Chapter 31 by John Smith

FUNDAMENTALS OF CIRCUITS

31.1. Solve: From Table 30.1, the resistivity of carbon is p = 3.5 x

R m. From Equation 31.3, the resistance

of lead from a mechanical pencil is

p~ p~ R=-=-= A m'

(3.5 x lo-'

R m)(0.06 m)

n(0.35~10-'m)'

= 5.5 R

31.2. Solve:

(a) From Table 30.1, the resistivity of aluminum is p = 2.8 x lo-' R m. From Equation 31.3, the length L of a wire with a cross-sectional area A and having a resistance R is

L=-=

(IO x lo4 m)z(lOOOR) = 3.57 m 2.8 x lo4 R m

(b) The number of turns is the length of the wire divided by the circumference of one turn.Thus,

3.57 m = 379 2n(1.5 x 10" m)

31.3. Solve: The current I in a wire when a potential difference is applied to the ends of the wire can be obtained from Ohm's law: I = A V / R . Using R = p L / A , (3.0 V)lc(0.40 x

m)'

(1.5~ lo4 R m)(0.50 m)

= 2.0 A

where we have used nichrome's resistivity from Table 30.1.

31.4. Solve: The potential difference between the ends of a copper wire that carries a current can be obtained from Ohm's law and the relationship R = p L/A . Finding the resistivity of copper from Table 30.1, L AV=fR=Ip-= A

(3.0 A)(1.7 x lo-* C2 m)(0.20 m)

~ ( 0 . x5

m)'

= 13 mV

31.5. Solve: We need an aluminum wire whose resistance and length are the same as that of a 0.50-mmdiameter copper wire. That is,

r . = E r C u= \1-(0.25 2.8 x 1.7 x IO-'

Rm Rm

mm) = 0.32 mm

We need a 0.64-mm-diameter aluminum wire.

31.6. Solve: The slope of the I versus AV graph, according to Ohm's law, is the resistance R . From the graph in Figure Ex3 I .6, the slope of the material is

31-1

~.-x...._...I

"... ._,

31-2

Chapter 31

i;.

31.7. Solve:

9 v f-

From the circuit in Figure Ex37.1, we see that 50 R and 100 R resistors are connected in series across the battery. Another resistor of 75 R is also connected across the battery.

31.8. Solve: In Figure Ex31.8, the positive terminal of the battery is connected to a resistor. The other end of that resistor is connected to resistor and a capacitor in parallel.

31.9. Model: Assume that the connecting wires are ideal. Visualize: Please refer to Figure Ex3 1.9. Solve: The current in the 2 R resistor is I, = 6 V/2 R = 3 A to the left. The current in the 5 R resistor is I, = 10 V/5 SZ = 2 A downward. Using Kirchhoff's junction law, we see that

I = I , i- I2 = 3 A -I-2 A = 5 A This current flows toward the junction, that is, downward.

31.10. Model: The batteries and the connecting wires are ideal. Visualize: Please refer to Figure Ex3 1.10. Solve: (a) Choose the current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the same for all elements in the circuit. With the 9 V battery being labeled 1 and the 6 V battery being labeled 2. Kirchhoff s loop law is

X A Y =A%at, +AVK +4V,,, =+&, - l R - & 2 = O

Note the signs: Potential is gained in battery I, but potential is lost both in the resistor and in battery 2. Because I is positive, we can say that I = 0.100 A flows from left to right through the resistor. (b) The graph shows 9 V gained in battery 1, 4 V K= -IR = 3 V lost in the resistor, and another 6 V lost in battery 2. The final potential is the same as the initial potential, as required.

v (V) I

Wie Battery I

Wue

Resistor

Wue

BdIIerq 2

Wlre

Fundamentals of Circuits

3 1-3

31.11. Model: Assume ideal connecting wires and an ideal battery for which AV,,, = E. Visualize: Please refer to Figure Ex31.11. We will choose a clockwise direction for I. Note that the choice of the current's direction is arbitrary because, with two batteries, we may not be sure of the actual current direction. The 3 V battery will be labeled 1 and the 6 V battery will be labeled 2. Solve: (a) Kirchhoff s loop law, going clockwise from the negative terminal of the 3-V battery is A~losedloop

=c(Av)i = '"ball

+Avba~2

9v I=-= 0.50A 18 Q Thus, the current through the 18 R resistor is 0.50 A. Because I is positive, the current is left to right (i.e., clockwise). *+3V-(18Q)I+6V=O*

(b)

VW)

Assess: The graph shows a 3 V gain in battery 1, a -9 V loss in the resistor, and a gain of 6 V in battery 2. The final potential is the same as the initial potential, as required.

31.12. Visualize: Please refer to Figure Ex 31.12. Define the current I as a clockwise flow. Solve: (a) There are no junctions, so conservation of current tells us that the same current flows through each circuit element. From Kirchhoff s loop law, CAV, = AVba,+ AV,, + AV20= 0 As we go around the circuit in the direction of the current, potential is gained in the battery (AV,,, = &, = +15 V) and potential is lost in the resistors (AV,,,,,, = -ZR).The loop law is

0 = I , ,- IR, - IRz = I,,- I(R, + R,)

A =l5 = 0.50 A R,+R, 30Q

Now that we know the current, we can find the potential difference across each resistor: AVl0= ZR, = (0.50 A)(10 Q) = 5.0 V

AV2, = ZR? = (0.50 A)(20 Q) = 10.0 V

(b)

wire

'IORresirtor

Wire

Wire ~

resistor

31.13. Model: The 1500 W rating is for operating at 120 V. Solve: The hair dryer dissipates 1500 W at AV, = 120 V. Thus, the hair dryer's resistance is (120 V)' R=--(AVR)? --PR 1500 W

- 9.6Q

31-4

Chapter 31

The current in the hair dryer when it is used is given by Ohm’s law:

31.14. Model: Assume ideal connecting wires and an ideal battery. Vialii: Please refer to Figure Ex3 1.14. Solve: The power dissipated by each resistor can be calculated from Equation 31.18, PR= 12R, provided we can find the current through the resistors. Let us choose a clockwise direction for the current and solve for the value of I by using Kirchhoff s loop law. Going clockwise from the negative terminal of the battery, C ( A V ) i = AV,,

+ AVR, + AV,

* +9 V - IR, - IR, = 0

*I=-=

9 v = -I A 12Q + 1 5 Q 3

9 v R,+R,

The power dissipated by resistors R , and R2 is:

PRI= 12R, = (3 A)*(12 Q) = 1.33 W

PR2= Z’&

= (3 A)’(15 Q) = 1.67 W

31.15. Model: The 100 W rating is for operating at 120 V. Solve: A standard bulb uses AV= 120 V. We can use the power dissipation to find the resistance of the filament: AV’ p = -AV‘ qR=-=-=

(120V)’ IOOW

144Q

But the resistance is related to the filament’s geometry:

(9.0 x lo-’ Q m)(0.070 m) R=e=g*r=g=/ A mz a(144a) = 1 . 1 8 ~ 1 0 -m~ =

11.8pm

The filament’s diameter is d = 2r = 23.6 pm.

31.16. Solve: Making use of the conversions 1 J/1 s = 1 W and 1 kW = IO00 J/s, we can write 1 k w h = (IO00 J/~)(3600S) = 3.6 x lo6J

31.17. Solve:

(a) The average power consumed by a typical American family is

pavg =

lcWh kWh =-‘Oo0 kW = 1.389kW 1000= 1000 month 3Ox24h 720

Because P = (AQI with AVas the voltage of the power line to the house,

1.389 kW - 1389 W - 11.6A 120V 120V

pavg =-= AV

(b) Because P = (AV)* / R ,

31.18. Solve: The power dissipated by the night light in 1 year is P = 60 watt x-

12hr

The cost of the night light in one year is $26.30.

day

365days = 263 kWh/year 1 year

x----

Fundamentals of Circuits

31-5

31.19. Solve: The copper wire and the iron wire are connected in series. The composite resistance is simply the equivalent resistance of R, and RFe:

Using the resistivity data from Table 30.1,

( 1 . 7 ~ 1 0 Rm)(0.20 -~ m)

~ ( 0 . x5

(9.7 X

R m)(0.60 m)

~ ( 0 . x5

m)'

R + 74.1 x

= 4.33 x

Q = 78.4 mR

m)*

31.20. Visualize: Please refer to Figure Ex32.20. Solve: The three resistors are in series. The total resistance of this combination is R,= R -t 50 R + R = 2R + 50 R Thus, R, > 50 Q as long as R > 0 R. 31.21.

Model: Assume ideal connecting wires and an ideal battery. Solve: As shown in Figure Ex31.21, a potential difference of 5.0 V causes a current of 100 mA through the three series resistors. The situation is the same if we replace the three resistors with an equivalent resistor Reg.That is, a potential difference of 5.0 V across R, causes a current of 100 mA through it. From Ohm's law, 5.0 V R, =-A" a R + 1 5 Q + l O Q = -* R + 2 5 R = 5 0 i 2 * R = 2 5 i Z I loo mA

121Fl ;"i

31.22. Model: Assume ideal connecting wires and an ideal power supply. Visualize:

75 W

The two light bulbs are basically two resistors in series. Solve: A 75 W (120 V) light bulb has a resistance of

The combined resistance of the two bulbs is

R,,=Rl+R2=192R+192R=384R The current I flowing through Re, is I=-=-AV R,

120V - 0.3125 A 384Q

Because R, is a series combination of R, and R2, the current 0.3125 A flows through R, and R?. Thus, PR,=I'R,=(0.3125A)'(192 Q)= 18.8 W = P ,

31.23. Model: Assume ideal connecting wires and an ideal power supply. Visualize: Corroded hV=120V

contact

u RCO",,

-I

12ov

Re,

Rbulb

-I

3 1-6

Chapter 31

Solve: We have two resistors in series such that R , = R,,, a 100 W (120 V) bulb:

+ R,,,.

R,, can be found from the fact that we have

AV2 (120V)' Rbdb= -= = 144 R P looW We have a total resistance of Re, = 1 4 4 Q + 5 Q = 149 Q. The current flowing through R, is

AV 120V I = -= = 0.8054 A Req 149Q Because R, is a series combination of R,

and R,,,, this current flows through both bulbs. Thus, P,,,,, = PR,, = (0.8054 A)'( 1 4 4 Q) = 93.4 W Assess: The corroded leads change the circuit's total resistance and reduce the current below that at which the bulb was rated. So, it makes sense for it to operate at less than full power.

31.24. Model: Assume ideal connecting wires but not an ideal battery. Visualize: Please refer to Figure 3 1.23. Solve: From Equation 31.25, the potential difference across the battery is AV,

=R+r

Assess: 1 R is a typical internal resistance for a battery. This causes the battery's terminal voltage in the circuit to be 0.5 V less than its emf.

31.25. Model: Assume ideal connecting wires but not an ideal battery. Visualize: The circuit for an ideal battery is the same as the circuit in Figure Ex3 1.25, except that the 1 Q resistor is not present. Solve: In the case of an ideal battery, we have a battery with E = 15 V connected to two series resistors of 10 R and 20 R resistance. Because the equivalent resistance is R, = 10 Q + 20 R = 30 Q and the potential difference across R, is 15 V, the current in the circuit is

' ' Rq

I=-=-=--

'''

- 0.5A

30Q

The potential difference across the 20 R resistor is

AV, = IR = (0.5 A)(20 R)= 10 V In the case of a real battery, we have a battery with E = 15 V connected to three series resistors: 10 R,20 R, and an internal resistance of 1.0 R.Now the equivalent resistance is R& = l O R

+ 20R

-k

1.0Q = 3 1 R

The potential difference across R, is the same as before (& = 15 V). Thus,

Therefore, the potential difference across the 20 Q resistor is

AV,b = I'R = (0.4839 A)(20 R)= 9.68 V That is, the potential difference across the 20 Q resistor is reduced from 10 V to 9.68 V due to the internal resistance of 1 R of the battery. The percentage change in the potential difference is (lo

c68

")xlOO = 3.23%

31.26. Model: Assume ideal connecting wires and an ideal ammeter but not an ideal battery. Visualize:

Please refer to Figure Ex3 1.26.

Fundamentals of Circuits

31-7

Solve: An ideal ammeter has zero resistance, so the battery is being short circuited. If I is the current flowing through the circuit, then

The power dissipated by the internal resistance is

P = I'r = (2.3 A)'(0.6522 Q) = 3.45 W 31.27. Visualize: Rl2

+*e-W/k Rl2

Res

Rf2

The figure shows a metal wire of resistance R that is cut into two pieces of equal length. This produces two wires each of resistance Rl2. Solve: Since these two wires are connected in parallel, 1 1 1 _ --+-=-+-=-

Reg

RJ2

2 R

4 R jR,=R 4

31.28. Visualize: The three resistors in Figure Ex31.28 are equivalent to a resistor of resistance R,. Solve: Because the three resistors are in parallel, 1 1 1 -=-+-+-=-+--

Reg

200Q

1 R

2 R

1 400R+R 2 0 0 R - (200Q)R

200 R R, = (200 R ) R ( 4 0 0 R + R ) - 1+(401R)

From this equation, we see that (i>R, = 0 D if R = 0 R and (ii) R, = 200 R if R -+

00.

Thus, R, < 200 R for R < W.

31.29. Model: Assume ideal connecting wires. Visualize: Please refer to Figure Ex3 1.29. Solve: The resistance R is given by Ohm's law, R = AV,/IR. To determine IR we use Kirchhoff's junction law. The input current I splits into the three currents I , , I,,, and I*. That is,

2.0 A=[], +I15

8V 10R

+ I R = -+-+I,

8V 15R

* I,=

2.0 A-0.8 A-0.533 A ~ 0 . 6 6 7A

Using this value of IR in Ohm's law,

R=-- 8 V - 12.0 n 0.667 A

31.30. Model: The connecting wires are ideal with zero resistance. Solve:

3 1-8

Chapter 31

For the first step, the 10 R and 30 R resistors are in series and the equivalent resistance is 40 R.For the second step, the 60 R and 4 0 R resistors are in parallel and the equivalent resistance is

+I-'[

1 60Q

40Q

-' = 2 4 R

For the third step, the 24 Q and 10 Q resistors are in series and the equivalent resistance is 34 0.

3131. Model: The connecting wires are ideal with zero resistance. Solve:

30 R +

flan

For the first step, the resistors 30 R and 45 Q are in parallel. Their equivalent resistance is 1 -=-

Re,,

1 +- 1 30 Q 45 Q

* R,,

= 18 R

For the second step, resistors 42 R and R, , = 18 R are in series. Therefore,

Rq2 = R,

i- 42

R = 18 R i- 42 R = 60 R

For the third step, the resistors 40 R and R, = 60 52 are in parallel. So, 1 -=-

Re,,

1 +- 1 60 i2 40 R

R,, = 24 Q

The equivalent resistance of the circuit is 24 R.

l00R

IOOR *b

200 R a*

loon

100 R

1 -=Re,

1 300Q

+-+- 1

The equivalent resistance of the circuit is 54.5 R.

200Q

lOOQ

+R,=54.5

54.5

a---ww--b

[24R

Fundamentals of Circuits

100 n

loo n

loon

son

33.3

.--a*

3 1-9

183.3 R

1 1 -=+- 1 3 Re,, = 50 R Re,, 10052 1 0 0 R The three 100 R resistors at the end are in parallel. Their equivalent resistance is 1 1 -=+-+- 1 Req2 100R 100R

100 1 3 Rq2= -R 1OOst 3

= 33.3 R

For the second step, the three resistors are in series, so their equivalent resistance is 100R +50 R + 33.3 R = 183.3 R The equivalent resistance of the circuit is 183.3 R.

31.34. Model: Grounding does not affect a circuitâ&#x20AC;&#x2122;s behavior. Visualize: Please refer to Figure Ex31.34. Solve: Because the earth has V,,, = 0 V, point d has a potential of zero. In going from point d to point a, the potential increases by 9 V. Thus, point a is at a potential of 9 V. Let us calculate the current I in the circuit before calculating the potentials at points b and c. Applying Kirchhoff s loop rule, starting clockwise from point d,

+9 V -1(2 R) - 3 V -1(4 R) = 0

6V 6R

Z = -= 1 A

There is a drop in potential from point a to point b by an amount IR = (1 A)(2 R) = 2 V. Thus, the potential at point b is 9 V - 2 V = 7 V. The potential decreases from 7 V at point b to 7 V - 3 V = 4 V at point c. There is a further decrease in potential across the 4 R resistor of ZR = (1 A)(4 R)= 4 V. That is, the potential of 4 V at c becomes 0 V at point d, as it must. In summary, the potentials at a, b, c, and d are 9 V, 7 V, 4 V, and 0 V.

31.35. Model: Grounding does not affect a circuitâ&#x20AC;&#x2122;s behavior. Visualize: Please refer to Figure Ex3 1.35. Solve: Let us first obtain the value of the current I in the circuit. Applying Kirchhoffâ&#x20AC;&#x2DC;s loop rule, starting clockwise from point c, C ( A V ) , = A V ? n + A ~ 5 V b a t + A V 4 R + A V 9 V=bOa t 6V 6Q Because the earth has V,, = 0 V, point c is at zero potential. There is a potential drop of IR = ( 1 A)(2 R) = 2 V across the 2 R resistor, so the potential at point d is -2 V. From point d to point a, there is an increase in potential of 15 V, thus the potential at point a is 15 V - 2 V = 13 V. The potential decreases from point a to point b by IR = ( 1 A)(4 R)= 4 V,SO the potential at point b is 13 V - 4 V = 9 V. The potential at point c is 9 V lower than the potential at b, so it 0 V,as it must be. In summary, the potentials at a, b, c, and d are 13 V, 9 V, 0 V, and -2V. 9

-I(2 R)+ 15 V -I(4 R)- 9 V = O *

I = -= 1 A

31.36. Solve: Noting that the unit of resistance is the ohm (VIA) and the unit of capacitance is the farad ( C N ) , the unit of RC is

31-10

Chapter 31

31.37. Model: Assume ideal wires as the capacitors discharge through the two 1 kR resistors. Visualize: The circuit in Figure Ex37.37 has an equivalent circuit with resistance R, and capacitance C,. Solve: The equivalent capacitance is

and the equivalent resistance is R, = 1 kR capacitors is

+ 1 kR = 2 kQ. Thus, the time constant for the discharge of the

o= RqC, = (2 kR)(1 pF) = 2 x

s = 2 ms

Assess: The capacitors will be almost entirely discharged 5.r= 5 x 2 ms = 10 ms after the switch is closed.

31.38. Model: Assume ideal wires as the capacitors discharge through the two 1 kR resistors. Visualize: The circuit in Figure Ex37.38 has an equivalent circuit with resistance Re, and capacitance C,. Solve: The equivalent capacitance is C i = 2 pF + 2 ,uF = 4 pF, and the equivalent resistance is

1 -=-

1wZ

i -

1wZ

aR,=500Q

Thus, the time constant for the discharge of the capacitors is T = R,C,

= (500 Q)(4 pF) = 2 x

s=2ms

31.39. Model: The capacitor discharges through a resistor. Assume that the wires are ideal. Solve: The decay of the capacitor charge is given by the Equation 3 1.38: Q = Q, e"*. The time constant is ~ = R C = ( ~ . O X I O ~ Q ) ( I O~Xo - ~ F ) = o . o ~ o ~ The initial charge on the capacitor is Q, = 20 pC and it decays to 10 pC in time t. That is,

31.40. Model: The capacitor discharges through a resistor. Assume ideal wires. Visualize: The switch in the circuit in Figure Ex3 1.40 is in position a. When the switch is in position b the circuit consists of a capacitor and a resistor. Solve: (a) The switch has been in position a for a long time. That means the capacitor is fully charged to a charge

e,,=

CAV= CE=(2 pF)(9 V) = 18 pC

Immediately after the switch is moved to the b position, the charge on the capacitor is Q, = 18 pC. The current through the resistor is

Io=%=9 v = 0 . 1 8 A = 18OmA R 50Q Note that as soon as the switch is closed, the potential difference across the capacitor AV, appears across the 50 SL resistor. (b) The charge Q, decays as Q = Qoedr, where

r= RC= (50 R)(2 pF) = loops Thus, the charge is

Q = (18 pC)e-" &'O0

= (18 pC)e4.' = 10.9 pC

The resistor current is

I = I0e-"' = (180d ) e - m w / ' w w = 109 mA (c) Likewise, the charge is Q = 2.44

pC and the current is I = 24.4 mA.

Fundamentals of Circuits

31.41. Model: A capacitor discharges through a resistor. Assume ideal wires. Solve: A capacitor initially charged to Q, decays as Q = Q, e&'. We wish to find R so that a 1.O

3 1-1 1

capacitor

will discharge to 10% of its initial value in 2.0 ms. That is,

(O.IO)Q, = ~,,e-' ms/R(l

x 10" s = 869Q @' In(0.10) = - R(l.O2 ~x110"0 -F)~ a R = ( 1 . -2.0 0 ~ 1 0F)ln(O.10) ~ 3

Assess: A time constant of z = RC = (869 a)(1.0 X

F) = 0.87 ms is reasonable.

31.42. Model: A capacitor discharges through a resistor. Assume ideal wires. Solve: The discharge current or the resistor current follows Equation 31.39: I = loe-i'Rc. We wish to find the capacitance C so that the resistor current will decrease to 25% of its initial value in 2.5 ms. That is,

31.43. Visualize: Please refer to Figure P3 1.43. Solve: Bulbs D and E are in series, so the same current will go through both and make them equally bright (D = E). Bulbs B and C are in parallel, so they have the same potential difference across them. Because they are identical bulbs with equal resistances, they will have equal currents and be equally bright (B = C). Now the equivalent resistance of B + C in parallel is less than the resistance of E, so the total resistance along the path through A is less than the total resistance along path through D. The two paths have the same total potential difference-the emf of the battery-so more current will flow through the A path than through the D path. Consequently, A will have more current than D and E and will be brighter than D and E (A 7 D =E). Bulbs B and C each have half the current of A. because the current splits at the junction, so A is also brighter than B and C (A > B = C). The remaining issue is how B and C compare to D and E. Suppose B and C were replaced by wires with zero resistance, leaving just bulb A in the middle path. Then the resistance of the path through A would be half of the resistance of the path through D. This would mean that the current through A would be twice the current through D, so IA = ?I1,. When B and C are present, their resistance adds to the resistance of A to lower the current through the middle path. So in reality, IA< ?ID. We already know that I , = I , = * I A , so we can conclude that I, = I, < I,. Since the current through B and C is less than the current through D and E, D and E are brighter than B and C. The final result of our analysis is A 7 D = E > B = C .

31.44.

Visualize: Please refer to Figure P3 1.44. Solve: Bulb A is in parallel with the battery and experiences the full potential difference E whereas the other 5 bulbs divide up the potential into smaller pieces. So A will be brightest. Stated another way, the resistance of the right path, with bulb D in series with several other bulbs, is greater than the resistance of the middle path, with only bulb A. Both paths experience the full potential difference of the battery, so the current starting down the middle path is larger than the current starting down the right path, causing A to be brighter than D (A > D). All of the current in the right path passes through D, then it divides up. So D is brighter than B, C, E, or F (D > B, C, E, F). C and E are in parallel and have the same potential difference. Because they are identical bulbs with equal resistances, they have the same current and are equally bright (C = E). The current through F is the sum of the currents through C and E, so it is brighter than they are (F > C = E). The three-resistor combination C + E + F is in parallel with B. The combination C + E + F has more resistance than B, so more current will flow through B than through C + E + F. Consequently, B is brighter than F (B > F). Putting all these pieces together, the final result is A > D > B > F > C = E.

31.45. Model: The connecting wires and the battery are ideal. Visualize: Please refer to Figure P3 1.45. Solve: Applying Kirchhoff's loop law around the outside edge of the circuit, D V , = & - I R - A V l Z = O =SAVI?= E -IR

That is, the potentia1,difference AV,? between points 1 and 2 is the potential supplied by the battery minus the potential lost in resistor R. When bulb B is in place, a current I flows through the resistor and the bulb. In that case, AV,? is less than E . But if bulb B is removed, the current no longer flows and I = 0 A. Thus, AV,?(nobulb) = E . The answer is that the potential difference AV,?increases.

3 1- 12

Chapter 31

Assess: 1) Removing bulb B means that there is no resistor between points 1 and 2. “No resistor” is not the same thing as “no resistance.” “No resistor” means an insulator with R = -, not R = 0 R. 2) Ohm’s law AV = IR applies to resistors, not to empty spaces. Although I = 0 A when bulb B is removed, you cannot use IR to conclude that AV,, = 0 V because there is no longer a resistor between points 1 and 2 to which Ohm’s law could be applied.

31.46. Solve: The resistivity of aluminum is 2.8 x 10“ Rm and we want the wire’s resistance to be 1.0 R. Using the formula for the resistance of a wire, L R = pA

* 1.0 R

= (2.8 x

Qm)?

= (1.122 x 10’ m-’)rz

We need another relation connecting L and r. Making use of the mass density of aluminum,

1*ox10-3kg = 2700kg/m3 * r 2 L = 1 . 1 7 9 ~ 1 0 - 7 m3 m2L Using the value of L obtained above, ~ 1(1.122x 108m-’)?= 1.179x1O-’m3~r4=1 . 0 5 0 7 ~lO-I5m4- r = 1 . 8 104m=0.18mm Thus, the diameter of the wire is 0.36 mm and the length is

L = (1.122 x 10’ m-’)(1.8 x lo“ m)* = 3.64 m Assess: It is reasonable to make a 3.64 m long wire with a diameter of 0.36 mm from an aluminum block of 1 .O g.

31.47. Solve: Measuring the diameter and thickness of a copper penny yields 2r = 1.9 cm and L = 1.4 mm. Using p = 1.7 x 1O-* !An, L L 1 . 4 ~ 1 0 -m~ = 8.4 x lo-* Q R = p- = p= (1.7~10-*k) A m2 n(i x 1.9 x lo-* m) Assess: This is a small resistance. Relatively large currents can pass through the penny without causing enough heating to melt the penny and hence blow out the fuse.

31.48.

Visualize:

(a)

12 n

12n

12 R

12n

12R

(C)

Solve: (a) The three resistors in parallel have an equivalent resistance of 1 1 -=+-+-1 Req=4.0R Req 12R 12Q 1 2 R

(b) One resistor in parallel with two series resistors has an equivalent resistance of 1

Reg

12R

+-=-1 12i2

1252

24R

+---1 12Q

R,=8.0 R.

1 -=12R R-4

1 + (71’R +-)

1 12Q

-I

=12R

+6R

= 18.0R

Fundamentals of Circuits

3 1- 13

(d) The three resistors in series have an equivalent resistance of

12 R + 12 R + 12 R = 36 R

31.49. Model: Use the laws of series and parallel resistances. Visualize:

Wire=OR 10 R

Solve: Despite the diagonal orientation of the 12 R resistor, the 6 R, 12 R, and 4 R resistors are in parallel because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is

The trickiest issue is the 10 R resistor. It is in parallel with a wire, which is the same thing as a resistor with R = 0 R . The equivalent resistance of 10 R in parallel with 0 R is

In other words, the wire is a short circuit around the 10 R,so all the current goes through the wire rather than the resistor. The 10 R resistor contributes nothing to the circuit. So the total circuit is equivalent to a 2 R resistor in series with the 2 R equivalent resistance in series with the final 3 R resistor. The equivalent resistance of these three series resistors is

R,,=2 R + 2 R + 3 R = l R

31.50. Model: Assume the batteries and the connecting wires are ideal. Visualize: Please refer to Figure P3 1S O . Solve: (a) The two batteries in this circuit are oriented to “oppose” each other. The direction of the current is counterclockwise because the 12 V battery “wins.” (b) There are no junctions, so the same current I flows through all circuit elements. Applying Kirchhoff s loop law in the counterclockwise direction and starting at the lower right comer, CAV, = 12 V - I( 12 R) - 1(6 R) - 6 V - IR = 0 Note that the IR terms are all negative because we’re applying the loop law in the direction of current flow, and the potential decreases as current flows through a resistor. We can easily solve to find the unknown resistance R: 6 V - I(18 R)- IR = 0 (c) The power is P = I’R = (0.25 A)’(6

6 V-(18 R)I - 6 V -(18 9 ) ( 0 . 2 5 A) =6R I 0.25 A

R)= 0.315 UT.

31-14

Chapter 31

,frjyh

The potential difference across a resistor is AV = IR, giving AV, = 1.5 V, and AV,, = 3 V. Starting from the lower left comer. the graph goes around the circuit clockwise, opposite from the direction in which we applied the loop law. In this direction, we speak of potential as lost in the batteries and gained in the resistors.

31.51. Model: Assume that the connecting wires are ideal but the battery is not ideal. Visualize:

R Variable resistor

Solve: The figure shows a variable resistor R connected across the terminals of a battery that has an emf & and an internal resistance r. Using Kirchhoff s loop law and starting from the lower left comer,

+E - I r

- IR = 0 jE = I(r + R )

From the point in Figure P31.5 1 that corresponds to R = 0 R, &= (6 A ) ( r + 0 R)= (6 A)r From the point that corresponds to R = 10 R, E = (3 A ) ( r + 10 R) Combining the two equations, (6 A ) r = (3 A)(r + 10 R) 2 r = r + 1 0 R * r =

lOQ

A l s o , & = ( 3 A ) ( l O R +1 0 R ) = 6 0 V . Assess: With & = 60 V and r = 10 R,the equation E = I(r + R ) is satisfied by all values of R and I on the graph in Figure P3 1.51.

31.52. Model: The connecting wires are ideal, but the battery is not. Visualize: Please refer to Fig. P31.53. We will designate the current in the 5 R resistor Is and the voltage drop AV5. Similar designations will be used for the other resistors. Solve: Since the 10 R resistor is dissipating 40 W,

p,, = I,;R,, = 40 w

I,, =

& L ! Jz =

- 2.0 A

= i$,, = (2.0 A)(10 R)= 20.0 V

Fundamentals of Circuits

3 1-15

The 20 Q resistor is in parallel with the 10 Q resistor, so they have the same potential difference: AV,, = AV,, = 20.0 V. From Ohm's law,

The combined current through the 10 Q and 20 Q resistors first passes through the 5 Q resistor. Applying Kirchhoff s junction law at the junction between the three resistors,

Zs= I , ,

= 1.0 A + 2.0 A = 3.0 A*AV,

= I$s = (3.0 A)(5 Q) = 15.0 V

Knowing the currents and potential differences, we can now find the power dissipated:

Ps= IsAVs = (3.0 A)(15.0 V) = 45.0 W

Pz0= Z2,,AVzo = (1.0 A)(20.0 V) = 20.0 W

31.53. Model: Assume that the connecting wires are ideal, but the battery is not. The battery has internal resistance. Also assume that the ammeter does not have any resistance. Visualize: Please refer to Figure P3 1.53. Solve: When the switch is open,

E-lr-I(5.0 Q ) = O V *

&=(1.636 A)(r+5.0Q)

where we applied Kirchhoff's loop law, starting from the lower left comer. When the switch is closed, the current I comes out of the battery and splits at the junction. The current I' = 1.565 A flows through the 5.0 Q resistor and the rest ( I - 1') flows through the 10.0 Q resistor. Because the potential differences across the two resistors are equal, 175.0 Q) = ( I -I? (10.0 Q ) +(1.565 A)(5.0 Q) = (I- 1.565 A)(10.0 Q) * I =

2.348 A

Applying Kirchhoff's loop law to the left loop of the closed circuit,

E - Ir - 175.0 Q) = 0 V

& = (2.348 A)r + (1.565 A)(5.0 Q)= (2.348 A)r + 7.825 V

Combining this equation for &withthe equation obtained from the circuit when the switch was open, (2.348 A)r i7.825 V = (1.636 A)r + 8.18 V

3 (0.712 A)r

= 0.355 V

r = 0.50 Q

We also have E = (1.636 A)(0.50 Q + 5.0 Q) = 9.0 V.

31.54. Model: Assume that the connecting wire and the battery are ideal. Visualize: Please refer to Figure P3 1.54. Solve: The middle and right branches are in parallel, so the potential difference across these two branches must be the same. The currents are known, so these potential differences are AV,,,

= (3.0 A)R = AV,,

= (2.0 A)(R + 10 Q)

This is easily solved to give R = 20 Q. The middle resistor R is connected directly across the battery, thus (for an equals the emf of the battery. That is ideal battery, with no internal resistance) the potential difference

&= AV,

= (3.0 A)(20 Q) = 60 V

31.55. Model: The connecting wires are ideal. Visualize: a

I b

Solve: Let the current in the circuit be I. The terminal voltage of the 2.5 V battery is V, - V,. This is also the terminal voltage of the 1.5 V battery. V, - V, can be obtained by noting that V,+2.5V-I(1R)=V, * V,-V,=2.5V-I(l n)

Chapter 31

31-16

To determine I, we apply Kirchhoff s loop law starting from the lower left comer: +2.5 V - I(I Q) -Z(1 Q) - 1.5 V = 0 V- I = 0.5 A Thus, V, - V, = 2.5 V - (0.5 A)(1 Q) = 2.0 V. That is, the terminal voltage of the 1.5 V and the 2.5 V batteries is 2.0 V.

31.56. Model: The connecting wires are ideal. Visualize: Please refer to Figure 31.23. Solve: (a) The internal resistance r and the load resistance R are in series, so the total resistance is R + r and the current flowing in the circuit, due to the emf E, is Z = E/(R + r). The power dissipated by the load resistance is

This is not the power EZ generated by the battery, but simply the power dissipated by the load R. The power is a function of R , so we can find the maximum power by setting dPldR = 0: dP -=---= dR

E‘ (R+r)’

2E2R (R+r)3

E Z ( R + r ) - 2 E Z R- & ’ ( I - - ) =O ( R + r)’ ( R + r)’

That is, the power dissipated by R is a maximum when R = r. (b) The load’s maximum power dissipation will occur when R = r = 1 Q, in which case

(c) When R is very small (R + 0 Q), the current is a maximum (I + &/r)but the potential difference across the load is very small (AV = IR + 0 V). So the power dissipation of the load is also very small ( P = ZAV -+ 0 W). When R is very large (R + -), the potential difference across the load is a maximum (AV -+ &) but the current is very small (I + 0 A). Once again, P is very small. If P is zero both for R + 0 and for R + -, there must be some intermediate value of R where P is a maximum.

31.57. Model: The batteries are ideal, the connecting wires are ideal, and the ammeter has a negligibly small resistance. Visualize: Please refer to Figure P31.57. Solve: Kirchhoff‘s junction law tells us that the current flowing through the 2.0 Q resistance in the middle branch is I, + 1’ = 3.0 A. We can therefore determine I , by applying Kirchhoff s loop law to the left loop. Starting clockwise from the lower left comer, +9.0 V - Z,(3.0 Q) - (3.0 A)(2.0 Q) = 0 V 3 I, = 1.O A 3 Z2 = (3.0 A - Z,) = (3.0 A - 1.O A) = 2.0 A Finally, to determine the emf E, we apply Kirchhoff s loop law to the right loop and start counterclockwise from the lower right comer of the loop: +& - I’(4.5 Q) - (3.0 A)(2.0 Q) = 0 V 3 E - (2.0 A)(4.5 Q) - 6.0 V = 0 V I =15.0 V

31.58. Model: Assume an ideal battery and ideal connecting wires. Visualize: V

8V-

4v.

8V-

4v-

-4v-

(b)

12v-

Fundamentals of Circuits

3 1- 17

Solve: (a) Grounding one point doesn’t affect the basic analysis of the circuit. In Figure P31.58, there is a single loop with a single current I flowing in the clockwise direction. Applying Kirchhoff s loop law clockwise from the lower right comer gives 12 v ~AV,=-8I+12V-4I-121=OV~I=-=O0.50A 24 Q Knowing the current, we can use AV= IR to find the potential difference across each resistor: AV, = 4 V AV, = 2 V AV,, = 6 V ’ The purpose of grounding one point in the circuit is to establish that point as the spec@ potential V = 0 V. Grounding point d makes that potential there V, = 0 V. Then we can use the known potential differences to find the potential at other points in the circuit. Point a is 4 V less than point d (because potential decreases in the direction of current flow), so V, = V, - 4 V = 4 V. Point b is 12 V more than point a because of the battery. So V, = V, + 12 V = 8 V. Point c is 2 V less than point b, so V, = V, - 2 V = 6 V. Point d is 6 V less than point c, so V, = V, - 6 V = 0 V. This is a consistency check-making one complete loop brings us back to the potential at which we started, namely 0 V. (b) The information about the potentials is shown in the graph above. (c) Moving the ground to point a doesn’t change the basic analysis of part (a, or the potential digerences found there. All that changes is that now V, = 0 V. Point b is 12 V more than point a because of the battery. So, V, = V, + 12 V = 12 V. Point c is 2 V less than point b, so V, = V, - 2 V = 10 V. Point d is 6 V less than point c, so V, = V, 6 V = 4 V. Point a is 4 V less than point d, so V, = V, - 4 V = 0 V. This brings us back to where we started. The information about the potentials is shown in the graph above.

31.59. Visualize: Please refer to Figure P3 1.59. Solve: Because the 4 R resistor is grounded at both ends, the potential difference across this resistor is zero. That is, no current flows through the 4 SZ resistor, and the negative terminals of both batteries are at zero potential. To determine the current in the 2 R resistor, we apply Kirchhoff s loop law. We assume that current I flows clockwise through the 2 R resistor. Starting from the lower left comer, the sum of the potential differences across various elements in the circuit is +9.0 V - 1(2 12) - 3 V = O V 3 I = 3.0 A

31.60. Solve: The cost of energy of a 60 W incandescent bulb over its lifetime is 1 kW $0 10 X 1000 h X -= $6.00 1000 w 1 kWh This means the life-cycle cost of the incandescent bulb is $6.50. The cost of energy of a 15 W compact fluorescent bulb over its lifetime is 1 kW $0 10 x 10,000 h x -= $15.00 15WX1000 w 1 kWh With the bulb’s cost of $15, the life-cycle cost is $30.00. To make a comparison of the cost effectiveness of the two bulbs, we note that you need ten incandescent bulbs to last as long as one fluorescent bulb. Thus, it will cost a consumer $65.00 to use incandescent bulbs for the same time span as a single fluorescent bulb that will cost only $30.00. The fluorescent bulb is cheaper. 60Wx-

31.61. Solve: (a) The cost per month of the 1000 W refrigerator is 1kW $0.10 X 30X 24 h X 0.20X -= $14.40 1000 w 1 kWh (b)The cost per month of a refrigerator with a 800 W compressor is $1 1.52. The difference in the running cost of the two refrigerators is $2.88 per month. So, the number of months before you recover the additional cost of $100 (of the energy efficient refrigerator) is $100/$2.88 = 34.7 months. 1ooowx-

31.62. Visualize: Please refer to Figure P3 1.62. Solve: (a) Only bulb A is in the circuit when the switch is open. The bulb’s resistance R is in series with the internal resistance r, giving a total resistance R, = R + r. The current is E 1.5 V I,, =-=-- 0.231 A R + r 6.5Q This is the current leaving the battery. But all of this current flows through bulb A, SO I, = Iba,= 0.231 A.

31-18

Chapter 31

(b) With the switch closed, bulbs A and B are in parallel with an equivalent resistance R, = + R = 3 R. Their equivalent resistance is in series with the battery’s internal resistance, so the current flowing .from the battery is & 1.5V Ibl = -= -= 0.428 A R,+r 3.5Q

But only half this current goes through bulb A, with the other half through bulb B, so I , = 3I,,= = 0.214 A . This is a decrease of 7.4%. (d) If r = 0 R,the current when the switch is open would be I, = I,, = 0.250 A. With the switch closed, the current would be Ibac = 0.500 A and the current through bulb A would be I , = +IMr= 0.250 A . The current through A would not change when the switch is closed. (c) The change in I, when the switch is closed is 0.017 A.

31.63. Model: The battery and the connecting wires are ideal. Visualize:

I --

$La 4

(b)

The figure shows the two circuits formed from the circuit in Figure P31.63 when the switch is open and when the switch is closed. Solve: (a) Using the rules of series and parallel resistors, we have simplified the circuit in two steps as shown in figure (a). A battery with emf E = 24 V is connected to an equivalent resistor of 3 Q. The current in this circuit is 24 V/3 C2 = 8 A . Thus, the current that flows through the battery is I,, = 8 A. To determine the potential difference AV,,, we will find the potentials at point a and point b and then take their difference. To do this, we need the currents I, and I,. We note that the potential difference across the 3 R-3 51 branch is the same as the potential difference across the 5 R-1 R branch. So, & = 2 4 V = I a ( 3 R + 3 R )3 I , = 4 A = I , Now, V , - 1,(3 R)= V,, and V, - 1,(5 R) = V,. Subtracting these two equations give us AVa,,: V, - V, = 1,(5 R)- 1,(3 Q) = (4 A)(5 R)- (4 A)(3 R)= +8 V (b) Using the rules of the series and the parallel resistors, we have simplified the circuit as shown in figure (b). A battery with emf & = 24 V is connected to an equivalent resistor of qQ.The current in this circuit is 24 Vf? R = 9.143 A. Thus, the current that flows through the battery is I,,, = 9.14 A. When the switch is closed, points a and b are connected by an ideal wire and must therefore be at the same potential. Thus V,, = 0 V.

31.64. Model: The voltage sourcelbattery and the connecting wires are ideal. Visualize: Please refer to Figure P3 1.64. Solve: Let us first apply Kirchhoff s loop law starting clockwise from the lower left comer: +V,,-IR-I(lOOQ)=OV

*I=

’’

R+100Q

Fundamentals of Circuits

The output voltage is

x”t= (loo R)Z = (100 a)(R+FOOR) v,,= j

For

V,,

3 1- 19

100 R R+100Q

= V,/lO, the above equation can be simplified to obtain R:

/lo = , R + 100Q=lOOOQ V, R+100R

L-

+R=900R

31.65. Model: Assume ideal connecting wires. Visualize: Please refer to Figure P31.65. Because the ammeter we have shows a full-scale deflection with a current of 500 pA = 0.500 mA, we must not allow a current more than 0.500 mA to pass through the ammeter. Since we wish to measure a maximum current of 50 mA, we must split the current in such a way that 0.500 mA flows through the ammeter and 49.500 mA flows through the resistor R. Solve: (a) The potential difference across the ammeter and the resistor is the same. Thus, V, = Vmer (49.500 x A)R = (0.500 x 10” A)(50 R) R = 0.505 SL 1 1 (b) Effective resistance is -= -+ -* R = O S O Q Reg 0.505 R 50 R

31.66. Model: Assume ideal connecting wires. Visualize: Please refer to Figure P31.66. Because the ammeter we have shows a full-scale deflection with a current of 500 pA,we must not pass a current greater than this through the ammeter. Solve: The maximum potential difference is 5 V and the maximum current is 500 pA. Using Ohm’s law, A V = Z,(R + R-) 5.0 V = (500 x 10” A)(R i-50 Q) * R = 9950 Q

31.67. Model: The battery and the connecting wires are ideal. Visualize:

dr.,vR +I

i T i R 2 3 4

- 24vL71

The figure shows how to simplify the circuit in Figure P31.67 using the laws of series and parallel resistances. We have labeled the resistors as R, = 6 ST, R2 = 15 R, R, = 6 R,and R4 = 4 R. Having reduced the circuit to a single equivalent resistance R,,, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: R, and R, are combined to get R,, = 10 R. and then R, and Rz are combined to obtain Rz,:

Next, R,,, and R , are combined to obtain

R,= R,,, + R , = 6 R + 6 R = 12 R From the final circuit, I = - =E- = Re,

24v 12 R

2.4

31-20

Chapter 31

Thus, the current through the battery and R, is I,, = 2 A and the potential difference across R, is Z(R,) = (2 A)(6 = 12 V As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference AV.

-2A

r-2A 11.2A

R , =613

R3=6R

R2=150

Step 3

Step 2

Step 1

R4=4Kl

In Step 1 of the above figure, R, = 12 SZ is returned to R , = 6 SZ and R2,,= 6 51 in series. Both resistors must have the same 2.0 A current as R,. We then use Ohm’s law to find AVR,= (2 A)(6 SZ) = 12 V

AV,,

=(2 A)(6 SZ) = 12 V

As a check, 12 V + 12 V = 24 V, which was AV of the R, resistor. In Step 2, the resistance R,, is returned to R, and R,, in parallel. Both resistors must have the same A V = 12 V as the resistor RZ3&Then from Ohm’s law, 12 v IR2=--1 2 V - 0 . 8 A IW=-- - 1.2 A 15 R 10 n As a check, IR? + I=, = 2.0 A, which was the current I of the R,, resistor. In Step 3, R,, is returned to R, and R, in series. Both resistors must have the same 1.2 A as the R,, resistor. We then use Ohm’s law to find

(Av), = (1.2 A)(6 $2)= 7.2 V

(AmR4= (1.2 A) (4 Q) = 4.8 V

As a check, 7.2 V + 4.8 V = 12 V, which was A V of the resistor R3,. Resistor Potential difference (V) R, 12 12 R, 7.2 R, R. 4.8

Current (A) 2 0.8 1.2 1.2

12v&n +Fw+

31.68. Model: The battery and the connecting wires are ideal. Visualize:

12v

12v+-+6a

The figure shows how to simplify the circuit in Figure P31.68 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: From the last circuit in the diagram,

Thus, the current through the battery is 2 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference AV.

Fundamentals of Circuits

3 1-21

3R,6V

16 R 6V

Step 2

Step 1

In Step 1, the 6 Q resistor is returned to a 3 Q and 3 R resistor in series. Both resistors must have the same 2.0 A current as the 6 R resistance. We then use Ohm's law to find AV3 = (2.0 A)(3 Q) = 6 V As a check, 6 V + 6 V = 12 V, which was AV of the 6 Q resistor. In Step 2, one of the two 3 SL resistances is returned to the 4 R,48 Q, and 16 Q resistors in parallel. The three resistors must have the same AV = 6 V. From Ohm's law, 6V I, =-=1.5A 4R Resistor

3R 4Q 48 c2 16 12

6V 1, == 0.125 A 48 R Potential difference (V) 6 6 6 6

I,, = 6 v - 0.375 A 16 SL Current (A) 2 1.5 0.125 0.375

31.69. Model: The battery and the connecting wires are ideal. Visualize:

24V

24 R

24 R 24 fl

8 R and 24 R in parallel = 6 R

12 R and 24 R in parallel = 8 R

6 Rand 6 in series = 12 R

4 n a n d 8 f2in series = 12 R

31-22

Chapter 31

The figure shows how to simplify the circuit in Figure P31.69 using the laws of series and parallel resistances. We will reverse the procedure and “build up” the circuit using the loop law and junction law to find the current and potential difference of each resistor. Solve: Having found R , = 12 the current from the battery is I = (24 V.1412 Q) = 2.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors musr have the same potential difference AV.

4R,8V

24V

:Tv{rFl 4R,SV

;Fv

24 V

4R,8V

413 A

213

413 A 213

24 16 V

Step 3

Step 2

Step 1

24 v

4R,8V I

6 0

I”,

24 V

Step 4

In Step 1 of the above figure, the 12 R resistor is returned to 4 C2 and 8 52 resistors in series. Both resistors must have the same 2.0 A as the 12 Q resistor. We use Ohm’s law to find AV4 = 8 V and AV, = 16 V. As a check, 8 V + 16 V = 24 V, which was AV of the 12 R resistor. In Step 2, the 8 R resistor is returned to the 12 Q and 24 R resistors in parallel. Both resistors must have the same AV = 16 V as the 8 Q resistor. From Ohm’s law, I,, = (16 V)/(12 a)= A and = A. As a check, I,, + JZ4 = 2.0 A, which was the current I of the 8 Q resistor. In Step 3, the 12 R resistor is returned to the two 6 Q resistors in series. Both resistors must have the same $ A as the 12 Q resistor. We use Ohm’s law to find AV, = 8 V and AV6 = 8 V. As a check, 8 V + 8 V = 16 V, which was AV of the 12 Q resistor. Finally, in Step 4, the 6 R resistor is returned to the 8 R and 24 R resistors in parallel. Both resistors must have the same A V = 8 V as the 6 R resistor. From Ohm’s law, I, = (8 V)/(8 Q) = 1 A and Z24 = 4 A . As a check, l8+ I , = 4 A , which was the current I of the 6 R resistor. Current (A) Resistor Potential difference (V) 4R 8 2 -4 6R 8 8R 8 1 Bottom 24 R 8 -31 Right 24 R 16 i

Fundamentals of Circuits

31.70. Model: The batteries and the connecting wires are ideal. Visualize: 24 R

12V

= L _ _ I , *

T'"

12VT I

auive:

..---.".1

UI,"

Prom the last circuit in the figure and from Kirchhoff - " ",~ ~ ~ ,

the current through the batteries is 1.o same current l a n d t h n t nn-nl1-l ---:---

I,=-- 6.0 V 8R

12 V - 3 V = 1.0A 9R

- 0.75 A

I,, = 6.0 V = 0.25 A 24 R

T'I "

3 1-23

31-24

Chapter 31

As a check, 0.75 A + 0.25 A = 1.0 A which was the current I of the 6 R resistor. In Step 3, the 8 R resistor is returned to the 3 R and 5 R (right) resistors in series, so the two resistors must have the same current of 0.828 A. We use Ohmâ&#x20AC;&#x2122;s law to find AV, = (0.75 A)(5 R)= 3.75 V AV, = (0.75 A)(3 R) = 2.25 V As a check, 2.25 V + 3.75 V = 6.0 V, which was AV of the 8 R resistor. In Step 4, the 3 R resistor is returned to 4 R (left) and 12 R resistors in parallel, so the two must have the same potential difference AV = 2.25 V. From Ohmâ&#x20AC;&#x2122;s law,

As a check, 0.56 A + 0.19 A = 0.75 A, which was the same as the current through the 3 R resistor. Resistor Potential difference (V) Current (A) 6.00 0.25 R 24 3R 3.00 1 .oo 5R 3.75 0.75 4R 2.25 0.56 2.25 0.19 12 n

31.71. Model: The battery and the connecting wires are ideal. Visualize: 2 0

Anhra

4 a

t-J-GzJ 4 R . 4 0 V

The currents through the various resistors are I? = 1, = 10 A, Is = 8 A, and Iz0 = 2 A. (b) The power dissipated by the 20 R resistor is 1:0(20 R)= (2 A)2(20 a)= 80 W .

Fundamentals of Circuits

3 1-25

(c) Starting with zero potential at the grounded point, we travel along the outside path to point a and addsubtract the potential differences on the way:

0 V + (20 R)I,,

+ (2 Q)Z,

= (20 R)(2 A) + ( 2 R)(lO A) = 60 V = V,

31.72. Model: Assume ideal wires. The capacitor discharges through the resistor. Solve: (a) The capacitor discharges through the resistor R as Q = Q0edr.For Q= Qd2,

-Qo_ - a e - c / l o m s

* t = -(0.010

s)ln(0.5) = 6.93 ms

(b) If the initial capacitor energy is U,, we want the time when the capacitor's energy will be U = U,,/2. Noting , that U,, = & / 2 C , this means Q = Q,, . Applying the equation for the discharging capacitor,

31.73. Model: The capacitor discharges through the resistor, and the wires are ideal. Solve: In a RC circuit, the charge at a given time is related to the original charge as Q = Qoe". For a capacitor Q = CAV. so AV= AV0e'": From the Figure P 71.73, we note that AVO= 30 V and AV= 10 V at t = 2 ms. So, -2 ms/R(50x1Od F)

10 V = (30 V ) e

aln-

2 ~ 1 0 s- ~ * R = R(50xlO" F>

= 36.4 fJ (SO x IO4 F)ln(+)

31.74. Model: The capacitor discharges through the resistors. The wires are ideal. Solve: The charge Q on the capacitor when it is connected across a 50 V source is Q = CAV = (0.25 pF)(50 V) = 12.5 pC As this fully charged capacitor is connected in series with a 25 R resistor and a 100 Q resistor, it will dissipate all its stored energy

(12.5 pC)'

=312.5X1O4 J C 0.25pF through the two resistors. Energy dissipated by the resistor is 12R, which means the 100 R resistor will dissipate 4 times more energy than the 25 Q resistor at any given time. Thus, the energy dissipated by the 25 R resistor is =1-=L

312.5 x lod(

25R

25R + 1 0 0 Q )=62.5xIO4 J

31.75. Model: The connecting wires are ideal. The capacitors discharge through the resistors, Visualize:

._.-

3 1-26

Chapter 31

Solve: The 30 R and 20 R resistors are in parallel and are equivalent to a 12 R resistor. This 12 R resistor is in series with the 8 SL resistor so the equivalent resistance of the circuit R, = 20 R.The two 60 pF capacitors are in series producing an equivalent capacitance of 30 pF. This 30 pF capacitor is in parallel with the 20 pF capacitor so the equivalent capacitance C, of the circuit is 50 pF. The time constant of this circuit is

r = RqC, = (20R)(50 pF) = 1.O ms The current due to the three capacitors through the 20 Q equivalent resistor is the same as through the 8 R resistor. So, the voltage across the 8 R resistor follows the decay equation V = V,eâ&#x20AC;&#x153;. For V = Vd2, we get

31.76. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure P3 1.76. Solve: (a) A very long time after the switch has closed, the potential difference AV, across the capacitor is E . This is because the capacitor charges until AV, = E , while the charging current approaches zero. (b) The full charge of the capacitor is Q,,, = C(AVC)- = C E . (c) In this circuit, I = +dQ/dr because the capacitor is charging, that is, because the charge on the capacitor is increasing. (d) From Equation 3 1.40, capacitor charge at time t is Q = Q(, I - e+?. So, dt

A graph of I as a function o f t is shown below. ll(â&#x201A;ŹlR)

31.77. Model: Assume the battery and the connecting wires are ideal. Visualize: Please refer to Figure P3 1.77. Solve: (a) If the switch has been closed for a long time, the capacitor is fully charged and there is no current flowing through the right branch that contains the capacitor. So, a voltage of 60 V appears across the 60 R resistor and a voltage of 40 V appears across the 40 R resistor. That is, maximum voltage across the capacitor is 40 V. Thus, the charge on the capacitor is Q, = I C = (40 V)(2.0 x IOT6F) = SO pC (b) Once the switch is opened, the battery is disconnected from the capacitor. The capacitor C has two resistances ( 1 0 R and 40 R) in series and discharges according to Q = Q#-OIPc.For Q = 0.10 Qo,

t = -(50 R)(2 pF)ln(O.lO) = 0.23 ms

Fundamentals of Circuits

3 1-27

31.78. Model: Assume the battery and the connecting wires are ideal. Visualize: Please refer to Figure CP31.78. Solve: Because the switch has been in position a for a very long time, the capacitor is fully charged to Q, = EC = (so v)(20.0 x io“ F) = 1.0 x io-’ c When the switch is placed in position b for 1.25 ms, its charge will decay through the resistor to 1.25~10-~ s)/(so ~ ) ( 2 0 ~ F) 1 0 ~

Q = QOe-‘IRC = (1.0 x

= 0.2865 x 10- C

C)e-(

The energy in the capacitor after 1.25 ms is 1 Q’

u=--=-

1 ( 0 . 2 8 6 5 ~ 1 0 “c)’

= 2.05 ml 2 C 2 20~10°F The energy in the capacitor at t = 0 s (when the switch was just flipped to position b) is

1 =--=2 C

1 ( 1 . 0 ~ 1 0 -c)’ ~

20x10dF

= 25ml

So, the energy dissipated by the 50 R resistor is 25 ml - 2.05 ml = 23 ml.

31.79. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure 3 1.38(a). Solve: After the switch closes at t = 0 s, the capacitor begins to charge. At time t , let the current and the charge in the circuit be i and q , respectively. Also, assume clockwise direction for the current i. Using Kirchhoff s loop law and starting clockwise from the lower left comer of the loop,

Integrating both sides,

Letting Q,, = I C and

RC, we get Q = Q , ( I

- e’”?.

31.80. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure 3 1.38a. Solve: (a) According to Equation 31.40, during charging the charge on the capacitor increases according to Q = Qma(1 - e-tir).Therefore, the current in the circuit behaves as

Using Equation 3 I. 12, the power supplied by the battery as the capacitor is being charged is

Because

eat= dU/dt ,we have

That is, the total energy which has been supplied by the battery when the capacitor is fully charged is I ’ C . (b) The power dissipated by the resistor as the capacitor is being charged is

E2 P,,,,, = 12R= R R?

31-28

Chapter 31

Because P,,,,

= dU/dt , we have

(a) For energy conservation, the energy delivered by battery is equal to the energy dissipated by the resistor R plus the energy stored in the capacitor C. This is indeed the case because U,, = E2C, U,,, = 3EZC,and U, = +E2C. 31.81. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure CP3 131. Solve: (a) During charging, when the neon gas behaves like an insulator, the charge on the capacitor increases according to Equation 3 1.40. That is, Q = Q,, (1 - e-'?. Because Q = CAV,, AV,=AV,,(l

- e 4 ? = &(l -e-''>

Let us say that the period of oscillation begins when AV = V, and it ends when AV = V,. Then,

Because the period T = ton- taE, we have

(b) Substituting the given values into the above expression and noting that and T = l/f = 0.10 s , 0.10s = R ( 1 0 ~ 1 0F)~ln

(z:

0.10 s = 5140R R=(lOxlOd F ) h 7