SM chapter 28 - Solucionario capitulo 28 Serway 7ma edición PDF

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28 Direct Current Circuits CHAPTER OUTLINE 28 28 28 28 28 28 Electromotive Force Resistors in Series and Parallel Kirchhoff’s Rules RC Circuits Electrical Meters Household Wiring and Electrical Safety ANSWERS TO QUESTIONS Q28 No. If there is one battery in a circuit, the current inside it will be fr...

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28 Direct Current Circuits ANSWERS TO QUESTIONS

CHAPTER OUTLINE 28.1 28.2 28.3 28.4 28.5 28.6

Electromotive Force Resistors in Series and Parallel Kirchhoff ’s Rules RC Circuits Electrical Meters Household Wiring and Electrical Safety

Q28.1

No. If there is one battery in a circuit, the current inside it will be from its negative terminal to its positive terminal. Whenever a battery is delivering energy to a circuit, it will carry current in this direction. On the other hand, when another source of emf is charging the battery in question, it will have a current pushed through it from its positive terminal to its negative terminal.

*Q28.2 The terminal potential difference is e – Ir where I is the current in the battery in the direction from its negative to its positive pole. So the answer to (i) is (d) and the answer to (ii) is (b). The current might be zero or an outside agent might push current backward through the battery from positive to negative terminal. *Q28.3

*Q28.4 Answers (b) and (d), as described by Kirchhoff’s junction rule. *Q28.5 Answer (a). Q28.6

The whole wire is very nearly at one uniform potential. There is essentially zero difference in potential between the bird’s feet. Then negligible current goes through the bird. The resistance through the bird’s body between its feet is much larger than the resistance through the wire between the same two points.

*Q28.7 Answer (b). Each headlight’s terminals are connected to the positive and negative terminals of the battery. Each headlight can operate if the other is burned out.

121

122

Q28.8

Chapter 28

Answer their question with a challenge. If the student is just looking at a diagram, provide the materials to build the circuit. If you are looking at a circuit where the second bulb really is fainter, get the student to unscrew them both and interchange them. But check that the student’s understanding of potential has not been impaired: if you patch past the first bulb to short it out, the second gets brighter.

*Q28.9 Answer (a). When the breaker trips to off, current does not go through the device. *Q28.10 (i) For both batteries to be delivering electric energy, currents are in the direction g to a to b, h to d to c, and so e to f. Points f, g, and h are all at zero potential. Points b, c, and e are at the same higher voltage, d still higher, and a highest of all. The ranking is a > d > b = c = e > f = g = h. (ii) The current in ef must be the sum of the other two currents. The ranking is e = f > g = a = b > h = d = c. *Q28.11 Closing the switch removes lamp C from the circuit, decreasing the resistance seen by the battery, and so increasing the current in the battery. (i) Answer (a). (ii) Answer (d). (iii) Answer (a). (iv) Answer (a). (v) Answer (d). (vi) Answer (a). *Q28.12 Closing the switch lights lamp C. The action increases the battery current so it decreases the terminal voltage of the battery. (i) Answer (b). (ii) Answer (a). (iii) Answer (a). (iv) Answer (b). (v) Answer (a). (vi) Answer (a). Q28.13 Two runs in series: of one lift and two runs:

. Three runs in parallel:

. Junction

.

Gustav Robert Kirchhoff, Professor of Physics at Heidelberg and Berlin, was master of the obvious. A junction rule: The number of skiers coming into any junction must be equal to the number of skiers leaving. A loop rule: the total change in altitude must be zero for any skier completing a closed path. Q28.14 The bulb will light up for a while immediately after the switch is closed. As the capacitor charges, the bulb gets progressively dimmer. When the capacitor is fully charged the current in the circuit is zero and the bulb does not glow at all. If the value of RC is small, this whole process might occupy a very short time interval. Q28.15 The hospital maintenance worker is right. A hospital room is full of electrical grounds, including the bed frame. If your grandmother touched the faulty knob and the bed frame at the same time, she could receive quite a jolt, as there would be a potential difference of 120 V across her. If the 120 V is DC, the shock could send her into ventricular fibrillation, and the hospital staff could use the defibrillator you read about in Chapter 26. If the 120 V is AC, which is most likely, the current could produce external and internal burns along the path of conduction. Likely no one got a shock from the radio back at home because her bedroom contained no electrical grounds—no conductors connected to zero volts. Just like the bird in Question 28.6, granny could touch the “hot” knob without getting a shock so long as there was no path to ground to supply a potential difference across her. A new appliance in the bedroom or a flood could make the radio lethal. Repair it or discard it. Enjoy the news from Lake Wobegon on the new plastic radio. Q28.16 Both 120-V and 240-V lines can deliver injurious or lethal shocks, but there is a somewhat better safety factor with the lower voltage. To say it a different way, the insulation on a 120-V line can be thinner. On the other hand, a 240-V device carries less current to operate a device with the same power, so the conductor itself can be thinner. Finally, as we will see in Chapter 33, the last step-down transformer can also be somewhat smaller if it has to go down only to 240 volts from the high voltage of the main power line.

Direct Current Circuits

123

SOLUTIONS TO PROBLEMS Section 28.1 P28.1

(a)

(b)

Electromotive Force

P=

( ∆V )2 R

(11.6 V )2

becomes

20.0 W =

so

R = 6.73 Ω

∆V = IR so

11.6 V = I ( 6.73 Ω)

and

I = 1.72 A

R

FIG. P28.1

ε = IR + Ir 15.0 V = 11.6 V + ( 1.72 A ) r

so

r = 1.97 Ω

P28.2

P28.3

The total resistance is R =

3.00 V = 5.00 Ω. 0.600 A

(a)

Rlamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω

(b)

P batteries ( 0.408 Ω ) I 2 = 0.08 1 6 = 8.16% = (5.00 Ω) I 2 P total

(a)

Here

ε = I ( R + r) ,

so

I=

ε R+ r

=

FIG. P28.2

12.6 V

( 5.00 Ω + 0.080 0 Ω)

= 2.48 A.

Then, ∆V = IR = ( 2.48 A) ( 5.00 Ω ) = 12.4 V . (b)

*P28.4

Let I1 and I 2 be the currents flowing through the battery and the headlights, respectively. Then, I 1 = I 2 + 35.0 A , and ε − I1 r − I2 r = 0 so ε = (I 2 + 35.0 A ) (0.080 0 Ω ) + I 2 ( 5.00 Ω ) = 12.6 V giving

I 2 = 1.93 A

Thus,

∆V2 = (1.93 A )( 5.00 Ω ) = 9.65 V

FIG. P28.3

(a)

At maximum power transfer, r = R. Equal powers are delivered to r and R. The efficiency is 50.0% .

(b)

For maximum fractional energy transfer to R, we want zero energy absorbed by r, so we want r = 0 .

(c)

High efficiency. The electric company’s economic interest is to minimize internal energy production in its power lines, so that it can sell a large fraction of the energy output of its generators to the customers.

(d)

High power transfer. Energy by electric transmission is so cheap compared to the sound system that she does not spend extra money to buy an efficient amplifier.

124

Chapter 28

Section 28.2 P28.5

(a)

Resistors in Series and Parallel Rp =

1

(1 7.00 Ω ) + (1 10 .0 Ω )

= 4 .12 Ω

Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω (b)

∆V = IR

FIG. P28.5

34.0 V = I (17.1 Ω) I = 1.99 A

for 4.00 Ω , 9.00 Ω resistors

(1.99 A) (4.12 Ω) = 8.18 V 8.18 V = I ( 7.00 Ω )

Applying ∆V = IR,

so

I = 1.17 A

for 7.00 Ω resistor

8.18 V = I ( 10.0 Ω ) so *P28.6

I = 0.818 A

for 10.0 Ω resistor

(a)

The conductors in the cord have resistance. There is a potential difference across each when current is flowing. The potential difference applied to the light bulb is less than 120 V, so it will carry less current than it is designed to, and will operate at lower power than 75 W.

(b)

If the temperature of the bulb does not change much between the design operating point and the actual operating point, we can take the resistance of the filament as constant. For the bulb in use as intended,

P

I= and

∆V

R=

=

75.0 W = 0.625 A 120 V

∆V 120 V = = 192 Ω I 0.625 A

Now, presuming the bulb resistance is unchanged, I=

120 V = 0.620 A 193.6 Ω

Across the bulb is

∆V = IR = 192 Ω ( 0.620 A) = 119 V

so its power is

P = I ∆V = 0.620 A (119 V ) = 73.8 W

FIG. P28.6

Direct Current Circuits

P28.7

125

If we turn the given diagram on its side, we find that it is the same as figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first reduction is shown in (b). In addition, since the 10.0 Ω, 5.00 Ω , and 25.0 Ω resistors are then in parallel, we can solve for their equivalent resistance as: Req =

1 = 2 .94 Ω ⎛ 1 1 1 ⎞ ⎜⎝ 10.0 Ω + 5 .00 Ω + 25 .0 Ω ⎟⎠

This is shown in figure (c), which in turn reduces to the circuit shown in figure (d). ∆V Next, we work backwards through the diagrams applying I = and R ∆V = IR alternately to every resistor, real and equivalent. The 12.94 Ω resistor is connected across 25.0 V, so the current through the battery in every diagram is I=

∆ V 25.0 V = = 1.93 A R 12.94 Ω

In figure (c), this 1.93 A goes through the 2.94 Ω equivalent resistor to give a potential difference of: ∆V = IR = (1.93 A ) (2.94 Ω ) = 5 .68 V From figure (b), we see that this potential difference is the same across ∆Vab , the 10 Ω resistor, and the 5.00 Ω resistor. Thus we have first found the answer to part (b), which is ∆Vab = 5.68 V . (a)

Since the current through the 20.0 Ω resistor is also the current through the 25.0 Ω line ab, I=

P28.8

FIG. P28.7

∆Vab 5.68 V = = 0.227 A = 227 mA Rab 25.0 Ω

We assume that the metal wand makes low-resistance contact with the person’s hand and that the resistance through the person’s body is negligible compared to the resistance Rshoes of the shoe soles. The equivalent resistance seen by the power supply is 1.00 MΩ + R shoes . The current through both resistors is

50.0 V . The voltmeter displays 1.00 MΩ + Rshoes ∆V = I (1 .00 MΩ ) =

(a)

We solve to obtain

50.0 V( 1.00 MΩ ) = ∆V ( 1.00 MΩ ) + ∆V ( Rshoes ) Rshoes =

(b)

50 .0 V( 1.00 MΩ ) = ∆V 1. 00 MΩ + Rshoes

1.00 MΩ ( 50.0 − ∆ V ) ∆V

With Rshoes → 0, the current through the person’s body is 50 .0 V = 50 .0 µ A 1.00 M Ω

The current will never exceed 50µ A.

126

P28.9

Chapter 28

(a)

Since all the current in the circuit must pass through the series 100 Ω resistor, P = I 2 R 2 Pmax = RI max

FIG. P28.9

I max =

so

P

25 .0 W = = 0 .500 A 100 Ω R

1 ⎞ −1 1 + Req = 100 Ω + ⎛ Ω = 150 Ω ⎝ 100 100 ⎠ ∆Vmax = Req I max = 75.0 V (b)

P1 = I ∆V = ( 0.500 A ) ( 75. 0 V ) = 37. 5 W total power P1 = 25.0 W P2 = P3 = RI 2 (100 Ω) ( 0. 250 A) 2 = 6. 25 W

P28.10

Using 2.00-Ω, 3.00-Ω, and 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series 2.00 Ω 3.00 Ω 4.00 Ω 5.00 Ω

6.00 Ω 7.00 Ω 9.00 Ω

Parallel 0.923 Ω 1.20 Ω 1.33 Ω 1.71 Ω

Mixed 1.56 Ω 2.00 Ω 2.22 Ω 3.71 Ω 4.33 Ω 5.20 Ω FIG. P28.10

P28.11

When S is open, R1 , R2 , R3 are in series with the battery. Thus: R1 + R2 + R3 =

6V = 6 kΩ − 10 3 A

(1)

When S is closed in position 1, the parallel combination of the two R2’s is in series with R1 , R3 , and the battery. Thus: 1 6V = 5 kΩ R1 + R2 + R3 = − 2 1. 2 × 10 3 A

(2)

When S is closed in position 2, R1 and R2 are in series with the battery. R3 is shorted. Thus: R1 + R2 =

6V = 3 kΩ − 2 × 10 3 A

From (1) and (3): R3 = 3 k Ω. Subtract (2) from (1): R2 = 2 kΩ. From (3): R1 = 1 kΩ. Answers: R1 = 1.00 kΩ, R 2 = 2.00 kΩ, R3 = 3.00 kΩ .

(3)

Direct Current Circuits

P28.12

127

Denoting the two resistors as x and y , 1 1 1 = + 150 x y

x + y = 690, and

1 1 1 (690 − x ) + x = + = x( 690 − x) 150 x 690 − x x 2 − 690 x + 103 500 = 0 x=

690 ± ( 690 )2 − 414 000 2

x = 470 Ω

y = 220 Ω

*P28.13 The resistance between a and b decreases . Closing the switch opens a new path with resistance of only 20 Ω. The original resistance is R +

1 1

+

1

= R +50 Ω.

90 + 10 10 + 90

1

The final resistance is R +

1

+

1

1

+

90 10

1 10

We require R + 50 Ω = 2(R + 18 Ω) *P28.14 (a)

+

1

= R +18 Ω.

90

so R = 14.0 Ω

The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series with resistor 1, with resistance R, so the equivalent resistance of the whole circuit is 3R. In 1 series, potential difference is shared in proportion to the resistance, so resistor 1 gets 3 2 of the battery voltage and the 2-3-4 parallel combination gets of the battery voltage. 3 This is the potential difference across resistor 4, but resistors 2 and 3 must share this 2 1 to 3. The ranking by potential difference voltage. In this branch goes to 2 and 3 3 is ∆V 4 > ∆V 3 > ∆V1 > ∆V 2 .

(b)

Based on the reasoning above the potential differences are ∆V1 =

ε , ∆V

3

2

=

2ε 4ε 2ε , ∆V3 = , ∆V 4 = . 9 9 3

(c)

All the current goes through resistor 1, so it gets the most. The current then splits at the parallel combination. Resistor 4 gets more than half, because the resistance in that branch is less than in the other branch. Resistors 2 and 3 have equal currents because they are in series. The ranking by current is I1 > I4 > I2 = I3 .

(d)

Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that of resistor 4, twice as much current goes through 4 as through 2 and 3. The currents through 2I I . the resistors are I1 = I, I2 = I3 = , I4 = 3 3

(e)

Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current in the battery, which is the current through resistor 1, decreases. This decreases the potential difference across resistor 1, increasing the potential difference across the parallel combination. With a larger potential difference the current through resistor 4 is increased. With continued on next page

128

Chapter 28

more current through 4, and less in the circuit to start with, the current through resistors 2 and 3 must decrease. To summarize, I4 increases and I1 , I2 , and I3 decrease . (f)

If resistor 3 has an infinite resistance it blocks any current from passing through that branch, and the circuit effectively is just resistor 1 and resistor 4 in series with the battery. The circuit now has an equivalent resistance of 4R. The current in the 3 4 of the original current because the resistance has increased by . circuit drops to 4 3 All this current passes through resistors 1 and 4, and none passes through 2 or 3. Therefore I 1 =

3I 3I , I 2 = I 3 = 0, I 4 = . 4 4

−1

P28.15

1 1 ⎞ Rp = ⎛ + = 0. 750 Ω ⎝ 3. 00 1. 00 ⎠ Rs = ( 2.00 + 0.750 + 4. 00) Ω = 6. 75 Ω I battery =

∆ V 18.0 V = = 2 .67 A Rs 6.75 Ω

P2 = (2.67 A )2 (2.00 Ω )

P2 = I 2R:

P2 = 14.2 W in 2.00 Ω P4 = ( 2. 67 A )2 ( 4. 00 A) = 28. 4 W in 4.00 Ω ∆V2 = ( 2.67 A ) (2 .00 Ω ) = 5 .33 V, ∆V4 = ( 2. 67 A ) ( 4. 00 Ω ) = 10.67 V ∆Vp = 18 .0 V − ∆V2 − ∆V4 = 2 .00 V( = ∆V3 = ∆V1 )

FIG. P28.15

( ∆V3) = ( 2. 00 V )2 = P3 = 0Ω 1.33 W in 3.00 R3 3.00 Ω 2

P1 =

Section 28.3 P28.16

( ∆V1 ) = ( 2. 00 V)2 R1

1.00 Ω

= 4.00 W in 1.00 Ω

Kirchhoff ’s Rules

+15 .0 − ( 7 .00 ) I1 − (2 .00) (5 .00) = 0 5.00 = 7.00 I1

so

I1 = 0.714 A

so

I 2 = 1.29 A

I 3 = I1 + I 2 = 2.00 A 0. 714 + I 2 = 2. 00

+ε − 2.00 (1.29) − 5.00( 2.00) = 0

ε = 12.6 V

FIG. P28.16

Direct Current Circuits

P28.17

We name currents I1 , I 2 , and I 3 as shown. From Kirchhoff’s current rule, I3 = I1 + I2 . Applying Kirchhoff’s voltage rule to the loop containing I 2 and I 3 , 12.0 V −( 4.00 ) I3 − ( 6.00) I2 − 4.00 V = 0 8. 00 = ( 4. 00) I 3 + (6.00 ) I 2 Applying Kirchhoff’s voltage rule to the loop containing I1 and I 2 , − ( 6. 00) I 2 − 4. 00 V + ( 8. 00 ) I 1 = 0

FIG. P28.17

(8.00 )I 1 = 4.00 + (6.00 ) I 2

Solving the above linear system, we proceed to the pair of simultaneous equations: ⎧8 = 4 I1 + 4 I2 + 6 I2 ⎨ ⎩8 I1 = 4 + 6 I2

⎧8 = 4 I 1 + 10 I 2 ⎨ ⎩I2 = 1 .33 I1 − 0 .667

or

and to the single equation 8 = 4 I1 + 13 .3 I1 − 6 .67 I1 = and

14. 7 V = 0. 846 A 17.3 Ω

I3 = I1 + I2

Then give

I 2 = 1.33( 0.846 A ) − 0.667 I 1 = 846 mA, I 2 = 462 mA, I 3 = 1.31 A

All currents are in the directions indicated by the arrows in the circuit diagram. P28.18

The solution figure is shown to the right.

FIG. P28.18

*P28.19 We use the results of Problem 28.17. (a)

(b)

(c)

By the 4.00-V battery:

∆U = (∆V ) I ∆t = (4.00 V ) ( −0.462 A )120 s = −222 J

By the 12.0-V battery:

(12.0 V ) (1.31 A) 120 s = 1.88 kJ

By the 8.00-Ω resistor:

I R∆ t = (0 .846 A ) (8 .00 Ω )120 s = 687 J

By the 5.00-Ω resistor:

2 (0.462 A ) ( 5. 00 Ω) 120 s = 128 J

By the 1.00-Ω resistor:

2 (0.462 A ) ( 1. 00 Ω) 120 s = 25.6 J

By the 3.00-Ω resistor:

2 (1.31 A) ( 3 .00 Ω) 120 s = 616 J

By the 1.00-Ω resistor:

2 (1.31 A) ( 1 .00 Ω) 120 s = 205 J

2

2

−222 J + 1. 88 kJ = 1. 66 kJ from chemical to electrically transmitted. Like a child counting his lunch money twice, we can count the same energy again, 687 J + 128 J + 25. 6 J + 616 J + 205 J = 1. 66 kJ, as it is transformed from electrically transmitted to internal. The net energy transformation is from chemical to internal .

129

130

Chapter 28

*P28.20 (a)

The first equation represents Kirchhoff’s loop theorem. We choose to think of it as describing a clockwise trip around the left-hand loop in a circuit; see Figure (a). For the right-hand loop see Figure (b). The junctions must be between the 5.8 V and the 370 Ω and between the 370 Ω and the 150 Ω. Then we have Figure (c). This is consistent with the ...