Solve:
From 32.1. the circuit in Figure EX32.1, we see that 50 and 100 resistors are connected in series across the battery. Another resistor of 75 is also connected across the battery.
32.2. Solve: In Figure EX32.2, 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.
32.3. Model: Assume that the connecting wires are ideal.
Visualize: Please refer to Figure EX32.3. Solve: The current in the 2 resistor is I1 = 6 V 2 = 3 A to the left. The current in the 5 resistor is
I 2 = 10 V 5 = 2 A downward. Using Kirchhoffs junction law, we see that
I = I1 + I2 = 3 A + 2 A = 5 A
This current flows toward the junction, that is, downward.
32.4. Model: The batteries and the connecting wires are ideal.
Visualize: Please refer to Figure EX32.4. 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. Kirchhoffs loop law is
V = V
i
bat 1
+ VR + Vbat 2 = +E1 IR E2 = 0
I=
E1 E2 9 V 6 V = = 0.10 A R 30
Note the signs: Potential is gained in battery 1, 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, VR = 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.
32.5. Model: Assume ideal connecting wires and an ideal battery for which Vbat = E .
Visualize: Please refer to Figure EX32.5. We will choose a clockwise direction for I. Note that the choice of the currents 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) Kirchhoffs loop law, going clockwise from the negative terminal of the 3-V battery is
Vclosed loop = ( V )i = Vbat 1 + VR + Vbat 2 = 0
i
+3 V (18 ) I + 6 V = 0 I =
9V = 0.5 A 18
Thus, the current through the 18 resistor is 0.5 A. Because I is positive, the current is left to right (i.e., clockwise). (b)
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.
32.6. Visualize: Please refer to Figure EX32.6. 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 Kirchhoffs loop law,
Vi = Vbat + V10 + V20 = 0 As we go around the circuit in the direction of the current, potential is gained in the battery (Vbat = Ebat = +15 V) and potential is lost in the resistors (Vresistor = IR). The loop law is 0 = Ebat IR1 IR2 = Ebat I(R1 + R2) I =
Ebat 15 V = = 0.50 A R1 + R2 30
Now that we know the current, we can find the potential difference across each resistor: V10 = IR1 = (0.50 A)(10 ) = 5.0 V
(b)
V20 = IR2 = (0.50 A)(20 ) = 10.0 V
32.7. Model: The 1500 W rating is for operating at 120 V.
Solve:
The hair dryer dissipates 1500 W at VR = 120 V. Thus, the hair dryers resistance is
= 9.60 PR 1500 W The current in the hair dryer when it is used is given by Ohms law:
I= VR 120 V = = 12.5 A 9.60 R
( VR ) R=
2
(120 V ) =
2
32.8. Model: Assume ideal connecting wires and an ideal battery.
Visualize: Please refer to Figure EX32.8. Solve: The power dissipated by each resistor can be calculated from Equation 32.14, PR = I2R, 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 Kirchhoffs loop law. Going clockwise from the negative terminal of the battery,
( V )
i
i
= Vbat + VR1 + VR2 = 0 +12 V IR1 IR2 = 0
I=
12 V 12 V 2 = = A = 0.40 A R1 + R2 12 + 18 5
PR2 = I 2 R2 = ( 0.40 A ) (18 ) = 2.9 W
2
The power dissipated by resistors R1 and R2 is:
PR1 = I 2 R1 = ( 0.40 A ) (12 ) = 1.92 W
2
Solve: A standard bulb uses V = 120 V. We can use the power dissipation to find the resistance of the filament:
32.9. Model: The 100 W rating is for operating at 120 V.
P=
V 2 V 2 (120 V ) R= = = 144 P 100 W R
2
But the resistance is related to the filaments geometry:
R=
L
A
=
L L = r= 2 r R
( 9.0 10
7
(144 )
m ) ( 0.070 m )
= 1.18 105 m = 11.8 m
The filaments diameter is d = 2r = 23.6 m.
32.10. Solve: Making use of the conversions 1 J 1 s = 1 W and 1 kW = 1000 J/s, we can write
1 kWh = (1000 J/s)(3600 s) = 3.6 106 J
32.11. Solve: (a) The average power consumed by a typical American family is
Pavg = 1000 kWh kWh 1000 = 1000 = kW = 1.389 kW month 30 24 h 720
Because P = (V )I with V as the voltage of the power line to the house,
I avg =
(b) Because P = ( V ) /R,
2
Pavg V
=
1.389 kW 1389 W = = 11.6 A 120 V 120 V
Ravg =
( V )
Pavg
2
=
(120 V )
1389 W
2
= 10.4
32.12. Solve: The cost of running the waterbed 35% of the time for a year is
24 hr kW $0.11 (0.35)(450 W)(365 days) = $152 day 1000 W kW hr
32.13. Visualize: Please refer to Figure EX32.13.
Solve: The three resistors are in series. The total resistance of this combination is Req = R + 50 + R = 2R + 50 Thus, Req > 50 as long as R > 0 .
32.14. Model: Assume ideal connecting wires and an ideal battery.
Solve: As shown in Figure EX32.14, 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 Req. That is, a potential difference of 5.0 V across Req causes a current of 100 mA through it. From Ohms law,
Req =
VR 5.0 V R + 15 + 10 = R + 25 = 50 R = 25 100 mA I
32.15. Model: Assume ideal connecting wires and an ideal power supply.
Visualize:
The two light bulbs are basically two resistors in series. Solve: A 75 W (120 V) light bulb has a resistance of
R= V 2 (120 V ) = = 192 P 75 W
2
The combined resistance of the two bulbs is Req = R1 + R2 = 192 + 192 = 384 The current I flowing through Req is
I=
V 120 V = = 0.3125 A Req 384
Because Req is a series combination of R1 and R2, the current 0.3125 A flows through R1 and R2. Thus, PR1 = I2R1 = (0.3125 A)2(192 ) = 18.8 W = PR2
32.16. Model: Assume ideal connecting wires and an ideal power supply.
Visualize:
Solve: We have two resistors in series such that Req = Rbulb + Rcontacts. Rbulb can be found from the fact that we have a 100 W (120 V) bulb:
Rbulb = V 2 (120 V ) = = 144 P 100 W
2
We have a total resistance of Req = 144 + 5.0 = 149 . The current flowing through Req is
I=
V 120 V = = 0.8054 A Req 149
Because Req is a series combination of Rbulb and Rcontacts, this current flows through both bulbs. Thus, Pbulb = I2Rbulb = (0.8054 A)2(144 ) = 93.4 W
Assess: The corroded leads change the circuits 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.
32.17. Model: Assume ideal connecting wires and an ideal ammeter but not an ideal battery.
Visualize: Please refer to Figure EX32.17. Solve: An ideal ammeter has zero resistance, so the battery is being short circuited. If I is the current flowing through the circuit, then
I=
E E 1.5 V r= = = 0.65 r I 2.3 A
The power dissipated by the internal resistance is P = I2r = (2.3 A)2(0.6522 ) = 3.5 W
Visualize: The circuit for an ideal battery is the same as the circuit in Figure EX32.18, except that the 1 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 and 20 resistance. Because the equivalent resistance is Req = 10 + 20 = 30 and the potential difference across Req is 15 V, the current in the circuit is
32.18. Model: Assume ideal connecting wires but not an ideal battery.
I=
V E 15 V = = = 0.50 A Req Req 30
The potential difference across the 20 resistor is V20 = IR = (0.50 A)(20 ) = 10.0 V In the case of a real battery, we have a battery with E = 15 V connected to three series resistors: 10 , 20 , and an internal resistance of 1.0 . Now the equivalent resistance is
Req = 10 + 20 + 1.0 = 31 The potential difference across Req is the same as before ( E = 15 V). Thus, I = V E 15 V = = = 0.4839 A Req Req 31
Therefore, the potential difference across the 20 resistor is V20 = I R = ( 0.4839 A ) ( 20 ) = 9.68 V That is, the potential difference across the 20 resistor is reduced from 10.0 V to 9.68 V due to the internal resistance of 1 of the battery. The percentage change in the potential difference is 10.0 V 9.68 V 100 = 3.2% 10.0 V
32.19. Model: Assume ideal connecting wires but not an ideal battery.
Visualize: Please refer to Figure EX32.20. Solve: From Equation 32.21, the potential difference across the battery is
Vbat =
E ER 9.0 V r = R 1 = ( 20 ) 1 = 1.18 R+r Vbat 8.5 V
Assess: 1 is a typical internal resistance for a battery. This causes the batterys terminal voltage in the circuit to be 0.5 V less than its emf.
32.20.
Visualize:
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 R/2. Solve: Since these two wires are connected in parallel,
1 1 1 224 R = + = + = Req = 4 Req R 2 R 2 R R R
32.21. Visualize: The three resistors in Figure EX32.21 are equivalent to a resistor of resistance Req = 75 .
Solve:
Because the three resistors are in parallel,
1 1 1 12 1 400 + R ( 200 ) R = Req = 75 = =+ +=+ = Req R 200 R R 200 ( 200 ) R ( 400 + R )
400 = 240 200 1 75
200 400 1+ R
R=
32.22. Model: Assume ideal connecting wires.
Visualize: Please refer to Figure EX32.22. Solve: The resistance R is given by Ohms law, R = VR I R . To determine IR we use Kirchhoffs junction law. The input current I splits into the three currents I10, I15, and IR. That is, 2.0 A = I10 + I15 + IR = 8 V + 8 V + I R IR = 2.0 A 0.80 A 0.533 A = 0.667 A 10 15
Using this value of IR in Ohms law, R= 8V = 12 0.667 A
32.23. Model: The connecting wires are ideal with zero resistance.
Solve:
For the first step, the 10 and 30 resistors are in series and the equivalent resistance is 40 . For the second step, the 60 and 40 resistors are in parallel and the equivalent resistance is
1 1 40 + 60 = 24 For the third step, the 24 and 10 resistors are in series and the equivalent resistance is 34 .
1
32.24. Model: The connecting wires are ideal with zero resistance.
Solve:
For the first step, the resistors 30 and 45 are in parallel. Their equivalent resistance is
1 1 1 = + Req 1 = 18 Req 1 30 45 For the second step, resistors 42 and Req 1 = 18 are in series. Therefore,
Req 2 = Req 1 + 42 = 18 + 42 = 60
For the third step, the resistors 40 and Req 2 = 60 are in parallel. So, 1 1 1 = + Req 3 = 24 Req 3 60 40 The equivalent resistance of the circuit is 24 .
32.25. Model: The connecting wires are ideal with zero resistance.
Solve:
In the first step, the resistors 100 , 100 , and 100 in the top branch are in series. Their combined resistance is 300 . In the middle branch, the two resistors, each 100 , are in series. So, their equivalent resistance is 200 . In the second step, the three resistors are in parallel. Their equivalent resistance is
1 1 1 1 Req = 54.5 = + + Req 300 200 100 The equivalent resistance of the circuit is 54.5 .
32.26. Model: The connecting wires are ideal with zero resistance.
Solve:
For the first step, the two resistors in the middle of the circuit are in parallel, so their equivalent resistance is
1 1 1 = + Req 1 = 50 Req 1 100 100 The three 100 resistors at the end are in parallel. Their equivalent resistance is 1 1 1 1 100 = 33.3 = + + Req 2 = Req 2 100 100 100 3
For the second step, the three resistors are in series, so their equivalent resistance is 100 + 50 + 33.3 = 183 The equivalent resistance of the circuit is 183 .
32.27. Model: Grounding does not affect a circuits behavior.
Visualize: Please refer to Figure EX32.27. Solve: Because the earth has Vearth = 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 Kirchhoffs loop rule, starting clockwise from point d,
( V )
i
i
= V9 V bat + V2 + V6 V bat + V1 = 0
3V =1 A 3
+9 V I(2 ) 6 V I(1 ) = 0 I =
There is a drop in potential from point a to point b by an amount IR = (1 A)(2 ) = 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 6 V = 1 V at point c. There is a further decrease in potential across the 1 resistor of IR = (1 A)(1 ) = 1 V. That is, the potential of 1 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, 1 V, and 0 V.
32.28. Model: Grounding does not affect a circuits behavior.
Visualize: Please refer to Figure EX32.28. Solve: Let us first obtain the value of the current I in the circuit. Applying Kirchhoffs loop rule, starting clockwise from point c,
( V )
i
i
= V1 + V15 V bat + V4 + V5 V bat = 0
I(1 ) + 15 V I(4 ) 5 V = 0 I =
10 V =2A 5
Because the earth has Vearth = 0 V, point c is at zero potential. There is a potential drop of IR = (2 A)(1 ) = 2 V across the 1 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 = (2 A)(4 ) = 8 V, so the potential at point b is 13 V 8 V = 5 V. The potential at point c is 5 V lower than the potential at b, so it is 0 V, as it must be. In summary, the potentials at a, b, c, and d are 13 V, 5 V, 0 V, and 2 V.
32.29. Solve: Noting that the unit of resistance is the ohm (V/A) and the unit of capacitance is the farad (C/V), the unit of RC is
RC = VCC C == =s A V A Cs
32.30. Model: Assume ideal wires as the capacitors discharge through the two 1 k resistors. Visualize: The circuit in Figure EX32.30 has an equivalent circuit with resistance Req and capacitance Ceq. Solve: The equivalent capacitance is
1 1 1 = + Ceq = 1 F Ceq 2 F 2 F and the equivalent resistance is Req = 1 k + 1 k = 2 k. Thus, the time constant for the discharge of the capacitors is
= ReqCeq = (2 k)(1 F) = 2 103 s = 2 ms
Assess:
The capacitors will be almost entirely discharged 5 = 5 2 ms = 10 ms after the switch is closed.
32.31. Model: Assume ideal wires as the capacitors discharge through the two 1 k resistors. Visualize: The circuit in Figure EX32.31 has an equivalent circuit with resistance Req and capacitance Ceq. Solve: The equivalent capacitance is Ceq = 2 F + 2 F = 4 F, and the equivalent resistance is
1 1 1 = + Req = 0.5 k Req 1 k 1 k Thus, the time constant for the discharge of the capacitors is
= ReqCeq = (0.5 k)(4 F) = 2 103 s = 2 ms
32.32. 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 32.34: Q = Q0 et/. The time constant is
= RC = (1.0 103 )(10 106 F) = 0.010 s
The initial charge on the capacitor is Q0 = 20 C and it decays to 10 C in time t. That is, 10 C t t t = (0.010 s)ln 2 = 6.9 ms 10 C = (20 C)e t / 0.010 s ln = 0.010 s 0.010 s 20 C
32.33. Model: The capacitor discharges through a resistor. Assume ideal wires.
Visualize: The switch in the circuit in Figure EX32.33 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 Q0 = CV = C E = (4 F)(9 V) = 36 C Immediately after the switch is moved to the b position, the charge on the capacitor is Q0 = 36 C. The current through the resistor is
I0 = VR 9V = = 0.36 A 25 R
Note that as soon as the switch is closed, the potential difference across the capacitor VC appears across the 25 resistor. (b) The charge Q0 decays as Q = Q0 et/, where
= RC = (25 )(4 F) = 100 s
Thus, the charge is
Q = ( 36 C ) e 50 s 100 s = ( 36 C ) e0.5 = 22 C The resistor current is I = I 0e t = ( 0.36 A ) e 50 s 100 s = 0.22 A
(c) Likewise, at t = 200 s, the charge is Q = 4.9 C and the current is I = 49 mA.
Solve: A capacitor initially charged to Q0 decays as Q = Q0 et/RC. We wish to find R so that a 1.0 F capacitor will discharge to 10% of its initial value in 2.0 ms. That is,
32.34. Model: A capacitor discharges through a resistor. Assume ideal wires.
( 0.10 ) Q0 = Q0e2.0 ms R (1.0 F) ln ( 0.10 ) =
Assess:
2.0 103 s 2.0 103 s R= = 0.87 k R (1.0 106 F ) (1.0 106 F) ln(0.10)
A time constant of = RC = (870 )(1.0 106 F) = 0.87 ms is reasonable.
32.35. Model: A capacitor discharges through a resistor. Assume ideal wires.
Solve: The discharge current or the resistor current follows Equation 32.35: I = I 0e t 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,
0.25 I 0 = I 0e 2.5 ms (100 )C ln ( 0.25 ) =
2.5 103 s C = 18.0 F (100 ) C
32.36. Visualize: Please refer to Figure P32.36. 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 differencethe emf of the batteryso 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 > 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 = 2ID. 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 < 2ID. We already know that I B = I C = 1 I A , so we can conclude that IB = IC < ID. Since 2 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 > D = E > B = C.
32.37. Visualize: Please refer to Figure P32.77. 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.
32.38. Solve: The resistivity of aluminum is 2.8 108 m and we want the wire to dissipate 7.5 W when connected to a 1.5 V battery. The resistance of the wire must be
P= V2 V 2 (1.5 V ) R= = = 0.30 R P 7.5 W
2
Using the formula for the resistance of a wire,
R=
L 0.30 = A
( 2.8 10
8
m )
L L = ( 3.366 107 m 1 ) r 2 r2
We need another relation connecting L and r. Making use of the mass density of aluminum,
1.0 103 kg = 2700 kg/m 3 r 2 L = 1.179 107 m3 r 2L
Using the value of L obtained above, r2(3.366 107 m1)r2 = 1.179 107 m3 r4 = 3.50 1015 m4 r = 2.43 104 m = 0.243 mm Thus, the diameter of the wire is 0.48 mm and the length is L = (3.366 107 m1)(2.43 104 m)2 = 1.99 m Assess: It is reasonable to make a 1.99 m long wire with a diameter of 0.48 mm from an aluminum block of 1.0 g.
32.39. Solve: The copper wire and the iron wire are connected in series. The composite resistance is simply the equivalent resistance of RCu and RFe:
R = RCu + RFe =
Using the resistivity data from Table 31.2,
Cu LCu
ACu
+
Fe LFe
AFe
R=
(1.7 10 m ) ( 0.20 m ) + ( 9.7 10 m ) ( 0.60 m ) = 4.3 10 ( 0.50 10 m ) ( 0.50 10 m )
8 8 3 2 3 2
3
+ 74 103 = 78 m
32.40. Model: The wires and battery are ideal.
Visualize:
Solve:
We can find the equivalent resistance necessary for the battery to deliver 9 W of power:
P=
= 4.0 R P 9.0 W The combination of the 2.0 , 3.0 , and 6.0 resistors that make 4.0 is shown in the figure. The 3.0 and 6.0 parallel combination has an equivalent resistance of 2.0 , which when added to the 2.0 resistor in series totals 4.0 equivalent resistance.
( V )
2
R=
( V )
2
=
( 6.0 V )
2
32.41.
Visualize:
Solve:
(a) The three resistors in parallel have an equivalent resistance of
1 1 1 1 = + + Req = 4.0 Req 12 12 12
(b) One resistor in parallel with two series resistors has an equivalent resistance of
1 1 1 1 1 1 Req = 8.0 = + = + = Req 12 + 12 12 24 12 8
(c) One resistor in series with two parallel resistors has an equivalent resistance of
1
1 1 1 = 12 + + = 12 + 6 = 18.0 Req 12 12 (d) The three resistors in series have an equivalent resistance of 12 + 12 + 12 = 36
32.42. Model: Use the laws of series and parallel resistances.
Visualize:
Solve: Despite the diagonal orientation of the 12 resistor, the 6 , 12 , and 4 resistors are in parallel because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is
1 1 1 + + Req = = 2 6 12 4 The trickiest issue is the 10 resistor. It is in parallel with a wire, which is the same thing as a resistor with R = 0 . The equivalent resistance of 10 in parallel with 0 is 1 1 1 1 + Req = = () = = 0 10 0 In other words, the wire is a short circuit around the 10 , so all the current goes through the wire rather than the resistor. The 10 resistor contributes nothing to the circuit. So the total circuit is equivalent to a 2 resistor in series with the 2 equivalent resistance in series with the final 3 resistor. The equivalent resistance of these three series resistors is
Rab = 2 + 2 + 3 = 7
1
1
32.43. Model: Assume the batteries and the connecting wires are ideal.
Visualize: Please refer to Figure P32.43. Solve: (a) The two batteries in this circuit are oriented to oppose each other. The curent will flow in the direction of the battery with the greater voltage. The direction of the current is counterclockwise because the 12 V battery is greater. (b) There are no junctions, so the same current I flows through all circuit elements. Applying Kirchhoffs loop law in the counterclockwise direction and starting at the lower right corner, Vi = 12 V I(12 ) I(6 ) 6 V IR = 0 Note that the IR terms are all negative because were 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 ) IR = 0 R =
6 V (18 ) I 6 V (18 ) ( 0.25 A ) = =6 0.25 A I
(c) The power is P = I2R = (0.25 A)2(6 ) = 0.38 W. (d)
The potential difference across a resistor is V = IR, giving V6 = 1.5 V, and V12 = 3 V. Starting from the lower left corner, 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.
32.44. Model: Assume that the connecting wires are ideal but the battery is not ideal.
Visualize:
Solve: The figure shows a variable resistor R connected across the terminals of a battery that has an emf E and an internal resistance r. Using Kirchhoffs loop law and starting from the lower left corner, + E Ir IR = 0 E = I(r + R) From the point in Figure P32.44 that corresponds to R = 0 ,
E = (6 A)(r + 0 ) = (6 A)r
From the point that corresponds to R = 10 ,
E = (3 A)(r + 10 )
Combining the two equations, (6 A)r = (3 A)(r + 10 ) 2r = r + 10 r = 10 Also, E = (3 A)(10 + 10 ) = 60 V. Assess: With E = 60 V and r = 10 , the equation E = I(r + R) is satisfied by all values of R and I on the graph in Figure P32.44.
32.45. Model: The connecting wires are ideal, but the battery is not.
Visualize: Please refer to Fig. P32.45. We will designate the current in the 5 resistor I5 and the voltage drop V5. Similar designations will be used for the other resistors. Solve: Since the 10 resistor is dissipating 40 W,
2 P = I10 R10 = 40 W I10 = 10
P 40 W 10 = = 2.0 A V10 = I10R10 = (2.0 A)(10 ) = 20 V R10 10
The 20 resistor is in parallel with the 10 resistor, so they have the same potential difference: V20 = V10 = 20 V. From Ohms law,
I 20 =
V20 20 V = = 1.0 A R20 20
The combined current through the 10 and 20 resistors first passes through the 5 resistor. Applying Kirchhoffs junction law at the junction between the three resistors, I5 = I10 + I20 = 1.0 A + 2.0 A = 3.0 A V5 = I5R5 = (3.0 A)(5 ) = 15 V Knowing the currents and potential differences, we can now find the power dissipated: P5 = I5V5 = (3.0 A)(15 V) = 45 W P20 = I20V20 = (1.0 A)(20 V) = 20 W
32.46. 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 P32.46. Solve: When the switch is open,
E Ir I(5.0 ) = 0 V E = (1.636 A)(r + 5.0 )
where we applied Kirchhoffs loop law, starting from the lower left corner. 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 resistor and the rest (I I) flows through the 10.0 resistor. Because the potential differences across the two resistors are equal, I(5.0 ) = (I I) (10.0 ) (1.565 A)(5.0 ) = (I 1.565 A)(10.0 ) I = 2.348 A Applying Kirchhoffs loop law to the left loop of the closed circuit,
E Ir I(5.0 ) = 0 V E = (2.348 A)r + (1.565 A)(5.0 )= (2.348 A)r + 7.825 V
Combining this equation for E with the equation obtained from the circuit when the switch was open, (2.348 A)r + 7.825 V = (1.636 A)r + 8.18 V (0.712 A)r = 0.355 V r = 0.50 We also have E = (1.636 A)(0.50 + 5.0 ) = 9.0 V.
32.47. Model: Assume that the connecting wire and the battery are ideal.
Visualize: Please refer to Figure P32.47. 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 Vmiddle = (3.0 A)R = Vright = (2.0 A)(R + 10 ) This is easily solved to give R = 20 . The middle resistor R is connected directly across the battery, thus (for an ideal battery, with no internal resistance) the potential difference Vmiddle equals the emf of the battery. That is
E = Vmiddle = (3.0 A)(20 ) = 60 V
32.48. Model: The connecting wires are ideal.
Visualize:
Solve: Let the current in the circuit be I. The terminal voltage of the 2.5 V battery is Va Vb. This is also the terminal voltage of the 1.5 V battery. Va Vb can be obtained by noting that Vb + 2.5 V I(1 ) = Va Va Vb = 2.5 V I(1 ) To determine I, we apply Kirchhoffs loop law starting from the lower left corner:
+2.5 V I(1 ) I(1 ) 1.5 V = 0 V I = 0.5 A
Thus, Va Vb = 2.5 V (0.5 A)(1 ) = 2.0 V. That is, the terminal voltage of the 1.5 V and the 2.5 V batteries is 2.0 V.
32.49. Model: The connecting wires are ideal.
Visualize: Please refer to Figure 32.20. 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 I = E/(R + r). The power dissipated by the load resistance is
P = I 2R =
E 2R
(R + r)
2
This is not the power EI 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 dP/dR = 0:
E E2 dP 2E 2 R = = 2 3 dR ( R + r ) ( R + r )
2
( R + r ) 2E 2 R = E 2 ( r R ) = 0 3 3 (R + r) (R + r)
That is, the power dissipated by R is a maximum when R = r. (b) The loads maximum power dissipation will occur when R = r = 1 , in which case
P=
E 2R
(R + r)
2
=
( 9 V ) (1 ) = 20 W 2 ( 2 )
2
(c) When R is very small (R 0 ), the current is a maximum (I E/r) but the potential difference across the load is very small (V = IR 0 V). So the power dissipation of the load is also very small (P = IV 0 W). When R is very large (R ), the potential difference across the load is a maximum (V E) 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.
32.50. Model: The batteries are ideal, the connecting wires are ideal, and the ammeter has a negligibly Visualize: small
resistance. Please refer to Figure P32.50. Solve: Kirchhoffs junction law tells us that the current flowing through the 2.0 resistance in the middle branch is I1 + I2 = 3.0 A. We can therefore determine I1 by applying Kirchhoffs loop law to the left loop. Starting clockwise from the lower left corner, +9.0 V I1(3.0 ) (3.0 A)(2.0 ) = 0 V I1 = 1.0 A I2 = (3.0 A I1) = (3.0 A 1.0 A) = 2.0 A Finally, to determine the emf E, we apply Kirchhoffs loop law to the right loop and start counterclockwise from the lower right corner of the loop:
+E I2(4.5 ) (3.0 A)(2.0 ) = 0 V E (2.0 A)(4.5 ) 6.0 V = 0 V E = 15 V
32.51. Model: Assume an ideal battery and ideal connecting wires.
Visualize:
Solve: (a) Grounding one point doesnt affect the basic analysis of the circuit. In Figure P32.51, there is a single loop with a single current I flowing in the clockwise direction. Applying Kirchhoffs loop law clockwise from the lower right corner gives
12 V = 0.50 A 24 Knowing the current, we can use V = IR to find the potential difference across each resistor:
Vi = 8I + 12 V 4I 12I = 0 V I =
V8 = 4 V
V4 = 2 V
V12 = 6 V
The purpose of grounding one point in the circuit is to establish that point as the specific potential V = 0 V. Grounding point d makes that potential there Vd = 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 = Vd 4 V = current flow), so Va 4 V. Point b is 12 V more than point a because of the battery. So Vb = Va + 12 V = 8 V. Point c is 2 V less than point b, so Vc = Vb 2 V = 6 V. Point d is 6 V less than point c, so Vd = Vc 6 V = 0 V. This is a consistency checkmaking 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 doesnt change the basic analysis of part (a) or the potential differences found there. All that changes is that now Va = 0 V. Point b is 12 V more than point a because of the battery. So, Vb = Va + 12 V = 12 V. Point c is 2 V less than point b, so Vc = Vb 2 V = 10 V. Point d is 6 V less than point c, so Vd = Vc 6 V = 4 V. Point a is 4 V less than point d, so Va = Vd 4 V = 0 V. This brings us back to where we started. The information about the potentials is shown in the graph above.
32.52. Visualize: Please refer to Figure P32.52. Solve: Because the 4 resistor is grounded at both ends, the potential difference across this resistor is zero. That is, no current flows through the 4 resistor, and the negative terminals of both batteries are at zero potential. To determine the current in the 2 resistor, we apply Kirchhoffs loop law. We assume that current I flows clockwise through the 2 resistor. Starting from the lower left corner, the sum of the potential differences across various elements in the circuit is
+9 V I(2 ) 3 V = 0 V I = 3 A
32.53. Solve: The cost of energy of a 60 W incandescent bulb over its lifetime is
1 kW $0.10 1000 h = $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 15 W 10,000 h = $15.00 1000 W 1 kWh With the bulbs 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. 60 W
32.54. Solve: (a) The cost per month of the 1000 W refrigerator is
1000 W 1 kW $0.10 30 24 h 0.20 = $14.40 1000 W 1 kWh
(b) The cost per month of a refrigerator with a 800 W compressor is $11.52. 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.
32.55. Visualize: Please refer to Figure P32.55. Solve: (a) Only bulb A is in the circuit when the switch is open. The bulbs resistance R is in series with the internal resistance r, giving a total resistance Req = R + r. The current is E 1.50 V I bat = = = 0.231 A R + r 6.50
This is the current leaving the battery. But all of this current flows through bulb A, so IA = Ibat = 0.231 A. (b) With the switch closed, bulbs A and B are in parallel with an equivalent resistance Req = 1 R = 3.00 . Their 2 equivalent resistance is in series with the batterys internal resistance, so the current flowing from the battery is E 1.50 V I bat = = = 0.428 A Req + r 3.50
But only half this current goes through bulb A, with the other half through bulb B, so I A = 1 I bat = 0.214 A . 2 (c) The change in IA when the switch is closed is 0.017 A. This is a decrease of 7.4%. (d) If r = 0 , the current when the switch is open would be IA = Ibat = 0.250 A. With the switch closed, the current would be Ibat = 0.500 A and the current through bulb A would be I A = 1 I bat = 0.250 A . The current 2 through A would not change when the switch is closed.
32.56. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows the two circuits formed from the circuit in Figure P32.56 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 . The current in this circuit is 24 V 3 = 8 A . Thus, the current that flows through the battery is Ibat = 8 A. To determine the potential difference Vab, we will find the potentials at point a and point b and then take their difference. To do this, we need the currents Ia and Ib. We note that the potential difference across the 3 -3 branch is the same as the potential difference across the 5 -1 branch. So, E = 24 V = Ia(3 + 3 ) Ia = 4 A = Ib
Now, Vc Ia(3 ) = Va, and Vc Ib(5 ) = Vb. Subtracting these two equations give us Vab:
Va Vb = Ib(5 ) Ia(3 ) = (4 A)(5 ) (4 A)(3 ) = +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 E = 24 V is connected to an equivalent resistor of 281 . The current in this circuit is
24 V
21 8
= 9.143 A. Thus, the current that flows through the battery is Ibat = 9 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 Vab = 0 V.
32.57. Model: The voltage source/battery and the connecting wires are ideal.
Visualize: Please refer to Figure P32.57. Solve: Let us first apply Kirchhoffs loop law starting clockwise from the lower left corner: +Vin IR I (100 ) = 0 V I =
Vin R + 100
The output voltage is
Vin Vout 100 Vout = (100 ) I = (100 ) = R + 100 Vin R + 100
For Vout = Vin 10 , the above equation can be simplified to obtain R:
Vin 10 100 = R + 100 = 1000 R = 900 Vin R + 100
32.58. Model: Assume ideal connecting wires.
Visualize: Please refer to Figure P32.58. Because the ammeter we have shows a full-scale deflection with a current of 500 A = 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, VR = Vammeter (49.500 103 A)R = (0.500 103 A)(50.0 ) R = 0.505 (b) Effective resistance is
1 1 1 = + Req = 0.500 Req 0.505 50.0
32.59. Model: Assume ideal connecting wires.
Visualize: Please refer to Figure P32.59. Because the ammeter we have shows a full-scale deflection with a current of 500 A, 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 A. Using Ohms law, V = IA(R + Rammeter) 5.0 V = (500 106 A)(R + 50.0 ) R = 9.95 k
32.60. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P32.60 using the laws of series and parallel resistances. We have labeled the resistors as R1 = 6 , R2 = 15 , R3 = 6 , and R4 = 4 . Having reduced the circuit to a single equivalent resistance Req, 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: R3 and R4 are combined to get R34 = 10 , and then R34 and R2 are combined to obtain R234:
1 1 1 1 1 R234 = 6 = + = + R234 R2 R34 15 10 Next, R234 and R1 are combined to obtain Req = R234 + R1 = 6 + 6 = 12 From the final circuit, E 24 V I= = = 2A Req 12
Thus, the current through the battery and R1 is IR1 = 2 A and the potential difference across R1 is I(R1) = (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 V.
In Step 1 of the above figure, Req = 12 is returned to R1 = 6 and R234 = 6 in series. Both resistors must have the same 2.0 A current as Req. We then use Ohms law to find VR1 = (2 A)(6 ) = 12 V VR234 =(2 A)(6 ) = 12 V As a check, 12 V + 12 V = 24 V, which was V of the Req resistor. In Step 2, the resistance R234 is returned to R2 and R34 in parallel. Both resistors must have the same V = 12 V as the resistor R234. Then from Ohms law,
I R2 =
12 V = 0.80 A 15
I R34 =
12 V = 1.2 A 10
As a check, IR2 + IR34 = 2.0 A, which was the current I of the R234 resistor. In Step 3, R34 is returned to R3 and R4 in series. Both resistors must have the same 1.2 A as the R34 resistor. We then use Ohms law to find (V )R3 = (1.2 A)(6 ) = 7.2 V (V )R4 = (1.2 A) (4 ) = 4.8 V Current (A) 2 0.80 1.2 1.2 As a check, 7.2 V + 4.8 V = 12 V, which was V of the resistor R34. Resistor Potential difference (V) 6 left 12 12 15 7.2 6 right 4.8 4
32.61. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P32.61 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,
12 V E = =2A 6 6 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 V. I=
In Step 1, the 6 resistor is returned to a 3 and 3 resistor in series. Both resistors must have the same 2 A current as the 6 resistance. We then use Ohms law to find
V3 = (2 A)(3 ) = 6 V
As a check, 6 V + 6 V = 12 V, which was V of the 6 resistor. In Step 2, one of the two 3 resistances is returned to the 4 , 48 , and 16 resistors in parallel. The three resistors must have the same V = 6 V. From Ohms law,
I4 = 6V = 1.5 A 4 I 48 = 6V 1 =A 48 8 I16 = 6V 3 =A 16 8
Resistor 3 4 48 16
Potential difference (V) 6 6 6 6
Current (A) 2 1.5 1.2 3.8
32.62. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P32.62 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 Req = 12 , the current from the battery is I = (24 V)/(12 ) = 2.0 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 V.
In Step 1 of the above figure, the 12 resistor is returned to 4 and 8 resistors in series. Both resistors must have the same 2.0 A as the 12 resistor. We use Ohms law to find V4 = 8 V and V8 = 16 V. As a check, 8 V + 16 V = 24 V, which was V of the 12 resistor. In Step 2, the 8 resistor is returned to the 12 and 24 resistors in parallel. Both resistors must have the same V = 16 V as the 8 resistor. From Ohms law, I12 = (16 V ) (12 ) = 4 A and I 24 = 2 A . As a check, I12 + I24 = 2.0 A, which was the current I of the 8 3 3
resistor. In Step 3, the 12 resistor is returned to the two 6 resistors in series. Both resistors must have the 4 same 3 A as the 12 resistor. We use Ohms law to find V6 = 8 V and V6 = 8 V. As a check, 8 V + 8 V = 16 V, which was V of the 12 resistor. Finally, in Step 4, the 6 resistor is returned to the 8 and 24 resistors in parallel. Both resistors must have the same V = 8 V as the 6 resistor. From Ohms law, I8 = (8 V)/(8 ) = 1 4 A and I 24 = 1 A . As a check, I 8 + I 24 = 3 A , which was the current I of the 6 resistor. 3 Resistor 4 6 8 Bottom 24 Right 24 Potential difference (V) 8 8 8 8 16 Current (A) 2 1.3 1 0.33 0.66
32.63. Model: The batteries and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P32.63 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 figure and from Kirchhoffs loop law,
12 V 3 V = 1.0 A 9 Thus, the current through the batteries is 1.0 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. I=
In Step 1 of the above figure, the 9 resistor is returned to the 6 and 3 resistors in series. Both resistors must have the same 1.0 A current as the 6 resistor. We use Ohms law to find
V3 = (1.0 A)(3 ) = 3.0 V V6 = 6.0 V As a check, 3.0 V + 6.0 V = 9 V, which was V = (12 V 3 V) = 9 V of the 9 resistor. In Step 2, the 6 resistor is returned to the 24 and 8 resistors in parallel. The two resistors must have the same potential difference V = 6.0 V. From Ohms law,
I7 =
6.0 V 3 = A 8 4
I 24 =
6.0 V 1 =A 24 4
As a check, 0.75 A + 0.25 A = 1.0 A which was the current I of the 6 resistor. In Step 3, the 8 resistor is returned to the 3 and 5 (right) resistors in series, so the two resistors must have the same current of 0.828 A. We use Ohms law to find V3 = (3/4 A)(3 ) = 9/4 V V4 = (3/4 A)(5 ) = 15/4 V As a check, 9/4 V + 15/4 V = 24/4 V = 6.0 V, which was V of the 8 resistor. In Step 4, the 3 resistor is returned to 4 (left) and 12 resistors in parallel, so the two must have the same potential difference V = 9/4 V. From Ohms law,
I4 = 9/4 V = 9 /16 A = 0.56 A 4 I12 = 9/4 V 9 = A = 0.19 A 12 48
As a check, 0.56 A + 0.19 A = 0.75 A, which was the same as the current through the 3 resistor. Resistor Potential difference (V) Current (A) 6.0 0.25 24 3.0 1.0 3 3.75 0.75 5 2.25 0.56 4 2.25 0.19 12
32.64. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P32.64 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 to find the current and potential difference of each resistor. Solve: (a) From the last circuit in the figure and from Kirchhoffs law, I = 100 V 10 = 10 A. Thus, the current through the battery is 10 A. Now 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.
In Step 1 of the above diagram, we return the 10 resistor to the 4 , 4 , and 2 resistors in series. These resistors must have the same 10 A current as the 10 resistance. That is, the current through the 2 and the 4 resistors is 10 A. The potential differences are
V2 = (10 A)(2 ) = 20 V V4 (left) = (10 A)(4 ) = 40 V V4 (left) = (10 A)(4 ) = 40 V
In Step 2, we return the left 4 resistor to the 20 and 5 resistors in parallel. The two resistors must have the same potential difference V = 40 V. From Ohms law,
I5 = 40 V = 8A 5 I 20 = 40 V = 2.0 A 20
The currents through the various resistors are I2 = I4 = 10 A, I5 = 8 A, and I20 = 2.0 A. 2 2 (b) The power dissipated by the 20 resistor is I 20 ( 20 ) = ( 2.0 A ) ( 20 ) = 80 W .
(c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract the potential differences on the way:
0 V + (20 )I20 + (2 )I2 = (20 )(2.0 A) + (2 )(10 A) = 60 V = Va
32.65. Model: The wires and batteries are ideal.
Visualize:
Solve: Assign currents I1, I2, and I3 as shown in the figure. If I3 turns out to be negative, well know it really flows right to left. Apply Kirchhoffs loop rule counterclockwise to the top loop from the top right corner: I1 (5 ) + 12 V I3 (10 ) = 0. Apply the loop rule counterclockwise to the bottom loop starting at the lower left corner: I2 (5 ) + 9 V + I3 (10 ) = 0. Note that since we went against the current direction through the (10 ) resistor the potential increased. Apply the junction rule to the right middle: I1 = I2 + I3. These three equations can be solved for the current I3 by subtracting the second equation from the first then 3 making the substitution I2 I1 = I3 which was derived from the third equation. The result is I3 = A = 0.12 25 A, left to right.
32.66. Model: The wires and batteries are ideal.
Visualize:
Solve: The circuit has been redrawn for clarity. Assign the currents I1, I2, and I3 as shown in the figure. To find the power dissipated by the 2 resistor, we must find the current through it. If I3 turns out to be negative, well know it really flows bottom to top. Apply Kirchhoffs loop rule clockwise to the left loop from the bottom left corner: + 12 V I1 (4 ) I3 (2 ) = 0. Apply the loop rule clockwise to the right loop starting at the top right corner: + 15 V I2 (4 ) + I3 (2 ) = 0 Note that since we went against the current direction through the 2 resistor the potential increased. Apply the junction rule to the right middle: I1 = I2 + I3 These three equations can be solved for the current I3 by subtracting the second equation from the first then 3 making the substitution I2 I1 = I3 which was derived from the third equation. The result is I3 = A. So the 8 3 current in the 2 resistor is A , bottom to top. The power dissipated in the resistor is 8
9 3 P2 = A ( 2 ) = W = 0.28 W 8 32
2
32.67. Model: The wires and batteries are ideal.
Visualize:
Solve: If no power is dissipated in the 200 resistor, the current through it must be zero. To see if this is possible, set up Kirchhoffs rules for the circuit, then assume the current through the 200 resistor is zero and see if there is a solution. Assume the unknown battery is oriented with its positive terminal at the top and currents I1, I2, I3 defined as shown in the figure above. Apply Kirchhoffs loop rule clockwise to the left loop: 50 V I1 (100 ) I3 (200 ) = 0 Again, counterclockwise to the right hand loop: E I2 (300 ) I3 (200 ) = 0 The junction rule yields I1 + I2 = I3. Now assume I3 = 0 and solve for E. In that case, the first equation gives 50 V 1 I1 = = A. 100 2 From the third equation, I2 = I1, so the second equation gives us 1 E = I 2 ( 300 ) = A ( 300 ) = 150 V 2 Thus E = 150 V and it is oriented with negative terminal on top, opposite to our guess.
32.68. Model: The wires are ideal, but the batteries are not.
Visualize:
Solve:
(a) The good battery alone can drive a current through the starter motor 12 V I= = 200 A ( 0.01 + 0.05 )
(b) Alone, the dead battery drives a current
I=
( 0.50 + 0.05 )
8.0 V
= 14.5 A
(c) Let I1, I2, I3 be defined as shown in the figure above. Kirchhoffs laws applied to the good battery and dead battery loop, good battery and starter motor loop, and the top middle junction yield three equations in the three unknown currents: 12 V I1 ( 0.01 ) I 3 ( 0.05 ) = 0
12 V I1 ( 0.01 ) I 2 ( 0.50 ) 8.0 V = 0
I1 = I 2 + I 3 Substituting for I1 from the third equation into the first and second equations gives 12 V I 2 ( 0.01 ) I 3 ( 0.06 ) = 0
4 V I 2 ( 0.51 ) I 3 ( 0.01 ) = 0 Solving for I2 from the first equation,
I2 =
12 V I 3 ( 0.06 ) ( 0.01 )
Substituting into the second equation and solving for I3 yields the current through the starter motor is 199 A. (d) Substituting the value for I3 into the expression for I2 yields the current through the dead battery is 3.9 A. Assess: The good battery is charging the dead battery as well as running the started motor. A total of 203 A flows through the good battery.
32.69. Model: The wires and batteries are ideal.
Visualize:
Solve: The circuit is redrawn above for clarity, and currents added. We must find I5. Repeatedly apply Kirchhoffs rules. The loop rule applied clockwise about the three triangles yields Left: 9 V I1 (6 ) I3 (12 ) = 0 I1 = 1.5 A 2I3
Center: I4 (24 ) + I3 (12 ) = 0I4 = I3 Right: 15 V I2 (10 ) I4 (24 ) = 0 I2 = 1.5 A 2.4 I4 The junction rule applied at the bottom corners gives equations into which the results above may be substituted: I1 = I3 + I5 1.5 A 2I3 = I3 +I5 I5 = 1.5 A 3I3
I4 = I2 + I5 I4 = 1.5 A 2.4I4 + I5 I5 = 3.4I4 1.5 A Using I4 = I3 and solving for I3, 30 1.5 A 3I3 = 3.4( I3) 1.5 A I3 = A 47 30 201 I5 = 1.5 A 3( A) = A = 2.1 A 47 94 Since the current is negative, 2.1 A flows from left to right through the bottom wire.
32.70. Model: Assume ideal wires. The capacitor discharges through the resistor.
Solve:
(a) The capacitor discharges through the resistor R as Q = Q0et/. For Q= Q0/2,
Q0 t 1 t = (0.010 s)ln(0.5) = 6.9 ms = Q0e t 10 ms ln = 2 2 0.010 s
(b) If the initial capacitor energy is U0, we want the time when the capacitors energy will be U = U 0 2 . Noting
that U 0 = Q02 2C , this means Q = Q0
2 . Applying the equation for the discharging capacitor,
Q0 t 1 1 = Q0e t 10 ms ln = 0.010 s t = ( 0.010 s ) ln = 3.5 ms 2 2 2
Solve: In an RC circuit, the charge at a given time is related to the original charge as Q = Q0et/. For a capacitor Q = CV, so V = V0et/. From the Figure P32.71, we note that V0 = 30 V and V = 10 V at t = 4 ms. So, 4 103 s 4 103 s 4 ms R ( 50106 F ) 10 V 10 V = ( 30 V ) e ln = R= = 73 R ( 50 106 F ) 30 V ( 50 106 F) ln ( 13 )
32.71. Model: The capacitor discharges through the resistor, and the wires are ideal.
32.72. 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 = CV = (0.25 F)(50 V) = 12.5 C As this fully charged capacitor is connected in series with a 25 resistor and a 100 resistor, it will dissipate all its stored energy
UC = 1 2
Q 2 1 (12.5 C ) = = 312.5 106 J C 2 0.25 F
2
through the two resistors. Energy dissipated by the resistor is I2R, which means the 100 resistor will dissipate four times more energy than the 25 resistor at any given time. Thus, the energy dissipated by the 25 resistor is
25 6 312.5 106 = 62.5 10 J = 63 J 25 + 100
32.73. Model: The battery and the connecting wires are ideal.
Visualize: Please refer to Figure P32.73. Solve: (a) A very long time after the switch has closed, the potential difference VC across the capacitor is E . This is because the capacitor charges until VC = E , while the charging current approaches zero. (b) The full charge of the capacitor is Qmax = C(VC)max = C E . (c) In this circuit, I = + dQ dt because the capacitor is charging, that is, because the charge on the capacitor is increasing. (d) From Equation 32.36, capacitor charge at time t is Q = Qmax(1 et/). So,
I=
dQ d 1 1 t E t = CE (1 e t ) = CE e t = CE e = e dt dt R RC
A graph of I as a function of t is shown below.
32.74. Model: Assume the battery and the connecting wires are ideal.
Visualize: Please refer to Figure P32.74. 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 resistor and a voltage of 40 V appears across the 40 resistor. That is, maximum voltage across the capacitor is 40 V. Thus, the charge on the capacitor is Q0 = EC = (40 V)(2.0 106 F) = 80 C (b) Once the switch is opened, the battery is disconnected from the capacitor. The capacitor C has two resistances (10 and 40 ) in series and discharges according to Q = Q0et/RC. For Q = 0.10 Q0,
0.10Q0 = Q0e t /(50 )(2.0 F) ln ( 0.10 ) =
t ( 50 )( 2.0 F )
t = ( 50 ) ( 2.0 F ) ln ( 0.10 ) = 0.23 ms
32.75. Model: Assume the battery and the connecting wires are ideal.
Visualize: Please refer to Figure CP32.75. Solve: Because the switch has been in position a for a very long time, the capacitor is fully charged to Q0 = EC = (50 V)(20 106 F) = 1.00 103 C When the switch is placed in position b for 1.25 ms, its charge will decay through the resistor to
Q = Q0e t RC = (1.00 103 C ) e The energy in the capacitor after 1.25 ms is
1.25103 s
(
)
( 50 ) ( 20106
F
)
= 0.2865 106 C
6 1 Q 2 1 ( 0.2865 10 C ) U= = = 2.05 mJ 2C 2 20 106 F 2
The energy in the capacitor at t = 0 s (when the switch was just flipped to position b) is
1 Q02 1 (1.0 10 C ) U0 = = = 25 mJ 2 C 2 20 106 F
3 2
So, the energy dissipated by the 50 resistor is 25 mJ 2.05 mJ = 23 mJ.
32.76. Model: The connecting wires are ideal. The capacitors discharge through the resistors.
Visualize:
The figure shows how to simplify the circuit in Figure P32.76 using the laws of series and parallel resistors and the laws of series and parallel capacitors. Solve: The 30 and 20 resistors are in parallel and are equivalent to a 12 resistor. This 12 resistor is in series with the 8 resistor so the equivalent resistance of the circuit Req = 20 . The two 60 F capacitors are in series producing an equivalent capacitance of 30 F. This 30 F capacitor is in parallel with the 20 F capacitor so the equivalent capacitance Ceq of the circuit is 50 F. The time constant of this circuit is
= ReqCeq = (20 )(50 F) = 1.0 ms
The current due to the three capacitors through the 20 equivalent resistor is the same as through the 8 resistor. So, the voltage across the 8 resistor follows the decay equation V = V0et/. For V = V0/2, we get t 1 V0/2 = V0et/1.0 ms ln 2 = 1.0 ms t = 0.69 ms
32.77. Model: The battery and the connecting wires are ideal.
Visualize: Please refer to Figure 32.38a. 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 Kirchhoffs loop law and starting clockwise from the lower left corner of the loop,
+E iR
Integrating both sides,
Q
q dq q dq dt = =0E = R + RC dq = ( EC q)dt EC q RC C dt C
t
Q dq dt t t EC q = RC ln ( EC q ) 0 = RC ln ( EC Q ) + ln ( EC ) = RC 0 0
t EC Q EC Q ln = e t RC Q = EC (1 e t RC ) = EC RC EC
Letting Qmax = EC and = RC, we get Q = Qmax (1 et/).
32.78. Model: The battery and the connecting wires are ideal.
Visualize: Please refer to Figure 32.38a. Solve: (a) According to Equation 32.36, during charging the charge on the capacitor increases according to Q = Qmax (1 e t ) . Therefore, the current in the circuit behaves as
I= dQ EC t RC E t RC 1 e = Qmax ( e t ) = =e dt RC R
Using Equation 32.8, the power supplied by the battery as the capacitor is being charged is E2 E Pbat = I E = e t RC E = e t RC R R Because Pbat = dU dt , we have
E 2 t RC E2 e dt = RC et RC = E 2C 0 R 0R
dU = Pbat dt dU = Pbat dt =
0
That is, the total energy which has been supplied by the battery when the capacitor is fully charged is E2C. (b) The power dissipated by the resistor as the capacitor is being charged is E2 Presistor = I 2 R = 2 R e 2t RC R Because Presistor = dU dt , we have dU = Presistordt dU =
E 2 2t RC E 2 RC 2t RC E 2C e dt U resistor = 2 e =2 R 0 0R
(c) The energy stored in the capacitor when it is fully charged is 2 1 Qmax 1 C 2E 2 1 2 UC = = = CE 2C 2C 2 (d) 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 Ubat = E2C, U resistor = 1 E 2C , and 2
U C = 1 E 2C . 2
32.79. Model: The battery and the connecting wires are ideal.
Visualize: Please refer to Figure CP32.79. Solve: (a) During charging, when the neon gas behaves like an insulator, the charge on the capacitor increases according to Equation 32.36. That is, Q = Qmax (1 et/). Because Q = CVC, VC = VC0(1 et/) = E(1 et/) Let us say that the period of oscillation begins when V = Voff and it ends when V = Von. Then,
Voff = E (1 e toff
Because the period T = ton toff, we have
) E EV
off
= e toff
toff
E = ln E Voff
E E E Voff E Voff T = ln ln = ln = RC ln E Von E Voff E Von E Von
(b) Substituting the given values into the above expression and noting that and T = 1 f = 0.10 s ,
0.10 s 90 V 20 V = 5.1 k 0.10 s = R (10 106 F ) ln R= 90 V 80 V (10 106 F) ln 7
32.80. Model: The wire can be thought of as a large number of small wire segments in series.
Solve: (a) The resistance R of a segment of the wire of length x is x x x 2 x l R = = = e 2 xl A r0 2 re
(
0
)
The resistance of the complete length of wire is
R = R
0 L
r0
2
e
2x
l
l 2x dx = 2 e l r0 2
L
=
0
2L l e l 1 2 2 r0
(
)
(b) The exponential decrease in radius would seem flat if l W L . Or, put another way, (c) The Taylor series expansion of the exponential function is 2 2L 2L 1 2L + e l =1+ + l 2 l 2L L 2L Keeping the expansion to first order in , e l 1 + . Thus l l l 2L L L R 1 = 2 = 1 + 2 r02 l A r0 Assess: This is the expected resistance for a wire of constant radius r0.
L V1 . l
32-1