationship bet'een the loa! impe!ance an! the internal impe!ance of the

energy source 'ill gi8e the po'er in the loa!1 Consi!er the circuit belo'1

The9e,i,* !i9ae,t Ci"&!it@

In our /he8enin e*ui8alent circuit abo8e. the ma9imum po'er transfer theorem states that ?the maximum amount

of power will e dissipated in the load resistance if it is equal in value to the $hevenin or %orton source

resistance of the network supplying the power?1

In other 'or!s. the loa! resistance resulting in greatest po'er !issipation must be e*ual in 8alue to the

e*ui8alent /he8enin source resistance. then R&= RSbut if the loa! resistance is lo'er or higher in 8alue than

the /he8enin source resistance of the net'or:. its !issipate! po'er 'ill be less than ma9imum1 #or e9ample. fin!

the 8alue of the loa! resistance. R&that 'ill gi8e the ma9imum po'er transfer in the follo'ing circuit1

>a3?e ;1@

(hereKRS= ;JR&is 8ariable bet'een 7 4 677VS= 6778

/hen by using the follo'ing OhmHs &a'e*uationsK


16/18

(e can no' complete the follo'ing table to !etermine the current an! po'er in the circuit for !ifferent 8alues of

loa! resistance1

Ta'e 5 C!""e,t aai,*t P6e"

R&25 I 2amps5 P 2'atts5

7 7 7

J 1 JJ

67 ;1

6J ;1J

;7 ;1;

R&25 I 2amps5 P 2'atts5

;J ;17 677

7 61

7 61J

B7 61;

677 71 B

Using the !ata from the table abo8e. 'e can plot a graph of loa! resistance. R&against po'er. Pfor !ifferent

8alues of loa! resistance1 Also notice that po'er is +ero for an open4circuit 2+ero current con!ition5 an! also for a

short4circuit 2+ero 8oltage con!ition51

G"a?h 5 P6e" aai,*t Lad Re*i*ta,&e

#rom the abo8e table an! graph 'e can see that the $a>i3!3 P6e" T"a,*5e"occurs in the loa! 'hen the

loa! resistance. R&is e*ual in 8alue to the source resistance.RSthat isK RS= R&= ;J1/his is calle! a

?matche! con!ition? an! as a general rule. ma9imum po'er is transferre! from an acti8e !e8ice such as a po'er

supply or battery to an e9ternal !e8ice 'hen the impe!ance of the e9ternal !e8ice e9actly matches the

impe!ance of the source1

One goo! e9ample of impe!ance matching is bet'een an au!io amplifier an! a lou!spea:er1 /he output

impe!ance. )OU/of the amplifier may be gi8en as bet'een an! . 'hile the nominal input impe!ance. )INof

the lou!spea:er may be gi8en as only1 /hen if the spea:er is attache! to the amplifiers output. the

amplifier 'ill see the spea:er as an loa!1 Connecting t'o spea:ers in parallel is e*ui8alent to the

amplifier !ri8ing one spea:er an! both configurations are 'ithin the output specifications of the amplifier1


17/18

Improper impe!ance matching can lea! to e9cessi8e po'er loss an! heat !issipation1 0ut ho' coul! you

impe!ance match an amplifier an! lou!spea:er 'hich ha8e 8ery !ifferent impe!ances1 (ell. there are

lou!spea:er impe!ance matching transformers a8ailable that can change impe!ances from to .or to

6BHsto allo' impe!ance matching of many lou!spea:ers connecte! together in 8arious combinations such as

in PA 2public a!!ress5 systems1

T"a,*5"3e" I3?eda,&e $at&hi,

One 8ery useful application of impe!ance matching in or!er to pro8i!e ma9imum po'er transfer bet'een the

source an! the loa! is in the output stages of amplifier circuits1 Signal transformers are use! to match the

lou!spea:ers higher or lo'er impe!ance 8alue to the amplifiers output impe!ance to obtain ma9imum soun!

po'er output1 /hese au!io signal transformers are calle! ?matching transformers? an! couple the loa! to the

amplifiers output as sho'n belo'1

T"a,*5"3e" I3?eda,&e $at&hi,

/he ma9imum po'er transfer can be obtaine! e8en if the output impe!ance is not the same as the loa!

impe!ance1 /his can be !one using a suitable ?turns ratio? on the transformer 'ith the correspon!ing ratio of loa!

impe!ance. )&OADto output impe!ance. )OU/matches that of the ratio of the transformers primary turns to

secon!ary turns as a resistance on one si!e of the transformer becomes a !ifferent 8alue on the other1 If the loa!

impe!ance. )&OADis purely resisti8e an! the source impe!ance is purely resisti8e. )OU/then the e*uation for

fin!ing the ma9imum po'er transfer is gi8en asK

(hereK NPis the number of primary turns an!NSthe number of secon!ary turns on the transformer1 /hen by

8arying the 8alue of the transformers turns ratio the output impe!ance can be ?matche!? to the source

impe!ance to achie8e ma9imum po'er transfer1 #or e9ample.

>a3?e ;2@

If an lou!spea:er is to be connecte! to an amplifier 'ith an output impe!ance of 6777. calculate the turns

ratio of the matching transformer re*uire! to pro8i!e ma9imum po'er transfer of the au!io signal1 Assume the

amplifier source impe!ance is )6. the loa! impe!ance is );'ith the turns ratio gi8en as N1


18/18

"enerally. small transformers use! in lo' po'er au!io amplifiers are usually regar!e! as i!eal so any losses can

be ignore!1

dc theory.doc

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