Physics 102, Chapter 20 Homework Solutions

Transcription

1 hysics 0, Chapter 0 Homework Solutions. EASONING The current I is defined in Equation 0. as the amount of charge q per unit of time t that flows in a wire. Therefore, the amount of charge is the product of the current and the time interval. The number of electrons is equal to the charge that flows divided by the magnitude of the charge on an electron. a. The amount of charge that flows is 3 q I t 8 A.0 0 s C b. The number of electrons N is equal to the amount of charge divided by e, the magnitude of the charge on an electron. q C N e.60 0 C 4. EASONING a. According to Ohm s law, the current is equal to the voltage between the cell walls divided by the resistance. b. The number of Na + ions that flow through the cell wall is the total charge that flows divided by the charge of each ion. The total charge is equal to the current multiplied by the time. 7 a. The current is A I 5.00 (0.) b. The number of Na + ions is the total charge q that flows divided by the charge +e on each ion, or q/ e. The charge is the product of the current I and the time t, according to Equation 0., so that Number of Na + q I t.50 A0.50 s 7 ions = 4.70 e e C 0. EASONING a. The resistance of a piece of material is related to its length L and cross-sectional area A by Equation 0.3, L / A, where is the resistivity of the material. In order to rank the resistances, we need to evaluate L and A for each configuration in terms of L 0, the unit of length. a b c esistance 4L0 L L L L0 L 4L 8L L0 L 4L L ank 3

2 Therefore, we expect that a has the largest resistance, followed by c, and then by b. pg. b. Equation 0. states that the current I is equal to the voltage divided by the resistance, I /. Since the current is inversely proportional to the resistance, the largest current arises when the resistance is smallest, and vice versa. Thus, we expect that b has the largest current, followed by c, and then by a. a. The resistances can be found by using the results from the EASONING:.50 0 m L m.50 0 m L m.50 0 m 0.50 L m a b c b. The current in each case is given by Equation 0., where the value of the resistance is obtained from part (a): 3.00 a I 5.00 A b I 80.0 A c I 0.0 A EASONING AND The resistance of the cable is Since A = r, the radius of the cable is L I A m0.4 m00 A LI 3 r m.6 0. EASONING AND According to Equation 0.6c, the power delivered to the iron is W 4. EASONING AND The power delivered is = I, so that we have a. bd = I bd = (0 )( A) = 300 W b. vc = I vc = (0 )(4.0 A) = 480 W

3 pg. 3 c. The energy is E = t, so that we have E E bd vc bd tbd (300 W)(5 min).4 t (480 W)(30.0 min) vc vc 7. SSM EASONING According to Equation 6.0b, the energy used is Energy = t, where is the power and t is the time. According to Equation 0.6a, the power is = I, where I is the current and is the voltage. Thus, Energy = It, and we apply this result first to the dryer and then to the computer. The energy used by the dryer is For the computer, we have Solving for t we find Energy t It (6 A)(40 )(45 min) Energy J It.7 A 60 s.00 min Converts m inutes to seconds t J.04 0 J.00 h t 3.0 s 3.0 s 8.9 h.7 A s 3. EASONING Substituting 0 into Equation 0.7 gives a result that can be solved directly for the desired time. From Equation 0.7 we have 0 0 sin f t or sin f t Using the inverse trigonometric sine function, we find ftsin 0.54 In this result, the value of 0.54 is in radians and corresponds to an angle of 30.0º. Thus we find that the smallest value of t is t.39 0 s f 60.0 Hz 33. EASONING a. The average power delivered to the copy machine is equal to the square of the rms-current I rms times the resistance, or rms I (Equation 0.5b). Both I rms and are known.

4 b. According to the discussion in Section 0.5, the peak power peak. peak a. The average power is b. The peak power is twice the average power, so pg. 4 is twice the average power, or I rms 6.50 A W (0.5b) peak 786 W 57 W 4. SSM EASONING Using Ohm's law (Equation 0.) we can write an expression for the voltage across the original circuit as I00. When the additional resistor is inserted in series, assuming that the battery remains the same, the voltage across the new combination is given by I( 0 ). Since is the same in both cases, we can write I00 I( 0 ). This expression can be solved for 0. Solving for 0, we have Therefore, we find that I I I or ( I I) I I (.0 A)(8.00 ) 3 I I 5.0 A.0 A EASONING Ohm s law relates the resistance of either resistor to the current I in it and the voltage across it: (0.) I Because the two resistors are in series, they must have the same current I. We will, therefore, apply Equation 0. to the 86- resistor to determine the current I. Following that, we will use Equation 0. again, to obtain the potential difference across the 67- resistor. Let = 86 be the resistance of the first resistor, which has a potential difference of = 7 across it. The current I in this resistor, from Equation 0., is I () Let = 67 be the resistance of the second resistor. Again employing Equation 0., the potential difference across this resistor is given by

5 pg. 5 I () Since the current in both resistors is the same, substituting Equation () into Equation () yields 7 I EASONING AND The rule for combining parallel resistors is which gives 5 or EASONING a. The three resistors are in series, so the same current goes through each resistor: I I I3. The voltage across each resistor is given by Equation 0. as = I. Because the current through each resistor is the same, the voltage across each is proportional to the resistance. Since, we expect the 3 ranking of the voltages to be. 3 b. The three resistors are in parallel, so the same voltage exists across each:. The current 3 through each resistor is given by Equation 0. as I = /. Because the voltage across each resistor is the same, the current through each is inversely proportional to the resistance. Since, we expect 3 the ranking of the currents to be I I I. 3 a. The current through the three resistors is given by I = / s, where s is the equivalent resistance of the series circuit. From Equation 0.6, the equivalent resistance is s = The current through each resistor is 4.0 I I I A 85.0 s The voltage across each resistor is 3 3 I 0.8 A I 0.8 A I 0.8 A b. The resistors are in parallel, so the voltage across each is the same as the voltage of the battery: 4.0 3

6 The current through each resistor is equal to the voltage across each divided by the resistance: 4.0 I A 50.0 pg. 6 I A 5.0 I A EASONING We will approach this problem in parts. The resistors that are in series will be combined according to Equation 0.6, and the resistors that are in parallel will be combined according to Equation 0.7. The and 3.00 resistors are in series with an equivalent resistance of s = This equivalent resistor of 6.00 Ω is in parallel with the resistor, so This new equivalent resistor of.00 Ω is in series with the resistor, so s ' = s ' is in parallel with the resistor, so Finally, p ' is in series with the.00-, so the total equivalent resistance is 4.67 Ω.

7 65. SSM EASONING When two or more resistors are in series, the equivalent resistance is given by Equation 0.6: s Likewise, when resistors are in parallel, the expression to be solved to find the equivalent resistance is given by Equation 0.7:.... We will p 3 successively apply these to the individual resistors in the figure in the text beginning with the resistors on the right side of the figure. Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the combination of those two resistors is 0.0. The 9.0- and 8.0- resistors are in parallel; their equivalent resistance is 4.4. The equivalent resistances of the parallel combination (9.0 and 8.0 ) and the series combination (4.0 and the 6.0 ) are in parallel; therefore, their equivalent resistance is.98. The.98- combination is in series with the 3.0- resistor, so that equivalent resistance is Finally, the combination and the 0.0- resistor are in parallel, so the equivalent resistance between the points A and B is EASONING The two resistors and are wired in series, so we can determine their equivalent resistance. The resistor 3 is wired in parallel with the equivalent resistance, so the equivalent resistance 3 can be found. Finally, the resistor 4 is wired in series with the equivalent resistance 3. With these observations, we can evaluate the equivalent resistance between the points A and B. pg. 7 = 6 = 8 Ω A 4 = 6 B 3 = 48 Since and are wired in series, the equivalent resistance is (0.6) The resistor 3 is wired in parallel with the equivalent resistor, so the equivalent resistance 3 of this combination is or (0.7) The resistance 4 is in series with the equivalent resistance 3, so the equivalent resistance AB between the points A and B is AB

8 68. EASONING The total power delivered by the battery is related to the equivalent resistance eq connected between the battery terminals and to the battery voltage according to Equation 0.6c: /. eq We note that the combination of resistors in circuit A is also present in circuits B and C (see the shaded part of these circuits in the following drawings). In circuit B an additional resistor is in parallel with the combination from A. The equivalent resistance of resistances in parallel is always less than any of the individual resistances alone. Therefore, the equivalent resistance of circuit B is less than that of A. In circuit C an additional resistor is in series with the combination from A. The equivalent resistance of resistances in series is always greater than any of the individual resistances alone. Therefore, the equivalent resistance of circuit C is greater than that of A. We conclude then that the equivalent resistances are ranked C, A, B, with C the greatest and B the smallest. pg. 8 + A + B Since the total power delivered by the battery is / eq, it is inversely proportional to the equivalent resistance. The battery voltage is the same in all three cases, so the power ranking is the reverse of the ranking deduced previously for eq. In other words, we expect that, from greatest to smallest, the total power delivered by the battery is B, A, C. The total power delivered by the battery is / eq. The voltage is given, but we must determine the equivalent resistance in each case. In circuit A each branch of the parallel combination consists of two resistances in series. Thus, the resistance of each branch is eq, according to Equation 0.6. The two parallel branches have an equivalent resistance that can be determined from Equation 0.7 as or A A + C

9 pg. 9 In circuit B the resistance of circuit A is in parallel with an additional resistance. According to Equation 0.7, the equivalent resistance of this combination is = + or B A B In circuit C the resistance of circuit A is in series with an additional resistance. Equation 0.6, the equivalent resistance of this combination is According to C A We can now use Circuit A / to determine the total power delivered by the battery in each case: eq W 9.0 Circuit B W 6.0 Circuit C.0 W SSM EASONING The current I can be found by using Kirchhoff's loop rule. Once the current is known, the voltage between points A and B can be determined. a. We assume that the current is directed clockwise around the circuit. Starting at the upper-left corner and going clockwise around the circuit, we set the potential drops equal to the potential rises: Solving for the current gives I 0.38 A. b. The voltage between points A and B is c. oint B is at the higher potential. (5.0 )I (7)I 0.0 ()I (8.0)I 30.0 otentialdrops AB 30.0 (0.38 A)(7 ).0 0 otentialrises

10 pg EASONING First, we draw a current I (directed to the right) in the 6.00-resistor. We can express I in terms of the other currents in the circuit, I and 3.00 A, by applying the junction rule to the junction on the left; the sum of the currents into the junction must equal the sum of the currents out of the junction. I I 3.00 A or I I 3.00 A Current into junction on left Current out of junction on left In order to obtain values for I and we apply the loop rule to the top and bottom loops of the circuit. Applying the loop rule to the top loop (going clockwise around the loop), we have 3.00 A A I 3.00 A6.00 otential drops This equation can be solved directly for the current; I 5.00 A. otential rises Applying the loop rule to the bottom loop (going counterclockwise around the loop), we have I 3.00 A I.00 otential drops Substituting I = 5.00 A into this equation and solving for gives otential rises EASONING This problem can be solved by using Kirchhoff s loop rule. We begin by drawing a current through each resistor. The drawing shows the directions chosen for the currents. The directions are arbitrary, and if any one of them is incorrect, then the analysis will show that the corresponding value for the current is negative. = 4.0 A + = B F =.0 + I C E = + I D We mark the two ends of each resistor with plus and minus signs that serve as an aid in identifying the potential drops and rises for the loop rule, recalling that conventional current is always directed from a higher potential (+) toward a lower potential ( ). Thus, given the directions chosen for I and I, the plus and minus signs must be those shown in the drawing. We then apply Kirchhoff's loop rule to the top loop (ABCF) and to the bottom loop (FCDE) to determine values for the currents I and I.

11 Applying Kirchhoff s loop rule to the top loop (ABCF) gives pg. I I otential otential rises drop Similarly, for the bottom loop (FCDE), I otential otential rise drop () () Solving Equation () for I gives I A Since I is a positive number, the current in the resistor goes from left to right, as shown in the drawing. Solving Equation () for I and substituting I = / into the resulting expression yields I I A 8.0 Since I is a positive number, the current in the resistor goes from left to right, as shown in the drawing.