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Electricity and Magnetism DC Circuits Resistance-Capacitance Circuits Lana Sheridan De Anza College Feb 12, 2018

Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

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Page 1: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Electricity and MagnetismDC Circuits

Resistance-Capacitance Circuits

Lana Sheridan

De Anza College

Feb 12, 2018

Page 2: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Last time

• using Kirchhoff’s laws

Page 3: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Overview

• two Kirchhoff trick problems

• resistance-capacitance circuits

Page 4: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Using Kirchhoff’s Laws examples

725QU E STION SPART 3

HALLIDAY REVISED

Pr at which energy is dissipated as thermal energy in the battery is

Pr ! i2r. (27-16)

The rate Pemf at which the chemical energy in the battery changes is

Pemf ! i!. (27-17)

Series Resistances When resistances are in series, they havethe same current. The equivalent resistance that can replace a se-ries combination of resistances is

(n resistances in series). (27-7)

Parallel Resistances When resistances are in parallel,they have the same potential difference. The equivalent resistancethat can replace a parallel combination of resistances is given by

(n resistances in parallel). (27-24)1

Req ! !

n

j!1

1Rj

Req ! !n

j!1 Rj

RC Circuits When an emf ! is applied to a resistance R and ca-pacitance C in series, as in Fig. 27-15 with the switch at a, the chargeon the capacitor increases according to

q ! C !(1 " e"t/RC) (charging a capacitor), (27-33)

in which C ! ! q0 is the equilibrium (final) charge and RC ! t isthe capacitive time constant of the circuit. During the charging, thecurrent is

(charging a capacitor). (27-34)

When a capacitor discharges through a resistance R, the charge onthe capacitor decays according to

q ! q0e"t/RC (discharging a capacitor). (27-39)

During the discharging, the current is

(discharging a capacitor). (27-40) i !dqdt

! "" q0

RC #e"t/RC

i !dqdt

! " !

R #e"t/RC

1 (a) In Fig. 27-18a, with R1 # R2, is the potential differenceacross R2 more than, less than, or equal to that across R1? (b) Is thecurrent through resistor R2 more than, less than, or equal to thatthrough resistor R1?

Fig. 27-18 Questions 1 and 2.

(a)

+–

R1 R2

R3

(b)

+–

R3

R1R2

(d)(c)

R2R1+–

R3

R3

+–

R1

R2

R

Fig. 27-21 Question 6.

2 (a) In Fig. 27-18a, are resistors R1 and R3 in series? (b) Are resistors R1 and R2 in parallel? (c) Rank the equivalent resistancesof the four circuits shown in Fig. 27-18, greatest first.

3 You are to connect resistors R1 and R2, with R1 # R2, to a bat-tery, first individually, then in series, and then in parallel. Rankthose arrangements according to the amount of current throughthe battery, greatest first.

4 In Fig. 27-19, a circuit consists ofa battery and two uniform resistors,and the section lying along an x axisis divided into five segments ofequal lengths. (a) Assume that R1 !R2 and rank the segments accordingto the magnitude of the averageelectric field in them, greatest first. (b) Now assume that R1 # R2

and then again rank the segments. (c) What is the direction of theelectric field along the x axis?

5 For each circuit in Fig. 27-20, are the resistors connected in series, in parallel, or neither?

6 Res-monster maze. In Fig. 27-21, all the resistors have aresistance of 4.0 $ and all the (ideal) batteries have an emf of 4.0V. What is the current through resistor R? (If you can find theproper loop through this maze, you can answer the question with afew seconds of mental calculation.)

Fig. 27-19 Question 4.

+ –

R1 R2

a b c d e

x

+–

+– +–

(a) (b) (c)

Fig. 27-20 Question 5.

7 A resistor R1 is wired to a battery, then resistor R2 is added inseries. Are (a) the potential difference across R1 and (b) the cur-

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 725

** View All Solutions Here **

** View All Solutions Here **

1Halliday, Resnick, Walker, 8th ed, page 725, question 6.

Page 5: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Using Kirchhoff’s Laws examples

726 CHAPTE R 27 CI RCU ITS

HALLIDAY REVISED

•4 Figure 27-27 shows a circuit of four resistors that are con-nected to a larger circuit.The graph below the circuit shows the elec-tric potential V(x) as a function of position x along the lower branchof the circuit, through resistor 4; the potential VA is 12.0 V. The graph

Fig. 27-24 Question 11.

(1) (2) (3)

sec. 27-6 Potential Difference Between Two Points•1 In Fig. 27-25, the idealbatteries have emfs !1 ! 12 V and !2 !6.0 V. What are (a) the current, the dissi-pation rate in (b) resistor 1 (4.0 ") and (c)resistor 2 (8.0 "), and the energy transferrate in (d) battery 1 and (e) battery 2? Isenergy being supplied or absorbed by (f)battery 1 and (g) battery 2?

•2 In Fig. 27-26, the ideal batteries have emfs !1 ! 150 V and !2 ! 50 Vand the resistances are R1 ! 3.0 "and R2 ! 2.0 ". If the potential at P is100 V, what is it at Q?

•3 A car battery with a 12 Vemf and an internal resistance of 0.040" is being charged with a current of 50A. What are (a) the potential differ-ence V across the terminals, (b) therate Pr of energy dissipation inside the battery, and (c) the rate Pemf

of energy conversion to chemical form? When the battery is used tosupply 50 A to the starter motor, what are (d) V and (e) Pr?

ILW

WWWSSM

–+

– +

12

R1

R2

–+

–+

Q

P

R1

R2

1 2

Fig. 27-26 Problem 2.

Fig. 27-27 Problem 4.

VA

0

0

x

x

4

1 2 3

∆VB ∆VC

V

V (V

)Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign

SSM Worked-out solution available in Student Solutions Manual

• – ••• Number of dots indicates level of problem difficulty

Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

WWW Worked-out solution is at

ILW Interactive solution is at http://www.wiley.com/college/halliday

R2 more than, less than, or equal to R1? (d) Is the total currentthrough R1 and R2 together more than, less than, or equal to thecurrent through R1 previously?

10 After the switch in Fig. 27-15 isclosed on point a, there is current ithrough resistance R. Figure 27-23gives that current for four sets ofvalues of R and capacitance C: (1) R0

and C0, (2) 2R0 and C0, (3) R0 and2C0, (4) 2R0 and 2C0. Which set goeswith which curve?

11 Figure 27-24 shows three sec-tions of circuit that are to be con-nected in turn to the same batteryvia a switch as in Fig. 27-15.The resistors are all identical, as are thecapacitors. Rank the sections according to (a) the final (equilib-rium) charge on the capacitor and (b) the time required for thecapacitor to reach 50% of its final charge, greatest first.

rent i1 through R1 now more than, less than, or the same as previ-ously? (c) Is the equivalent resistance R12 of R1 and R2 more than,less than, or equal to R1?

8 Cap-monster maze. In Fig. 27-22, all the capacitors have acapacitance of 6.0 mF, and all the batteries have an emf of 10 V.What is the charge on capacitor C? (If you can find the proper loopthrough this maze, you can answer the question with a few secondsof mental calculation.)

9 Initially, a single resistor R1 is wired to a battery. Then resistorR2 is added in parallel. Are (a) the potential difference across R1

and (b) the current i1 through R1 now more than, less than, or thesame as previously? (c) Is the equivalent resistance R12 of R1 and

C

Fig. 27-22 Question 8.

i

dc

a

b

t

Fig. 27-23 Question 10.

Fig. 27-25Problem 1.

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 726

** View All Solutions Here **

** View All Solutions Here **

1Halliday, Resnick, Walker, 8th ed, page 726, question 8.

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Time Varying Circuits

In circuits charge is not static, but moving.

Current does not necessarily have to remain constant in time.

Capacitors take some time to charge and discharge due toresistances in the wires.

Other components also cause current to behave differently atdifferent times, but for now, we will concentrate on circuits withresistors and capacitors.

Page 7: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits

Circuits with resistors and capacitors are called “RC circuits.”

720 CHAPTE R 27 CI RCU ITS

HALLIDAY REVISED

+– R2

A R1

r

a

b

c

d

V

i

i

Fig. 27-14 A single-loop circuit, show-ing how to connect an ammeter (A) and avoltmeter (V).

Fig. 27-15 When switch S is closed ona, the capacitor is charged through the re-sistor.When the switch is afterward closedon b, the capacitor discharges through theresistor.

C+–

S

Rb

a

Additional examples, video, and practice available at WileyPLUS

With Eq. 27-26 we then find that

i3 ! i1 " i2 ! #0.50 A " 0.25 A! #0.25 A.

The positive answer we obtained for i2 signals that ourchoice of direction for that current is correct. However, thenegative answers for i1 and i3 indicate that our choices for

those currents are wrong.Thus, as a last step here, we correctthe answers by reversing the arrows for i1 and i3 in Fig. 27-13and then writing

i1 = 0.50 A and i3 = 0.25 A. (Answer)

Caution: Always make any such correction as the last stepand not before calculating all the currents.

27-8 The Ammeter and the VoltmeterAn instrument used to measure currents is called an ammeter. To measure thecurrent in a wire, you usually have to break or cut the wire and insert the amme-ter so that the current to be measured passes through the meter. (In Fig. 27-14,ammeter A is set up to measure current i.)

It is essential that the resistance RA of the ammeter be very much smallerthan other resistances in the circuit. Otherwise, the very presence of the meterwill change the current to be measured.

A meter used to measure potential differences is called a voltmeter. To findthe potential difference between any two points in the circuit, the voltmeter ter-minals are connected between those points without breaking or cutting the wire.(In Fig. 27-14, voltmeter V is set up to measure the voltage across R1.)

It is essential that the resistance RV of a voltmeter be very much larger thanthe resistance of any circuit element across which the voltmeter is connected.Otherwise, the meter itself becomes an important circuit element and alters thepotential difference that is to be measured.

Often a single meter is packaged so that, by means of a switch, it can be madeto serve as either an ammeter or a voltmeter—and usually also as an ohmmeter,designed to measure the resistance of any element connected between its termi-nals. Such a versatile unit is called a multimeter.

27-9 RC CircuitsIn preceding sections we dealt only with circuits in which the currents did notvary with time. Here we begin a discussion of time-varying currents.

Charging a CapacitorThe capacitor of capacitance C in Fig. 27-15 is initially uncharged.To charge it, weclose switch S on point a. This completes an RC series circuit consisting of thecapacitor, an ideal battery of emf !, and a resistance R.

From Section 25-2, we already know that as soon as the circuit is complete,charge begins to flow (current exists) between a capacitor plate and a batteryterminal on each side of the capacitor. This current increases the charge q on theplates and the potential difference VC (! q/C) across the capacitor. When thatpotential difference equals the potential difference across the battery (whichhere is equal to the emf !), the current is zero. From Eq. 25-1 (q ! CV), the equi-librium (final) charge on the then fully charged capacitor is equal to C !.

Here we want to examine the charging process. In particular we want toknow how the charge q(t) on the capacitor plates, the potential difference VC(t)across the capacitor, and the current i(t) in the circuit vary with time during thecharging process. We begin by applying the loop rule to the circuit, traversing it

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 720

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Charging a Capacitor

When an uncharged capacitor is first connected to an electricalpotential difference, a current will flow.

Once the capacitor is fully charged however, q = C (∆V ), currenthas no where to flow and stops.

The capacitor gently “switches off” the current.

Page 9: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Charge varies with time

The charge on the capacitor changes with time.

72127-9 RC CI RCU ITSPART 3

HALLIDAY REVISED

clockwise from the negative terminal of the battery.We find

(27-30)

The last term on the left side represents the potential difference across the capac-itor. The term is negative because the capacitor’s top plate, which is connected tothe battery’s positive terminal, is at a higher potential than the lower plate. Thus,there is a drop in potential as we move down through the capacitor.

We cannot immediately solve Eq. 27-30 because it contains two variables,i and q. However, those variables are not independent but are related by

(27-31)

Substituting this for i in Eq. 27-30 and rearranging, we find

(charging equation). (27-32)

This differential equation describes the time variation of the charge q on thecapacitor in Fig. 27-15. To solve it, we need to find the function q(t) that satisfiesthis equation and also satisfies the condition that the capacitor be initiallyuncharged; that is, q ! 0 at t ! 0.

We shall soon show that the solution to Eq. 27-32 is

q ! C !(1 " e"t/RC) (charging a capacitor). (27-33)

(Here e is the exponential base, 2.718 . . . , and not the elementary charge.) Notethat Eq. 27-33 does indeed satisfy our required initial condition, because at t ! 0the term e"t/RC is unity; so the equation gives q ! 0. Note also that as t goes toinfinity (that is, a long time later), the term e"t/RC goes to zero; so the equationgives the proper value for the full (equilibrium) charge on the capacitor—namely, q ! C !.A plot of q(t) for the charging process is given in Fig. 27-16a.

The derivative of q(t) is the current i(t) charging the capacitor:

(charging a capacitor). (27-34)

A plot of i(t) for the charging process is given in Fig. 27-16b. Note that the currenthas the initial value !/R and that it decreases to zero as the capacitor becomesfully charged.

i !dqdt

! ! !

R "e"t/RC

R dqdt

#qC

! !

i !dqdt

.

! " iR "qC

! 0.

A capacitor that is being charged initially acts like ordinary connecting wire relativeto the charging current.A long time later, it acts like a broken wire.

By combining Eq. 25-1 (q ! CV ) and Eq. 27-33, we find that the potentialdifference VC(t) across the capacitor during the charging process is

(charging a capacitor). (27-35)

This tells us that VC ! 0 at t ! 0 and that VC ! ! when the capacitor becomesfully charged as t : $.

The Time ConstantThe product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensionsof time (both because the argument of an exponential must be dimensionless and

VC !qC

! !(1 " e"t/RC)Fig. 27-16 (a) A plot of Eq. 27-33,which shows the buildup of charge on thecapacitor of Fig. 27-15. (b) A plot of Eq.27-34, which shows the decline of thecharging current in the circuit of Fig. 27-15.The curves are plotted for R ! 2000 %,C ! 1 mF, and ! ! 10 V; the small trian-gles represent successive intervals of onetime constant t.

q (

C)

12

i (m

A)

6

8

4

4

2

0

0

2 4 6 8 10

2 4 6 8

Time (ms)

Time (ms)

(a)

(b)

10

µ

C

/R

The capacitor’s chargegrows as the resistor'scurrent dies out.

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 721

It is possible to determine how if changes by considering the looprule for a resistor in series with a capacitor:

E− iR −q

C= 0

Current is the rate of charge flow with time: i = dqdt .

Page 10: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Charge varies with time

If we replace i in our equation with the derivative:

E− Rdq

dt−q

C= 0

This is a differential equation. There is a way to solve suchequations to find solutions for how q depends on time.

Here, separation of the variables q and t is possible.

Page 11: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Charge varies with time

E− Rdq

dt−q

C= 0

Rearranging:

dq

dt=

CE

RC−

q

RC∫1

CE− qdq =

∫1

RCdt

The limits of our integral will be determined by the initialconditions for the situation we are considering.

Page 12: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Charging Capacitor

When charging an initially uncharged capacitor: q = 0 at t = 0

∫q0

1

CE− qdq =

∫ t0

1

RCdt

− ln(CE− q) + ln(CE− 0) =t

RC

ln

(CE

CE− q

)=

t

RC

CE

CE− q= et/RC

The solution is:

q(t) = CE(

1 − e−t/RC)

Page 13: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Charging Capacitor

q(t) = CE(

1 − e−t/RC)

This solution could also bewritten in a different way.

Notice Qmax = CE.

(E takes the place of thepotential difference.)

q(t) = Qmax

(1 − e−t/RC

)

72127-9 RC CI RCU ITSPART 3

HALLIDAY REVISED

clockwise from the negative terminal of the battery.We find

(27-30)

The last term on the left side represents the potential difference across the capac-itor. The term is negative because the capacitor’s top plate, which is connected tothe battery’s positive terminal, is at a higher potential than the lower plate. Thus,there is a drop in potential as we move down through the capacitor.

We cannot immediately solve Eq. 27-30 because it contains two variables,i and q. However, those variables are not independent but are related by

(27-31)

Substituting this for i in Eq. 27-30 and rearranging, we find

(charging equation). (27-32)

This differential equation describes the time variation of the charge q on thecapacitor in Fig. 27-15. To solve it, we need to find the function q(t) that satisfiesthis equation and also satisfies the condition that the capacitor be initiallyuncharged; that is, q ! 0 at t ! 0.

We shall soon show that the solution to Eq. 27-32 is

q ! C !(1 " e"t/RC) (charging a capacitor). (27-33)

(Here e is the exponential base, 2.718 . . . , and not the elementary charge.) Notethat Eq. 27-33 does indeed satisfy our required initial condition, because at t ! 0the term e"t/RC is unity; so the equation gives q ! 0. Note also that as t goes toinfinity (that is, a long time later), the term e"t/RC goes to zero; so the equationgives the proper value for the full (equilibrium) charge on the capacitor—namely, q ! C !.A plot of q(t) for the charging process is given in Fig. 27-16a.

The derivative of q(t) is the current i(t) charging the capacitor:

(charging a capacitor). (27-34)

A plot of i(t) for the charging process is given in Fig. 27-16b. Note that the currenthas the initial value !/R and that it decreases to zero as the capacitor becomesfully charged.

i !dqdt

! ! !

R "e"t/RC

R dqdt

#qC

! !

i !dqdt

.

! " iR "qC

! 0.

A capacitor that is being charged initially acts like ordinary connecting wire relativeto the charging current.A long time later, it acts like a broken wire.

By combining Eq. 25-1 (q ! CV ) and Eq. 27-33, we find that the potentialdifference VC(t) across the capacitor during the charging process is

(charging a capacitor). (27-35)

This tells us that VC ! 0 at t ! 0 and that VC ! ! when the capacitor becomesfully charged as t : $.

The Time ConstantThe product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensionsof time (both because the argument of an exponential must be dimensionless and

VC !qC

! !(1 " e"t/RC)Fig. 27-16 (a) A plot of Eq. 27-33,which shows the buildup of charge on thecapacitor of Fig. 27-15. (b) A plot of Eq.27-34, which shows the decline of thecharging current in the circuit of Fig. 27-15.The curves are plotted for R ! 2000 %,C ! 1 mF, and ! ! 10 V; the small trian-gles represent successive intervals of onetime constant t.

q (

C)

12

i (m

A)

6

8

4

4

2

0

0

2 4 6 8 10

2 4 6 8

Time (ms)

Time (ms)

(a)

(b)

10

µ

C

/R

The capacitor’s chargegrows as the resistor'scurrent dies out.

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 721

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RC Circuits: Charging Capacitor

Using the equation for q, an equation for current can also befound, since i = dq

dt :

i(t) =

(E

R

)e−t/RC

Dividing the charge by the capacitance C , we can also find thepotential difference across the capacitor:

|∆VC (t)| = E(1 − e−t/RC )

Page 15: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Charging Capacitor

How the solutions appear with time:

Charge:

q = Qmax (1 − e−t/RC )

72127-9 RC CI RCU ITSPART 3

HALLIDAY REVISED

clockwise from the negative terminal of the battery.We find

(27-30)

The last term on the left side represents the potential difference across the capac-itor. The term is negative because the capacitor’s top plate, which is connected tothe battery’s positive terminal, is at a higher potential than the lower plate. Thus,there is a drop in potential as we move down through the capacitor.

We cannot immediately solve Eq. 27-30 because it contains two variables,i and q. However, those variables are not independent but are related by

(27-31)

Substituting this for i in Eq. 27-30 and rearranging, we find

(charging equation). (27-32)

This differential equation describes the time variation of the charge q on thecapacitor in Fig. 27-15. To solve it, we need to find the function q(t) that satisfiesthis equation and also satisfies the condition that the capacitor be initiallyuncharged; that is, q ! 0 at t ! 0.

We shall soon show that the solution to Eq. 27-32 is

q ! C !(1 " e"t/RC) (charging a capacitor). (27-33)

(Here e is the exponential base, 2.718 . . . , and not the elementary charge.) Notethat Eq. 27-33 does indeed satisfy our required initial condition, because at t ! 0the term e"t/RC is unity; so the equation gives q ! 0. Note also that as t goes toinfinity (that is, a long time later), the term e"t/RC goes to zero; so the equationgives the proper value for the full (equilibrium) charge on the capacitor—namely, q ! C !.A plot of q(t) for the charging process is given in Fig. 27-16a.

The derivative of q(t) is the current i(t) charging the capacitor:

(charging a capacitor). (27-34)

A plot of i(t) for the charging process is given in Fig. 27-16b. Note that the currenthas the initial value !/R and that it decreases to zero as the capacitor becomesfully charged.

i !dqdt

! ! !

R "e"t/RC

R dqdt

#qC

! !

i !dqdt

.

! " iR "qC

! 0.

A capacitor that is being charged initially acts like ordinary connecting wire relativeto the charging current.A long time later, it acts like a broken wire.

By combining Eq. 25-1 (q ! CV ) and Eq. 27-33, we find that the potentialdifference VC(t) across the capacitor during the charging process is

(charging a capacitor). (27-35)

This tells us that VC ! 0 at t ! 0 and that VC ! ! when the capacitor becomesfully charged as t : $.

The Time ConstantThe product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensionsof time (both because the argument of an exponential must be dimensionless and

VC !qC

! !(1 " e"t/RC)Fig. 27-16 (a) A plot of Eq. 27-33,which shows the buildup of charge on thecapacitor of Fig. 27-15. (b) A plot of Eq.27-34, which shows the decline of thecharging current in the circuit of Fig. 27-15.The curves are plotted for R ! 2000 %,C ! 1 mF, and ! ! 10 V; the small trian-gles represent successive intervals of onetime constant t.

q (

C)

12

i (m

A)

6

8

4

4

2

0

0

2 4 6 8 10

2 4 6 8

Time (ms)

Time (ms)

(a)

(b)

10

µ

C

/R

The capacitor’s chargegrows as the resistor'scurrent dies out.

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 721

where for this circuitQmax = CE

Current:

i = Ii e−t/RC

72127-9 RC CI RCU ITSPART 3

HALLIDAY REVISED

clockwise from the negative terminal of the battery.We find

(27-30)

The last term on the left side represents the potential difference across the capac-itor. The term is negative because the capacitor’s top plate, which is connected tothe battery’s positive terminal, is at a higher potential than the lower plate. Thus,there is a drop in potential as we move down through the capacitor.

We cannot immediately solve Eq. 27-30 because it contains two variables,i and q. However, those variables are not independent but are related by

(27-31)

Substituting this for i in Eq. 27-30 and rearranging, we find

(charging equation). (27-32)

This differential equation describes the time variation of the charge q on thecapacitor in Fig. 27-15. To solve it, we need to find the function q(t) that satisfiesthis equation and also satisfies the condition that the capacitor be initiallyuncharged; that is, q ! 0 at t ! 0.

We shall soon show that the solution to Eq. 27-32 is

q ! C !(1 " e"t/RC) (charging a capacitor). (27-33)

(Here e is the exponential base, 2.718 . . . , and not the elementary charge.) Notethat Eq. 27-33 does indeed satisfy our required initial condition, because at t ! 0the term e"t/RC is unity; so the equation gives q ! 0. Note also that as t goes toinfinity (that is, a long time later), the term e"t/RC goes to zero; so the equationgives the proper value for the full (equilibrium) charge on the capacitor—namely, q ! C !.A plot of q(t) for the charging process is given in Fig. 27-16a.

The derivative of q(t) is the current i(t) charging the capacitor:

(charging a capacitor). (27-34)

A plot of i(t) for the charging process is given in Fig. 27-16b. Note that the currenthas the initial value !/R and that it decreases to zero as the capacitor becomesfully charged.

i !dqdt

! ! !

R "e"t/RC

R dqdt

#qC

! !

i !dqdt

.

! " iR "qC

! 0.

A capacitor that is being charged initially acts like ordinary connecting wire relativeto the charging current.A long time later, it acts like a broken wire.

By combining Eq. 25-1 (q ! CV ) and Eq. 27-33, we find that the potentialdifference VC(t) across the capacitor during the charging process is

(charging a capacitor). (27-35)

This tells us that VC ! 0 at t ! 0 and that VC ! ! when the capacitor becomesfully charged as t : $.

The Time ConstantThe product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensionsof time (both because the argument of an exponential must be dimensionless and

VC !qC

! !(1 " e"t/RC)Fig. 27-16 (a) A plot of Eq. 27-33,which shows the buildup of charge on thecapacitor of Fig. 27-15. (b) A plot of Eq.27-34, which shows the decline of thecharging current in the circuit of Fig. 27-15.The curves are plotted for R ! 2000 %,C ! 1 mF, and ! ! 10 V; the small trian-gles represent successive intervals of onetime constant t.

q (

C)

12

i (m

A)

6

8

4

4

2

0

0

2 4 6 8 10

2 4 6 8

Time (ms)

Time (ms)

(a)

(b)

10 µ

C

/R

The capacitor’s chargegrows as the resistor'scurrent dies out.

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 721

where for this circuit Ii =ER

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RC Circuits: Time Constant

τ = RC

τ is called the time constant of the circuit.

This gives the time for the current in the circuit to fall to 1/e ofits initial value.

It is useful for comparing the “relaxation time” of differentRC-circuits.

Page 17: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Discharging Capacitor

Imagine that we have charged up the capacitor, so that the chargeon it is Qi .

Now we flip the switch to b, the battery is disconnected, butcharge flows off the capacitor, creating a current:

720 CHAPTE R 27 CI RCU ITS

HALLIDAY REVISED

+– R2

A R1

r

a

b

c

d

V

i

i

Fig. 27-14 A single-loop circuit, show-ing how to connect an ammeter (A) and avoltmeter (V).

Fig. 27-15 When switch S is closed ona, the capacitor is charged through the re-sistor.When the switch is afterward closedon b, the capacitor discharges through theresistor.

C+–

S

Rb

a

Additional examples, video, and practice available at WileyPLUS

With Eq. 27-26 we then find that

i3 ! i1 " i2 ! #0.50 A " 0.25 A! #0.25 A.

The positive answer we obtained for i2 signals that ourchoice of direction for that current is correct. However, thenegative answers for i1 and i3 indicate that our choices for

those currents are wrong.Thus, as a last step here, we correctthe answers by reversing the arrows for i1 and i3 in Fig. 27-13and then writing

i1 = 0.50 A and i3 = 0.25 A. (Answer)

Caution: Always make any such correction as the last stepand not before calculating all the currents.

27-8 The Ammeter and the VoltmeterAn instrument used to measure currents is called an ammeter. To measure thecurrent in a wire, you usually have to break or cut the wire and insert the amme-ter so that the current to be measured passes through the meter. (In Fig. 27-14,ammeter A is set up to measure current i.)

It is essential that the resistance RA of the ammeter be very much smallerthan other resistances in the circuit. Otherwise, the very presence of the meterwill change the current to be measured.

A meter used to measure potential differences is called a voltmeter. To findthe potential difference between any two points in the circuit, the voltmeter ter-minals are connected between those points without breaking or cutting the wire.(In Fig. 27-14, voltmeter V is set up to measure the voltage across R1.)

It is essential that the resistance RV of a voltmeter be very much larger thanthe resistance of any circuit element across which the voltmeter is connected.Otherwise, the meter itself becomes an important circuit element and alters thepotential difference that is to be measured.

Often a single meter is packaged so that, by means of a switch, it can be madeto serve as either an ammeter or a voltmeter—and usually also as an ohmmeter,designed to measure the resistance of any element connected between its termi-nals. Such a versatile unit is called a multimeter.

27-9 RC CircuitsIn preceding sections we dealt only with circuits in which the currents did notvary with time. Here we begin a discussion of time-varying currents.

Charging a CapacitorThe capacitor of capacitance C in Fig. 27-15 is initially uncharged.To charge it, weclose switch S on point a. This completes an RC series circuit consisting of thecapacitor, an ideal battery of emf !, and a resistance R.

From Section 25-2, we already know that as soon as the circuit is complete,charge begins to flow (current exists) between a capacitor plate and a batteryterminal on each side of the capacitor. This current increases the charge q on theplates and the potential difference VC (! q/C) across the capacitor. When thatpotential difference equals the potential difference across the battery (whichhere is equal to the emf !), the current is zero. From Eq. 25-1 (q ! CV), the equi-librium (final) charge on the then fully charged capacitor is equal to C !.

Here we want to examine the charging process. In particular we want toknow how the charge q(t) on the capacitor plates, the potential difference VC(t)across the capacitor, and the current i(t) in the circuit vary with time during thecharging process. We begin by applying the loop rule to the circuit, traversing it

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 720

−Rdq

dt−q

C= 0

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RC Circuits: Discharging Capacitor

What happens to the charge on the capacitor?

q(t) = Qi e−t/RC

It decreases exponentially with time.

Page 19: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Discharging Capacitor

What happens to the charge on the capacitor?

q(t) = Qi e−t/RC

It decreases exponentially with time.

Page 20: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Discharging Capacitor

−Rdq

dt−q

C= 0

When discharging an initially charged capacitor: q = Qi at t = 0

∫qQi

1

qdq = −

∫ t0

1

RCdt

ln(q) − ln(Qi ) = −t

RC

ln

(q

Qi

)= −

t

RC

The solution is:

q(t) = Qi e−t/RC

Page 21: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits: Discharging CapacitorWhat happens to the current?

i(t) = −Ii e−t/RC

where Ii =Qi

RC

The negative sign means the current flows in the opposite directionthrough the resistor when discharging as compared with charging.

72127-9 RC CI RCU ITSPART 3

HALLIDAY REVISED

clockwise from the negative terminal of the battery.We find

(27-30)

The last term on the left side represents the potential difference across the capac-itor. The term is negative because the capacitor’s top plate, which is connected tothe battery’s positive terminal, is at a higher potential than the lower plate. Thus,there is a drop in potential as we move down through the capacitor.

We cannot immediately solve Eq. 27-30 because it contains two variables,i and q. However, those variables are not independent but are related by

(27-31)

Substituting this for i in Eq. 27-30 and rearranging, we find

(charging equation). (27-32)

This differential equation describes the time variation of the charge q on thecapacitor in Fig. 27-15. To solve it, we need to find the function q(t) that satisfiesthis equation and also satisfies the condition that the capacitor be initiallyuncharged; that is, q ! 0 at t ! 0.

We shall soon show that the solution to Eq. 27-32 is

q ! C !(1 " e"t/RC) (charging a capacitor). (27-33)

(Here e is the exponential base, 2.718 . . . , and not the elementary charge.) Notethat Eq. 27-33 does indeed satisfy our required initial condition, because at t ! 0the term e"t/RC is unity; so the equation gives q ! 0. Note also that as t goes toinfinity (that is, a long time later), the term e"t/RC goes to zero; so the equationgives the proper value for the full (equilibrium) charge on the capacitor—namely, q ! C !.A plot of q(t) for the charging process is given in Fig. 27-16a.

The derivative of q(t) is the current i(t) charging the capacitor:

(charging a capacitor). (27-34)

A plot of i(t) for the charging process is given in Fig. 27-16b. Note that the currenthas the initial value !/R and that it decreases to zero as the capacitor becomesfully charged.

i !dqdt

! ! !

R "e"t/RC

R dqdt

#qC

! !

i !dqdt

.

! " iR "qC

! 0.

A capacitor that is being charged initially acts like ordinary connecting wire relativeto the charging current.A long time later, it acts like a broken wire.

By combining Eq. 25-1 (q ! CV ) and Eq. 27-33, we find that the potentialdifference VC(t) across the capacitor during the charging process is

(charging a capacitor). (27-35)

This tells us that VC ! 0 at t ! 0 and that VC ! ! when the capacitor becomesfully charged as t : $.

The Time ConstantThe product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensionsof time (both because the argument of an exponential must be dimensionless and

VC !qC

! !(1 " e"t/RC)Fig. 27-16 (a) A plot of Eq. 27-33,which shows the buildup of charge on thecapacitor of Fig. 27-15. (b) A plot of Eq.27-34, which shows the decline of thecharging current in the circuit of Fig. 27-15.The curves are plotted for R ! 2000 %,C ! 1 mF, and ! ! 10 V; the small trian-gles represent successive intervals of onetime constant t.

q (

C)

12

i (m

A)

6

8

4

4

2

0

0

2 4 6 8 10

2 4 6 8

Time (ms)

Time (ms)

(a)

(b)

10

µ

C

/R

The capacitor’s chargegrows as the resistor'scurrent dies out.

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 721

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RC Circuits: Discharging Capacitor

Multiplying the current by the resistance R gives the potentialdifference across the resistor:

|∆VR(t)| = (∆V )i e−t/RC

The same expression describes the potential difference across thecapacitor!

|∆VC (t)| = (∆V )i e−t/RC

where (∆V )i = IiR = QiC .

Page 23: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits QuestionQuick Quiz 28.5: Consider the circuit shown and assume thebattery has no internal resistance.

28.4 RC Circuits 849

The following dimensional analysis shows that t has units of time:

3t 4 5 3RC 4 5 c aDVIb a Q

DVb d 5 c Q

Q /Dtd 5 3Dt 4 5 T

Because t 5 RC has units of time, the combination t/RC is dimensionless, as it must be to be an exponent of e in Equations 28.14 and 28.15. The energy supplied by the battery during the time interval required to fully charge the capacitor is Q maxe 5 Ce2. After the capacitor is fully charged, the energy stored in the capacitor is 12Q maxe 5 12Ce2, which is only half the energy out-put of the battery. It is left as a problem (Problem 68) to show that the remaining half of the energy supplied by the battery appears as internal energy in the resistor.

Discharging a CapacitorImagine that the capacitor in Figure 28.16b is completely charged. An initial poten-tial difference Q i/C exists across the capacitor, and there is zero potential differ-ence across the resistor because i 5 0. If the switch is now thrown to position b at t 5 0 (Fig. 28.16c), the capacitor begins to discharge through the resistor. At some time t during the discharge, the current in the circuit is i and the charge on the capacitor is q. The circuit in Figure 28.16c is the same as the circuit in Figure 28.16b except for the absence of the battery. Therefore, we eliminate the emf e from Equa-tion 28.11 to obtain the appropriate loop equation for the circuit in Figure 28.16c:

2qC

2 iR 5 0 (28.17)

When we substitute i 5 dq/dt into this expression, it becomes

2R dqdt

5qC

dqq 5 2

1RC

dt

Integrating this expression using q 5 Q i at t 5 0 gives

3q

Q i

dqq 5 2

1RC

3t

0 dt

ln a qQ i

b 5 2t

RC

q 1t 2 5 Q ie2t/RC (28.18)

Differentiating Equation 28.18 with respect to time gives the instantaneous current as a function of time:

i 1t 2 5 2Q i

RC e2t/RC (28.19)

where Q i /RC 5 Ii is the initial current. The negative sign indicates that as the capacitor discharges, the current direction is opposite its direction when the capaci-tor was being charged. (Compare the current directions in Figs. 28.16b and 28.16c.) Both the charge on the capacitor and the current decay exponentially at a rate characterized by the time constant t 5 RC.

Q uick Quiz 28.5 Consider the circuit in Figure 28.18 and assume the battery has no internal resistance. (i) Just after the switch is closed, what is the current in the battery? (a) 0 (b) e/2R (c) 2e/R (d) e/R (e) impossible to determine (ii) After a very long time, what is the current in the battery? Choose from the same choices.

�W Charge as a function of time for a discharging capacitor

�W Current as a function of time for a discharging capacitor

C

RRe

!

"

Figure 28.18 (Quick Quiz 28.5) How does the current vary after the switch is closed?

(i) Just after the switch is closed, what is the current in thebattery?

(A) 0

(B) E/2R

(C) 2E/R

(D) E/R

Page 24: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits QuestionQuick Quiz 28.5: Consider the circuit shown and assume thebattery has no internal resistance.

28.4 RC Circuits 849

The following dimensional analysis shows that t has units of time:

3t 4 5 3RC 4 5 c aDVIb a Q

DVb d 5 c Q

Q /Dtd 5 3Dt 4 5 T

Because t 5 RC has units of time, the combination t/RC is dimensionless, as it must be to be an exponent of e in Equations 28.14 and 28.15. The energy supplied by the battery during the time interval required to fully charge the capacitor is Q maxe 5 Ce2. After the capacitor is fully charged, the energy stored in the capacitor is 12Q maxe 5 12Ce2, which is only half the energy out-put of the battery. It is left as a problem (Problem 68) to show that the remaining half of the energy supplied by the battery appears as internal energy in the resistor.

Discharging a CapacitorImagine that the capacitor in Figure 28.16b is completely charged. An initial poten-tial difference Q i/C exists across the capacitor, and there is zero potential differ-ence across the resistor because i 5 0. If the switch is now thrown to position b at t 5 0 (Fig. 28.16c), the capacitor begins to discharge through the resistor. At some time t during the discharge, the current in the circuit is i and the charge on the capacitor is q. The circuit in Figure 28.16c is the same as the circuit in Figure 28.16b except for the absence of the battery. Therefore, we eliminate the emf e from Equa-tion 28.11 to obtain the appropriate loop equation for the circuit in Figure 28.16c:

2qC

2 iR 5 0 (28.17)

When we substitute i 5 dq/dt into this expression, it becomes

2R dqdt

5qC

dqq 5 2

1RC

dt

Integrating this expression using q 5 Q i at t 5 0 gives

3q

Q i

dqq 5 2

1RC

3t

0 dt

ln a qQ i

b 5 2t

RC

q 1t 2 5 Q ie2t/RC (28.18)

Differentiating Equation 28.18 with respect to time gives the instantaneous current as a function of time:

i 1t 2 5 2Q i

RC e2t/RC (28.19)

where Q i /RC 5 Ii is the initial current. The negative sign indicates that as the capacitor discharges, the current direction is opposite its direction when the capaci-tor was being charged. (Compare the current directions in Figs. 28.16b and 28.16c.) Both the charge on the capacitor and the current decay exponentially at a rate characterized by the time constant t 5 RC.

Q uick Quiz 28.5 Consider the circuit in Figure 28.18 and assume the battery has no internal resistance. (i) Just after the switch is closed, what is the current in the battery? (a) 0 (b) e/2R (c) 2e/R (d) e/R (e) impossible to determine (ii) After a very long time, what is the current in the battery? Choose from the same choices.

�W Charge as a function of time for a discharging capacitor

�W Current as a function of time for a discharging capacitor

C

RRe

!

"

Figure 28.18 (Quick Quiz 28.5) How does the current vary after the switch is closed?

(ii) After a very long time, what is the current in the battery?

(A) 0

(B) E/2R

(C) 2E/R

(D) E/R

Page 25: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits Example 28.10

Consider a capacitor of capacitance C that is being dischargedthrough a resistor of resistance R as shown in the figure.

28.4 RC Circuits 847

Charging a CapacitorFigure 28.16 shows a simple series RC circuit. Let’s assume the capacitor in this cir-cuit is initially uncharged. There is no current while the switch is open (Fig. 28.16a). If the switch is thrown to position a at t 5 0 (Fig. 28.16b), however, charge begins to flow, setting up a current in the circuit, and the capacitor begins to charge.3 Notice that during charging, charges do not jump across the capacitor plates because the gap between the plates represents an open circuit. Instead, charge is transferred between each plate and its connecting wires due to the electric field established in the wires by the battery until the capacitor is fully charged. As the plates are being charged, the potential difference across the capacitor increases. The value of the maximum charge on the plates depends on the voltage of the battery. Once the maximum charge is reached, the current in the circuit is zero because the potential difference across the capacitor matches that supplied by the battery. To analyze this circuit quantitatively, let’s apply Kirchhoff’s loop rule to the cir-cuit after the switch is thrown to position a. Traversing the loop in Figure 28.16b clockwise gives

e 2qC

2 iR 5 0 (28.11)

where q/C is the potential difference across the capacitor and iR is the potential difference across the resistor. We have used the sign conventions discussed earlier for the signs on e and iR. The capacitor is traversed in the direction from the posi-tive plate to the negative plate, which represents a decrease in potential. Therefore, we use a negative sign for this potential difference in Equation 28.11. Note that lowercase q and i are instantaneous values that depend on time (as opposed to steady-state values) as the capacitor is being charged. We can use Equation 28.11 to find the initial current Ii in the circuit and the maximum charge Q max on the capacitor. At the instant the switch is thrown to posi-tion a (t 5 0), the charge on the capacitor is zero. Equation 28.11 shows that the initial current Ii in the circuit is a maximum and is given by

Ii 5eR 1current at t 5 0 2 (28.12)

At this time, the potential difference from the battery terminals appears entirely across the resistor. Later, when the capacitor is charged to its maximum value Q max, charges cease to flow, the current in the circuit is zero, and the potential difference from the battery terminals appears entirely across the capacitor. Substituting i 5 0 into Equation 28.11 gives the maximum charge on the capacitor:

Q max 5 Ce (maximum charge) (28.13)

To determine analytical expressions for the time dependence of the charge and current, we must solve Equation 28.11, a single equation containing two variables q and i. The current in all parts of the series circuit must be the same. Therefore, the current in the resistance R must be the same as the current between each capacitor plate and the wire connected to it. This current is equal to the time rate of change of the charge on the capacitor plates. Therefore, we substitute i 5 dq/dt into Equa-tion 28.11 and rearrange the equation:

dqdt

5eR

2q

RC

To find an expression for q, we solve this separable differential equation as follows. First combine the terms on the right-hand side:

dqdt

5CeRC

2q

RC5 2

q 2 CeRC

3In previous discussions of capacitors, we assumed a steady-state situation, in which no current was present in any branch of the circuit containing a capacitor. Now we are considering the case before the steady-state condition is real-ized; in this situation, charges are moving and a current exists in the wires connected to the capacitor.

R

C

b

a

e

Ri

C

b

a

e

Ri

C

b

a

! "

! "

e! "

! "

! "

When the switch is thrown to position a, the capacitor begins to charge up.

When the switch is thrown to position b, the capacitor discharges.

a

b

c

Figure 28.16 A capacitor in series with a resistor, switch, and battery.

After how many time constants is the charge on the capacitorone-fourth its initial value?

Page 26: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits Example 28.10

After how many time constants is the charge on the capacitorone-fourth its initial value?

q(t) = Qi e−t/RC

Let T be the time when the charge is 1/4 of the initial charge.

q(T )

Qi=

1

4

e−T/τ =1

4T

τ= ln 4

So,T = (ln 4)τ = 1.39τ

Page 27: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits Example 28.10

After how many time constants is the charge on the capacitorone-fourth its initial value?

q(t) = Qi e−t/RC

Let T be the time when the charge is 1/4 of the initial charge.

q(T )

Qi=

1

4

e−T/τ =1

4T

τ= ln 4

So,T = (ln 4)τ = 1.39τ

Page 28: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits Example 28.10

Consider a capacitor of capacitance C that is being dischargedthrough a resistor of resistance R as shown in the figure.

28.4 RC Circuits 847

Charging a CapacitorFigure 28.16 shows a simple series RC circuit. Let’s assume the capacitor in this cir-cuit is initially uncharged. There is no current while the switch is open (Fig. 28.16a). If the switch is thrown to position a at t 5 0 (Fig. 28.16b), however, charge begins to flow, setting up a current in the circuit, and the capacitor begins to charge.3 Notice that during charging, charges do not jump across the capacitor plates because the gap between the plates represents an open circuit. Instead, charge is transferred between each plate and its connecting wires due to the electric field established in the wires by the battery until the capacitor is fully charged. As the plates are being charged, the potential difference across the capacitor increases. The value of the maximum charge on the plates depends on the voltage of the battery. Once the maximum charge is reached, the current in the circuit is zero because the potential difference across the capacitor matches that supplied by the battery. To analyze this circuit quantitatively, let’s apply Kirchhoff’s loop rule to the cir-cuit after the switch is thrown to position a. Traversing the loop in Figure 28.16b clockwise gives

e 2qC

2 iR 5 0 (28.11)

where q/C is the potential difference across the capacitor and iR is the potential difference across the resistor. We have used the sign conventions discussed earlier for the signs on e and iR. The capacitor is traversed in the direction from the posi-tive plate to the negative plate, which represents a decrease in potential. Therefore, we use a negative sign for this potential difference in Equation 28.11. Note that lowercase q and i are instantaneous values that depend on time (as opposed to steady-state values) as the capacitor is being charged. We can use Equation 28.11 to find the initial current Ii in the circuit and the maximum charge Q max on the capacitor. At the instant the switch is thrown to posi-tion a (t 5 0), the charge on the capacitor is zero. Equation 28.11 shows that the initial current Ii in the circuit is a maximum and is given by

Ii 5eR 1current at t 5 0 2 (28.12)

At this time, the potential difference from the battery terminals appears entirely across the resistor. Later, when the capacitor is charged to its maximum value Q max, charges cease to flow, the current in the circuit is zero, and the potential difference from the battery terminals appears entirely across the capacitor. Substituting i 5 0 into Equation 28.11 gives the maximum charge on the capacitor:

Q max 5 Ce (maximum charge) (28.13)

To determine analytical expressions for the time dependence of the charge and current, we must solve Equation 28.11, a single equation containing two variables q and i. The current in all parts of the series circuit must be the same. Therefore, the current in the resistance R must be the same as the current between each capacitor plate and the wire connected to it. This current is equal to the time rate of change of the charge on the capacitor plates. Therefore, we substitute i 5 dq/dt into Equa-tion 28.11 and rearrange the equation:

dqdt

5eR

2q

RC

To find an expression for q, we solve this separable differential equation as follows. First combine the terms on the right-hand side:

dqdt

5CeRC

2q

RC5 2

q 2 CeRC

3In previous discussions of capacitors, we assumed a steady-state situation, in which no current was present in any branch of the circuit containing a capacitor. Now we are considering the case before the steady-state condition is real-ized; in this situation, charges are moving and a current exists in the wires connected to the capacitor.

R

C

b

a

e

Ri

C

b

a

e

Ri

C

b

a

! "

! "

e! "

! "

! "

When the switch is thrown to position a, the capacitor begins to charge up.

When the switch is thrown to position b, the capacitor discharges.

a

b

c

Figure 28.16 A capacitor in series with a resistor, switch, and battery.

The energy stored in the capacitor decreases with time as thecapacitor discharges. After how many time constants is this storedenergy one-fourth its initial value?

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RC Circuits Example 28.10

After how many time constants is this stored energy one-fourth itsinitial value?

U =q2

2C

Now let T be the time when the energy stored is 1/4 of the initialenergy.

U(T )

Ui=

1

4

q(T )2

Q2i

=1

4

e−T/τ =1

2

So,T = (ln 2)τ = 0.693τ

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RC Circuits Example 28.10

After how many time constants is this stored energy one-fourth itsinitial value?

U =q2

2C

Now let T be the time when the energy stored is 1/4 of the initialenergy.

U(T )

Ui=

1

4

q(T )2

Q2i

=1

4

e−T/τ =1

2

So,T = (ln 2)τ = 0.693τ

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RC Circuits Example 28.11: Energy delivered

A 5.00 µF capacitor is charged to a potential difference of 800 Vand then discharged through a resistor. How much energy isdelivered to the resistor in the time interval required to fullydischarge the capacitor?

28.4 RC Circuits 847

Charging a CapacitorFigure 28.16 shows a simple series RC circuit. Let’s assume the capacitor in this cir-cuit is initially uncharged. There is no current while the switch is open (Fig. 28.16a). If the switch is thrown to position a at t 5 0 (Fig. 28.16b), however, charge begins to flow, setting up a current in the circuit, and the capacitor begins to charge.3 Notice that during charging, charges do not jump across the capacitor plates because the gap between the plates represents an open circuit. Instead, charge is transferred between each plate and its connecting wires due to the electric field established in the wires by the battery until the capacitor is fully charged. As the plates are being charged, the potential difference across the capacitor increases. The value of the maximum charge on the plates depends on the voltage of the battery. Once the maximum charge is reached, the current in the circuit is zero because the potential difference across the capacitor matches that supplied by the battery. To analyze this circuit quantitatively, let’s apply Kirchhoff’s loop rule to the cir-cuit after the switch is thrown to position a. Traversing the loop in Figure 28.16b clockwise gives

e 2qC

2 iR 5 0 (28.11)

where q/C is the potential difference across the capacitor and iR is the potential difference across the resistor. We have used the sign conventions discussed earlier for the signs on e and iR. The capacitor is traversed in the direction from the posi-tive plate to the negative plate, which represents a decrease in potential. Therefore, we use a negative sign for this potential difference in Equation 28.11. Note that lowercase q and i are instantaneous values that depend on time (as opposed to steady-state values) as the capacitor is being charged. We can use Equation 28.11 to find the initial current Ii in the circuit and the maximum charge Q max on the capacitor. At the instant the switch is thrown to posi-tion a (t 5 0), the charge on the capacitor is zero. Equation 28.11 shows that the initial current Ii in the circuit is a maximum and is given by

Ii 5eR 1current at t 5 0 2 (28.12)

At this time, the potential difference from the battery terminals appears entirely across the resistor. Later, when the capacitor is charged to its maximum value Q max, charges cease to flow, the current in the circuit is zero, and the potential difference from the battery terminals appears entirely across the capacitor. Substituting i 5 0 into Equation 28.11 gives the maximum charge on the capacitor:

Q max 5 Ce (maximum charge) (28.13)

To determine analytical expressions for the time dependence of the charge and current, we must solve Equation 28.11, a single equation containing two variables q and i. The current in all parts of the series circuit must be the same. Therefore, the current in the resistance R must be the same as the current between each capacitor plate and the wire connected to it. This current is equal to the time rate of change of the charge on the capacitor plates. Therefore, we substitute i 5 dq/dt into Equa-tion 28.11 and rearrange the equation:

dqdt

5eR

2q

RC

To find an expression for q, we solve this separable differential equation as follows. First combine the terms on the right-hand side:

dqdt

5CeRC

2q

RC5 2

q 2 CeRC

3In previous discussions of capacitors, we assumed a steady-state situation, in which no current was present in any branch of the circuit containing a capacitor. Now we are considering the case before the steady-state condition is real-ized; in this situation, charges are moving and a current exists in the wires connected to the capacitor.

R

C

b

a

e

Ri

C

b

a

e

Ri

C

b

a

! "

! "

e! "

! "

! "

When the switch is thrown to position a, the capacitor begins to charge up.

When the switch is thrown to position b, the capacitor discharges.

a

b

c

Figure 28.16 A capacitor in series with a resistor, switch, and battery.

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RC Circuits Example 28.11: Energy delivered

C = 5.00 µF, ∆V = 800 V. How much energy is delivered to theresistor in the time interval required to fully discharge thecapacitor?

Two ways: (1) Energy conservation. (2) “sum up” the powerdelivered over the time.

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RC Circuits Example 28.11: Energy delivered

Way (1): Energy not stored in the resistor must have beendelivered to the resistor.

∆UC + ∆ER = 0

∆ER = UC ,i − UC ,f

=Q2

i

2C− 0

=(C ∆V )2

2C− 0

=C (∆V )2

2

∆ER = 1.60 J

Page 34: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits Example 28.11: Energy delivered

Way (1): Energy not stored in the resistor must have beendelivered to the resistor.

∆UC + ∆ER = 0

∆ER = UC ,i − UC ,f

=Q2

i

2C− 0

=(C ∆V )2

2C− 0

=C (∆V )2

2

∆ER = 1.60 J

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RC Circuits Example 28.11: Energy deliveredWay (2): integrate the power delivered over the time

P =dW

dt

∆ER =

∫P dt

=

∫i2R dt

= R

∫∞0

I2i e−2t/RC dt

= I2i R

[−RC

2e−2t/RC

]∞0

=

(∆V

R

)2 R2C

2

=C (∆V )2

2

Same! ∆ER = 1.60 J

Page 36: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

RC Circuits Example 28.11: Energy deliveredWay (2): integrate the power delivered over the time

P =dW

dt

∆ER =

∫P dt

=

∫i2R dt

= R

∫∞0

I2i e−2t/RC dt

= I2i R

[−RC

2e−2t/RC

]∞0

=

(∆V

R

)2 R2C

2

=C (∆V )2

2

Same! ∆ER = 1.60 J

Page 37: Electricity and Magnetism DC Circuits Resistance ...nebula2.deanza.edu/~lanasheridan/4B/Phys4B-Lecture24-san.pdfElectricity and Magnetism DC Circuits Resistance-Capacitance Circuits

Summary

• resistance-capacitance circuits

Next Test on Feb 15.

HomeworkSerway & Jewett:

• NEW: Ch 28, onward from page 857. Problems: 37, 41, 43,45, 65, 71