Tài liệu Physics exercises_solution: Chapter 26 ppt

f Potential drops as move through each resistor in the direction of the current; false.. g Potential drops as move through each resistor in the direction of the current, so V b Vc; fal

26.1: a) 12.3

20

132

V240

;A5.732

V240

2 1 eq 1

2 1

2 1 1

2 1 eq

R R

R R R

R R

R R R

1 2 eq 1

2 1

2 1

R R

R R R R

R R

R R

26.3: For resistors in series, the currents are the same and the voltages add a) true

b) false c) P I2R i same, R different so P different; false d) true e) V = IR I same, R different; false f) Potential drops as move through each resistor in the

direction of the current; false g) Potential drops as move through each resistor in the direction of the current, so V b Vc; false h) true

26.4: a) False, current divides at junction a.

b) True by charge conservation

c) True

R I V

V1  2,so  1

d) False PIV.V1V2,butI1 I2,soP1P2

e) False 2.Since 2 1, 2 1

P P R R IV

f) True Potential is independent of path

g) True Charges lose potential energy (as heat) in R1

h) False See answer to (g)

i) False They are at the same potential.

26.5: a) 0.8

8.4

16

.1

14

.2

2 6 1

2 4

2

;

W

8 4

2 8

26.6: a) Req R i 2.41.64.88.8

b) The current in each resistor is the same and is 3.18A

8.8

V28

c) The current through the battery equals the current of (b), 3.18 A

d) V2.4  IR2.4 (3.18A)(2.4)7.64V;V1.6 IR1.6 (3.18A)(1.6)

.V3.15)8.4)(

A18.3(

2 6 1

2 4

2

2 8

000,9

)V120

10

.12

100

.6

100

.3

eq

A0.12)00.5()V00.6

12

;A00.3)0.12(412

3

;A00.8)0.12(63

.700.5

100

.100.3

eq

A0.16)00.3()V0.48

12

;A00.4)0.16(124

4

3 1 7

.4

150

.1

120

.8

11

1

4 3 2

V0.6

1 1 1 eq

33.1V

33.1)99.0()A34.1(

2 2 234

I

.A296.050.4

V33.1and

A887.050.1

V33.1

4

4 3

V

26.11: Using the same circuit as in Problem 27.10, with all resistances the same:

.4

350

.41

1

4 3 2 1 234 1

50.100.6

V00.9

1 4 3 2 eq

W13.10)50.4()A50.1

.4

250

.41

3 2 1 23 1

And so:

.A667.02

1A,

33.175.6

V00.9

1 3 2 eq

4

1,

W96.7)50.4()A33.1

e) So R2 and R3 are brighter than before, while R is fainter The amount of current 1

flow is all that determines the power output of these bulbs since their resistances are equal

26.12: From Ohm’s law, the voltage drop across the 6.00  resistor is V = IR =

V

24.0)

= (7.00 A)( 25.0 ) = 175 V, and total voltage drop across the top branch of the circuit is

175 + 24.0 = 199 V, which is also the voltage drop across the 20.0  resistor The

current through the 20.0  resistor is then I V R199V 209.95A

26.13: Current through 2.00-  resistor is 6.00 A Current through 1.00-  resistor also

is

6.00 A and the voltage is 6.00 V Voltage across the 6.00-  resistor is 12.0 V + 6.0 V = 18.0 V Current through the 6.00-  resistor is (18.0V) (6.00)3.00A The battery voltage is 18.0 V

26.14: a) The filaments must be connected such that the current can flow through each

separately, and also through both in parallel, yielding three possible current flows The parallel situation always has less resistance than any of the individual members, so it will give the highest power output of 180 W, while the other two must give power outputs of

60 W and

120 W

.120W

120

)V120(W

120and,240W

60

)V120(W

60

2 2

2

2 2

1 1

)V120()

(

2 1

1201

2401

2 1

1 1 2

2 1

V P

b) If R burns out, the 120 W setting stays the same, the 60 W setting does not work 1

and the 180 W setting goes to 120 W: brightnesses of zero, medium and medium

c) If R burns out, the 60 W setting stays the same, the 120 W setting does not work, 2

and the 180 W setting is now 60 W: brightnesses of low, zero and low

)800400

b)  (0.100A) (400)4.0W;  2 (0.100A)2(800)

800 2

2

P

.W12W8W4W

V120267

800

1400

1

eq total

R

.A150.0)A449.0(800400

400

;A30.0)A449.0(800400

2

P

.W54W18W

26.16: a) 72

W200

)V120(

;240W

60

)V120

W 200

2 2

W

P

V R

P

V R

.A769.0)72240

(

V240ε

W 200 W

I

b)

W.6.42)72()A769.0(

;W142)240()A769.0

W 200 2

I R

P

givesandQ mc T Pt

s1001.1W

0.20

)C0.40()KkgJ4190()kg100.0

t

1

R in parallel:

A1

A)2()5()10(

So I2 0.50A.R1 and R2 are in parallel, so

)5()A2()A50.0

)A00.2()00.3()A00.6(V0

c) Using a counterclockwise loop in the bottom half of the circuit:

.V0.420

)00.6()A00.4()00.3()A00.6

ε d) If the circuit is broken at point x, then the current in the 28 V battery is:

.A50.35.003.00

V0

26.20: From the given currents in the diagram, the current through the middle branch

of the circuit must be 1.00 A (the difference between 2.00 A and 1.00 A) We now use Kirchoff’s Rules, passing counterclockwise around the top loop:

A)(1.00

26.21: a) The sum of the currents that enter the junction below the 3- resistor equals 3.00 A + 5.00 A = 8.00 A.

b) Using the lower left loop:

.V0.36

0A00.800.3A00.300.4

Using the lower right loop:

.V0.54

0A00.800.3A00.500.6

c) Using the top loop:

A00.2

V0.180

V0.36A00.2V0

26.22: From the circuit in Fig 26.42, we use Kirchhoff’s Rules to find the currents, I1

to the left through the 10 V battery, I to the right through 5 V battery, and 2 I3to the right through the10 resistor:

Upper loop:

V0.5

0V00.500

.400.100

.300.2V0.10

2 1 2

1

2 1

I

I I

Lower loop:5.00V1.004.00I2 10.0I3 0

V00

Along with I1 I2 I3, we can solve for the three currents and find:

I1 0.800A,I2 0.200A,I3 0.600A b) V ab 0.200A4.00  0.800A3.003.20V

26.23: After reversing the polarity of the 10-V battery in the circuit of Fig 26.42, the

only change in the equations from Problem 26.22 is the upper loop where the 10 V battery is:

26.24: After switching the 5-V battery for a 20-V battery in the circuit of Fig 26.42,

there is a change in the equations from Problem 26.22 in both the upper and lower loops:Upper loop: 10.0V2.003.00I11.004.00I2 20.00V0

A6.1,

A4

2 2

2 4

2

P

Ptotal P2P3P4 P7 4W

26.26: a) If the 12-V battery is removed and then replaced with the opposite polarity,

the current will flow in the clockwise direction, with magnitude;

.A116

V4V12

b) V ab R4 R7Iε4 47 1A 4V7V

26.27: a) Since all the external resistors are equal, the current must be symmetrical

through them That is, there can be no current through the resistor R for that would imply

an imbalance

in currents through the other resistors

With no current going through R, the circuit is like that shown below at right.

So the equivalent resistance of the circuit is

.A131

V131

2

12

,A5.62

1

leg

I I total and no current passes through R.

b) As worked out above, Req  1

c) V ab 0, since no current flows

d) R does not show up since no current flows through it.

26.28: Given that the full-scale deflection current is 500 A and the coil resistance is

A10500A10200

.25A10

s

s s

s c c s

c

R

R R

I R I V

V

b) For a 500-m voltmeter, the resistances are in series:

.9750

.25A10500

V10500

R

R I

V R R R I V

26.29: The full-scale deflection current is 0.0224 A, and we wish a full-scale reading for

20.0 A

.9.1236

.9A0224.0

A499.0

0250.0A0224.0A0.2036

.9A0224.0

r R

r

ε R

r

εR R

r

εr ε Ir ε

V

V V

V

V V

Now if V is to be off by no more than 4% it requires: 1 0.0416

4.86

90  



V

R r

26.31: a) When the galvanometer reading is zero:

R ε ε IR ε IR

ε

ab

cb ab

 b) The value of the galvanometer’s resistance is unimportant since no current flows through it

m000.1

m365.0V15.9

26.32: Two voltmeters with different resistances are connected in series across a 120-V

line So the current flowing is 1.20 10 A

10100

required for full-scale deflection for each voltmeter is:

000,90

V150and

A0150.0000,10

V

k 90 )

1067.1

A1020.1V150and

V12A

0150.0

A1020.1V

26.33: A half-scale reading occurs with R600 So the current through the

galvanometer is half the full-scale current

2

A1060.3V50

52

.65

V52.1:

x

R R

V I R

5430

.65

V52

V52.1A

1025.64

I

.608608

A1025.1

V52.1A

1025.12

I

.203608

A10875.1

V52.1A

10875.14

I

t Q

Q I

Q V

Q I

26.36: An uncharged capacitor is placed into a circuit.

a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge stored

b) All the voltage of the battery is lost over the resistor, so V R  ε 125V

c) There is no charge on the capacitor

d) The current through the resistor is 0.0167A

e) After a long time has passed:

The voltage over the capacitor balances the emf: V c 125V

The voltage over the resister is zero

The capacitor’s charge is qCv c (4.6010 6 F)(125V)5.7510 4C

The current in the circuit is zero.

)F1055.4()1028.1(

C1055

10 6

b) τ  RC(1.28106 )(4.551010 F)5.82104s.

26.38:

.F1049.8))3/12((ln)1040.3(

s00.4)

/ln(

7 6

0

/ 0

τ C

e

v

v τ RC

26.39: a) The time constant RC (0.895106 )(12.410 6 F)11.1s.Soat:

.0)1

(:

)1

()V0.60()F104.12()1

(:

s5

4

) s 1 11 /(

) s 0 5 ( 6

)1

()V0.60()F104.12()1

(:

s10

4

) s 1 11 /(

) s 0 10 ( 6

)1

()V0.60()F104.12()1

(:

s20

4

) s 1 11 /(

) s 0 20 ( 6

)1

()V0.60()F104.12()1

(:

s100

4

) s 1 11 /(

) s 100 ( 6

95.8

V0.60:

95.8

V0.60:

95.8

V0.60:

8.95

V60.0:

s20

95.8

V0.60:

c) Charge against time:

Current against time:

26.40: a) Originally,   RC0.870s.The combined capacitance of the two identical capacitors in series is given by

2

;2111

tot tot

C C C C C

,V72)0.80()A900.0(,

1)

1

Q q C

t R

Q

q e

e Q

)F1090.5(

s103:

s10

)F1090.5()463(

)/1ln(

)1

26.43: a) The time constant RC (980)(1.5010 5 F)0.0147s

C.1033.1)1

()V0.18()F1050.1()1

(:

V11.9)980()A1030.9

26.44: a) 17.1A.

V240

)V240

2 2

P

V R R

W900V

120

W1500:

W900With

V120V

A20bulbsofNumberA

0.75V

120

W90

49.3A34.4

V120

.4.34

V I

26.48: a)

.If

11

2 1

2 1 2

1

2 1 1

3 1 eq

2

2 1 3

1

2 1 3 eq

1

R R

R R

R

R R R

R R R

R R

R R R

R R R R

2 1 3 1

3 2 1 eq

)(

11

R R R

R R R R

R R

)(

)(

If Req R1 R1 R1 R2 R3 R3 R1 R2 R3 R1 R1  R2 R2

26.49: a) We wanted a total resistance of400Ωandpowerof 2.4W from a

combination of individual resistors of 400Ωand1.2Wpower-rating

b) The current is given by:I  P/R  2.4W/400 0.077A.In each leg half the current flows, so the power in each resistor in each resistor in each combination is the same: P(I/2)2R(0.039A)2(400)0.6W

26.50: a) First realize that the Cu and Ni cables are in parallel.

Cu Ni Cable

111

R R

)(

/

/

2 2 Cu Cu

Cu

2 Ni Ni

Ni

a b π

L ρ

A L ρ R

πa

L ρ A L ρ R

a b π L ρ

πa

2 2

Ni

2

cable

)(

)m050.0(m

2 8

2 Cu

2 2

Ni 2

π

ρ

a b ρ

a L π

L ρ

m1014

2

m20

)106.13()m10.0(

8

6 2

R πb

ρ

26.51: Let R1.00,the resistance of one wire Each half of the wire has Rh R 2.

The equivalent resistance is  2  5 2 52(0.500)1.25

h h h

theA4.480.00

1

V0.8

4.4)80.0()A4.4(V0

9.280.00.2

V0

so the remaining bulb is brighter than before

26.53: The maximum allowed power is when the total current is the maximum allowed

value of I  P/R  36W/2.43.9A.Then half the current flows through the parallel resistors and the maximum power is:

.W54)4.2()A9.3(2

32

3)

2/()2/

P

26.54: a) 4.0 ;

16

116

18

1)16,16,8(

1)18,9(

20

16

2.4A

.42)

that of the top, since its resistance is half Therefore the potential of point a relative to point xisV ax  IR eq(9,18)(9.6A)(6.00)58V

26.55: Circuit (a)

toequivalentis

network

The

16.67resistanceequivalent

haveandparallelin

areresistorsΩ

50.0andΩ25.0

The

26.09resistanceequivalent

haveandparallelin

areresistorsΩ

40.0andΩ75.0

10

.100

11

eq eq

R R

Circuit (b)

The

0.18resistanceequivalent

haveandparallelin

areresistorsΩ

45.0andΩ

10

.10

11

eq eq

R R

26.56: Recognize that the ohmmeter measures the equivalent parallel resistance, not just

85

1130

1115

112

.201

X X

26.57: Topleftloop: 125(I2 I3)1I2 0126I2 5I3 0

.0103

01191012:loopBottom

.089901)(

89:loopright

Top

3 2 1 2

1 3

3 1 1

3 1

I I

I I I

I I

Solving these three equations for the currents yields:

.A171.0andA,14.2,848

)0.2(2)8.1(7:

loop

26.59:Leftloop: 20142I1 4(I2 I1)066I1 4I2 0

.094360)(

4536

,A32.6,A21

I

26.60: a) Using the currents as defined on the circuit diagram below we obtain three

equations to solve for the currents:

.043

0)

(2)(

:loopBottom

.03

2

0)

(2:loopTop

.1423

0)(

214

:loopLeft

2 1

2 2 1 2

1

2 1

1 2 1

2 1

2 1 1

I I I I

I I

I I I

I I I I

I I

I I I

Solving these equations for the currents we find:

A

0.2A;

0.6

;A0

0.4A;

0

R

26.61: a) Going around the complete loop, we have:

V

22.0

)112()A44.0(V0.10V0.12

.A44.00

)0.9(V0.8V0.12

V

I I

IR ε

ab

b) If now the points a and b are connected by a wire, the circuit becomes equivalent to

the diagram shown below The two loop equations for currents are (leaving out the units):

5.00

4410

12  I1  I2  I2 I1 and

A

464.0

05.255)24(2

0)(

54254810

1

1 1 1

2 1 2 3

I I I I

I

Thus the current through the 12-V battery is 0.464 A

26.62: a) First do series/parallel reduction:

Now apply Kirchhoff’s laws and solve for 

A25.4)A25.2(A2A

2

A25.2

0)20(V5)A2)(

20(:0

1 2

1 2

2 adefa

I I

I V

reversed

beshouldpolarity

V;

109

0)A25.2()20(A)

25.4()15(:0

b) Parallel branch has a 10resistance

V20)A2(10(

V RI

Current in upper part: 32 A

30 V

J60)10(A3

U Rt I U Pt

V7.12

;V706.12

V0.12)0.10(

c

c d

V V V V V

V

V I

V

26.64: First recognize that if the 40  resistor is safe, all the other resistors are also safe

A0.158I

W1)40(

2 2

Now use series / parallel reduction to simplify the circuit The upper parallel branch is 6.38  and the lower one is 25  The series sum is now 126  Ohm’s law gives

V19.9A)158.0(126

26.66: For three identical resistors in series,

2 2

V R

V P

1 1

b) When the resistors are connected in series the equivalent resistance is

2 1 2

2 1 2 2

2 1

2 tot

P P

P P P P

R R

26.68: a) Ignoring the capacitor for the moment, the equivalent resistance of the two

300

.3

100

.6

11

eq eq

R R

In the absence of the capacitor, the total current in the circuit (the current through the

V0

e R

6.00(8.00

/V)0

42



 R

resistor and the capacitor is thus V  iR(3.00A (6.00)18.0V (There is no

current through the 3.00 resistor and so no voltage drop across it.) The change on the capacitor is

C107.2V)(18.0farad)10

00.4



 CV

Q

26.69: a) When the switch is open, only the outer resistances have current through them

So the equivalent resistance of them is:

V

0.12)00.6(A00.82

1)00.3(A00.821

A00.84.50

V0.3650

.46

3

13

b) If the switch is closed, the circuit geometry and resistance ratios become identical

to that of Problem 26.60 and the same analysis can be carried out However, we can also use symmetry to infer the following:

.and

switch 3

1A

14.50

)3()6(3

2V

V0.36A

57.83

53

2

battery eq 3

3 3

I R I

I I

26.70: a) With an open switch: V ab  18.0V, since equilibrium has been reached b) Point “a” is at a higher potential since it is directly connected to the positive terminal of the battery.

c) When the switch is closed:

.V00.6)00.3(A00.2(A

00.2)

00.300.6(V

.C105.40V)0.18(F1000.3(

4 6

6

5 6

CV Q

After the switch is closed:

.C107.20V)

.06V0.18(F1000.6(

.C101.80V)12.0V0.18(F1000.3(

5 6

6

5 6

CV Q

So both capacitors lose 3.6010 5 C

26.71: a) With an open switch:

.C10.603V)0.18(F1000.2

eq 3

V0

C106.3

6

5 6

6 6

F

ab 

b) Point “b” is at the higher potential

c) If the switch is closed:

V

6.00)

(3.00A)00.2



 a

b V V

d) New charges are:

C

10.603C)107.20(C1060.3

C

101.80C)

10(1.80C

1060.3

.C100.7V)0.12)(

F1000.6(

.C10.801V)0.6(F1000.3(

5 5

5 6

5 5

5 3

5 6

6

5 6

CV Q

CV Q

So the total charge flowing through the switch is 5.4010 5 C