Direct-Current Circuits

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1 Direct-Current Circuits

2 A'.3/.". 39 '- )232.-/ 32,+/" 7+3(5-.)232.-/ )'03,.5B )*+,"- &'&./( 0-1* *7 2829*4-& )"< 35- )*+,"-= A0.5.C2/'-231).D')232.')2-1 < 2'+)/ /1 /-26,4" &'( )*+,-. /- & ' )232.')2 3-2)326 " *4 7+3(5-. 2<16"-0335-"-4*43"4& )232.-/ "2+,'9."')3*3-5)232.-/2: Combination of Resistors Parallel =35-19-""35- )"64,"2-9*35-<? Each resistor individually satisfies Ohm s law, A'.3/.". 39 '- )232.-/ 32,+/" 5)*)."-." '.") )232.-/2 :."-." " V1 = I1R1 and V2= I2R2. 7+3(5-.)232.-/ 3244)'03,.5B7+5.'."5)*2.)232.') 6 " The potential across the resistors are the same, V 1 =V2 =V. -1.".)/'--/.13-4)555/)'22('5.* 2'+)/ Current conservation then implies: &'&( *73-" "-4*43672-*4"-C,*"-0/> "1 &'&./( V V 1 )*+," * *7-8-23"* * "6:-" *"2,*3 I = I1 + I 2 = + = V *4-& = V )"< 35- )*+,"-= ' 5*+5-43 :836+- *4 *<<-0*63-8> 6?3-" 35- R1 R2 Req R1 & "&'(')*+'(,-&+(&'.,/(,0.+.11& *4 /-26, *"-4 56: *+*/8> 4<688 "-4* ". N 2<16"-0335-"-4*43"4& ")'+*" )2 3-.' "D670 /+))-..". = = )"64,"2-9*35-<? =35-19-""35-" *25-7-"+>*40-8*:-"-0*4 E>2,"" ":63*7=35-46<-2,""-73 Req R1 R2 R3 )232.') i=1 Ri < 6 F/" )232.') 3-13(31+55B 2.3 " ( ' " ' &'&B( 7 < " " " 7 < Series I'D():." 4'.-.35 /)'22." )232. GB/+))-./'-2)(.3'-:."/+))-..".4222.")'+*"."('5.*2'+)/,+2.13(31 3-.' /+))-.." ")'+*" )232.') -1 /+))-. <." ")'+*" )232.') < 6 F/" )232.') 3-13(31+55B H",C2 5D: -1 < < < 6 *73-" "-4*43672-*4"-C,*"-0/>-7-"+>274-":63*7& I'D():." 4'.-.35 /)'22." )232.')2 )." 2,: < 6 J+))-. /'-2)(.3'-."-3,4532 "&'(')*+'(,-&+(&'.,/(,0.+.11&1 < < < < E>2,"" ":63*7=35-46<-2,""-73*4?89*7+35",+5-625"-4*43"& & )45/1 &B '- 7+3(5-. )232.') 7 D3." ".D' )232.')2 3-4)555 " /-.D' )232.')2 3-4)555 /- & ) By current conservation, the same current I is flowing through each resistor. The total voltage drop from a to c across both elements is the sum of the voltage drops across the individual resistors, so V = IR1 + IR2 = I ( R1 + R2 ) = IReq J',4)3-*."2 )2+5.2:." 7+3(5-. )232.-/ 9').D' 7 =E3*+) 2(34+&""56(F-4*43"4* J',4)3-*." 7 =E3*+) )232.')2.".)/'--/.13-4)55532*3(-&B i )232.')2.".)/'--/.13-4)55532*3 i=1 &3 < 7 62"4435-*70*:*0,68"-4*43"4H "32) B*-)53M2.' )232.')2/'--/.13-4) (34+&""56(F-4*43"4*74-"*-4&/(GC,*:68-732*"2,*3& "D " 7 <? " N Req = R1 + R2 + = R &3 "32) B*-)53M2.')232.')2 62"4435-*70*:*0,68"-4*43"4H

3 Example (1) What is the equivalent resistance between points a and b?

4 Electromotive Force (emf) Electromotive force, emf () refers to voltage generated by a power supply to maintain a constant current in a closed circuit and is measured in volts. Note: The electromotive force" is not a force in the classical physics sense as can be seen it is measured in volt, not newton. Batteries, solar cells and thermocouples are some examples of emf source. They can be thought of as a charge pump that moves charges from lower potential to the higher one (in a direction opposite to the E-field). The emf,, of a source is the work done per unit charge, or energy per unit electric charge. We indicate EMF with this symbol: Long side: + terminal Short side: - terminal The current flows from + to Counterintuitive if you think about it in = V = IR I V + -

5 Internal Resistance A real battery always carries an internal resistance r, so the potential difference across the battery terminals becomes V = " Ir The potential V, called the terminal potential, is less than emf because of the term Ir, representing the potential drop across the internal resistance r. Since there is no net change in potential difference around a closed loop, we have IR = " Ir " Ir " IR = 0 I = R + r

6 Kirchhoff s Rules Junction Rule (Kirchhoff's Current Law, KCL) The sum of the currents flowing into a junction is equal to the sum of the currents flowing out of the junction: I in = I out " I in I out = 0 " I = 0 Loop Rule (Kirchhoff's Voltage Law, KVL) The sum of the voltage drops V, across any circuit elements that form a closed circuit is zero: " V = 0 " V = 0 closed loop

7 Potential Rise & Drop in a circuit The sum of the changes in potential (voltage drops) around any closed loop of a circuit must be zero. Loop Rule: V = 0

8 Problem Solving Strategy (I) (1) Draw a circuit diagram, and label all the quantities, both known and unknown. (2) Assign a direction to the current in each branch of the circuit. (If the actual direction is opposite to what you have assumed, your result at the end will be a negative number.) (3) Apply the junction and loop rules until the number of independent equations obtained is the same as the number of unknowns (4) Solve the simultaneous equations to obtain the solutions for the unknowns.

9 Problem Solving Strategy (II) &'B4&74,54288>7+7*1.,54)81B41,*813428(<8& H &47*7,8& 4/< 78+&)4 )'>')*,8& 54 7'/4 46+',*81 *7 83,'*14- (54,54&,54 ) > *7,&'B4&74- )28)9(*74 8& )8+1,4&)28)9(*74: 54 4G>&477*817 '),+'22= -*<<4& 3= '1 8B4&'22 14.',*B4 7*.1: I8(4B4&0+7*1.,54288>&+240(4'&424-,8J J 0'1-541)4,547'/446+',*81:

10 Example (2) Consider the circuit shown in the figure. Find the current I in the circuit. V = 0

11 Example (3)

12 Example (4) Consider the circuit shown in the figure. The emf sources 1 and 2, and the resistances R 1, R 2 and R 3 are all given. Find the currents through each resistor. ' ' &'()*)+'"&'(()*

13 RC Circuits Consider the circuit shown below. The capacitor is connected to a DC voltage source of emf. At time t < 0, there is no voltage across the capacitor so the capacitor acts like a short circuit. At t = 0, the switch is closed and current begins to flow according to I 0 = R At this instant, the potential difference from the battery terminals is the same as that across the resistor. This initiates the charging of the capacitor. As the capacitor starts to charge, the voltage across the capacitor increases in time V C (t) = q(t) C Using Kirchhoff s loop rule for capacitors and traversing the loop CW, t < 0 0 = " V C (t) " I(t)R I(t)R = " V C (t) I(t) = dq dt ' t > 0 Solve the equation by the method of separation of variables. Let s separate terms involving charge and time dq " q & C ' ( & & dq dt = 1 ( R " q + ) * C, - = 1 R dt ) dq q " C = " 1 RC dt

14 Charging Capacitor Now we can integrate both sides of the equation which yields dq q C = 1 RC dt integrate """ ln q C" & C" ' ( = t RC q dq' = 1 q' C RC This can now be exponentiated using the fact that exp(ln x) = x to yield "t RC ( ) = Q( 1" e ) "t RC q(t) = C 1" e where Q=C is the maximum amount of charge stored on the plates. The plot below, shows the time dependence of q(t) i.e., charge as a function of time during the charging process. Once we know the charge on the capacitor we also can determine the voltage across the capacitor, V C (t) = q(t) C 0 = 1" e"t RC ( ) The graph of voltage as a function of time has the same form as charge. After a long time the charge on the capacitor approaches the value Q and the voltage across the capacitor becomes equal to. t 0 dt '

15 "&'()*)+'"&'()(*&+',-&-*',./,&0'1+1(0,&0 31+&',&'1+1(0,&0+,+,'/+())42-&&',(556,&',.7 8",)+(55/'1"& *&+',-&,)-*'&<,''&() 8"9 : <1 =>&,')-* The current that flows in the circuit is equal to the derivative in time of the charge, A BACBD8AEB AFB98A"B AEB9DA"B AFBDA"B 8A"B A)B9D dq " (t RC I(t) = = 'e = I 0 e(t RC & ',. +-&)'(&',) (.() -* '1 /+(6 ',. *- '1 42-&&', dt R31 The coefficient in front of the exponential is equal to the initial the switch was closed at t = 0. current that flows in the circuit when G As shown in the graph, the currentin the charging circuit decreases8" 9 8" 9 exponentially in time, <1,+1)1-<)'1('(*'-&',.+-&)'(&' 1()5(2)/7'1+&'*(55) t RC t " I(t) = I 0 e G :MIN = I 7(),&/,+('/,&C,0HIJ(?-K=,.,5(567'1K-5'(0(+ 0e "&'()*)+'"&'()(*&+',-&-*',./,&0'1+1( " " &' (" )*+&),) *)-,.( -/ 0"*1 '(-2 -. (" 34*('5 6" (&) 8C,0HIL?5-<9+(&(5)-?42))/,&'.)-*'1',.+-&)'(&' where = RC is called the time constant. The time constant is a measure of the 31+&',&'1+1(0,&0+,+,'/+())42-&&',(556,&', / = > &'34-((2&.7&1,8595:4-< " decay time for the exponential function. The SI units of are seconds. *&+',-&,)-*'&<,''&() <1,)+(55/' / 88"9 " 9 :8G 9 =>&,')-* (()+-&/)7()+(&?)&*-.'1/,.&),-&(5(&(5 Time Constant A BACBD8AEB AFB98A"B AEB9DA"B AFBDA"B 8A"B 31 ',. +-&)'(&',) (.() -* '1 /+(6 ',. *- '1 8" 9 8" 9 G <1,+1)1-<)'1('(*'-&',.+-&)'(&' 1()5(2)/7'1+& G :MIN 7(),&/,+('/,&C,0HIJ(?-K=,.,5(567'1K-5'( ' "&'()*)+?"*1*'*/,.0(&-.-/(&)2,&.1("0"*1&.13-0''5 E-5'(0(+-))+(2(+,'-()(*&+',-&-*',./,&0'1+ "&'()*),' &'()*)+'"&'()(*&+',-&-*',./,&0'1+1(0,&02-+)) ' 8C,0HIL?5-<9+(&(5)-?42))/,&'.)-*'1',.+-& / 8" 9 8G " 9 " : / 8 " :9 : F*'-&',.+-&)'(&' O-',+'1(',&,',(556('', (0,&0+,+,'/+())42-&&',(556,&',.7 8"9 : 31,)

16 Discharging Capacitor Suppose initially the capacitor has been charged to some value Q. For t < 0, the switch is open and the potential difference across the capacitor is given by VC = Q/C. On the other hand, the potential difference across the resistor is zero because there is no current flow, that is, I = 0. Now suppose at t = 0 the switch is closed. The capacitor will begin to discharge. The charged capacitor is now acting like a voltage source to drive current around the circuit..'fg/(0/(* 3'-4,5"" When the capacitor discharges (electrons flow from the negative plate through the wire to the positive plate), the voltage across the capacitor decreases. The capacitor is losing strength as a voltage source. Applying the loop rule by traversing the loop CCW, the equations that describe the discharging process are q C IR = 0 " q I = dq C + R dq dt = 0 & dq q = 1 RC dt dt integrate '''( t RC q(t) = Qe q ) dq' = 1 q' RC ) dt ' * & ln q - +, Q. / = t RC Q t 0

17 8" 9 8" 9 G 8HIGJ9,-./ <1,+1)1-<)'1('(*'-&',.+-&)'(&' 1()5(2)/7'1+&'*(55)-**?6(*(+'--* &" + " & G &'()(*9 (& &" + ' :MIN 7(),&/,+('/,&C,0HIJ(?-K=,.,5(567'1K-5'(0(+-))'1+(2(+,'- & 8C,0HIL?5-<9+(&(5)-?42))/,&'.)-*'1',.+-&)'(&' :3527-0/ /6(18<.051-.= >8 / 8" 9 8G " 9 RC Circuits Charging Discharging Voltage ' : 7 / 8 " :9 : F*'-&',.+-&)'(&' 1()5(2)/7'1 O-',+'1(',&,',(556('',. " 2-'&',(5/,**&+(+-))'1+(2(+,'-25(')1(),&+()/?6(*(+'-8 G G 9 :IMP &'()(*E ' " & ) " & / 8 9 8G G 9 :IMP 8HIGI9 Current Charge HQGM

18 Example (5) Consider a series RC circuit for which C = 6.0mF, R = 2.0 x 10 6 W, and E = 20V. Find, (a) the time constant of the circuit. (b) the maximum charge on the capacitor after a switch in the circuit closes. (c) If the battery is immediately taken out of the circuit once the capacitor reaches its maximum charge, how long would it take for the capacitor to drain to one-tenth its maximum charge?

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