CHAPTE 0 ELECTIC CICUITS ANSWES TO FOCUS ON CONCEPTS QUESTIONS..5 A. (c) Ohm s law states that the voltage is directly proportional to the current I, according to = I, where is the resistance. Thus, a plot of voltage versus current is a straight line that passes through the origin. 3. (e) Since both wires are made from the same material, the resistivity ρ is the same for each. L The resistance is given by (Equation 0.3), where L is the length and A is the A cross-sectional area of the wire. With twice the length and one-half the radius (one-fourth L the cross-sectional area), the second wire has 8 A / 4 times the resistance as the first wire. 4. 50 C 5. (a) Power P is the current I times the voltage or P I (Equation 0.6a). However, since Ohm s law applies to a resistance, the power is also P I (Equation 0.6b) and P (Equation 0.6c). Therefore, all three of the changes specified leave the power unchanged. 6. 7 W 7. 0.9 A 8. (d) According to Ohm s law, the voltage across the resistance is I. The two resistors are connected in series, and their equivalent resistance is, therefore, +. According to Ohm s law, the current in the circuit is expression into the expression for gives. I. Substituting this
054 ELECTIC CICUITS 9. (b) The series connection has an equivalent resistance of S. The parallel connection has an equivalent resistance that can be determined from Therefore, it follows that P. The ratio of these values is S P. P 4. 0. (e) Since the two resistors are connected in parallel across the battery terminals, the same voltage is applied to each. Thus, according to Ohm s law, the current in each resistor is I / inversely proportional to the resistance, so that. I /. 0.09 A. (c) In arrangement B the two resistors in series have a combined resistance of. Each series combination is in parallel, so that the reciprocal of the equivalent resistance is, or eq, B. Following the technique outlined in Section 0.8, we eq, B also find that 3 and. eq, C 4 eq, A 5 3. (d) The internal resistance r of the battery and the resistance are in series, so that the current from the battery can be calculated via Ohm s law as I. The voltage between r the terminals is, then, I. This result can be solved to show that = r. r 4. (b) Kirchhoff s junction rule states that the sum of the magnitudes of the currents directed into a junction equals the sum of the magnitudes of the currents directed out of the junction. Here, this rule implies that I I 3 I. 5. (d) Once the current in the resistor is drawn, the markings of the plus and minus sign are predetermined. This is because conventional current always flows from a higher toward a lower potential. Thus, since the current I 4 is directed from right to left, the right side of 4 must be marked plus and the left side minus. 6. (b) Kirchhoff s loop rule states that, around any closed loop, the sum of the potential drops equals the sum of the potential rises. 7. (a) The ammeter must be connected so that the current that flows through the resistor also flows through the ammeter. The voltmeter must be connected across the resistor.
Chapter 0 Answers to Focus on Concepts Questions 055 8. (e) The two capacitors in series have an equivalent capacitance C S that can be determined from, so that C S C. This capacitance is in parallel with a capacitance C, so C C C S that the total equivalent capacitance is C C C 3 C. eq 9. (c) The time constant is given by the product of the resistance and the capacitance. Therefore, when the resistance is reduced to one-third of its initial value, the capacitance must be tripled, in order that the time constant remains unchanged. 0..8 s
056 ELECTIC CICUITS CHAPTE 0 ELECTIC CICUITS POBLEMS. 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. SOLUTION a. The amount of charge that flows is 3 q I t 8 A.0 0 s 3.6 0 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 3.6 0 C 7 N.3 0 9 e.60 0 C. EASONING We are given the average current I and its duration Δt. We will employ q I (Equation 0.) to determine the amount Δq of charge delivered to the ground by t the lightning flash. SOLUTION Solving I q t (Equation 0.) for Δq, we obtain 3 q It.60 A 0.38 s 74 C 3. SSM EASONING Electric current is the amount of charge flowing per unit time (see Equation 0.). Thus, the amount of charge is the current times the time. Furthermore, the potential difference is the difference in electric potential energy per unit charge (see Equation 9.4), so that, once the amount of charge has been calculated, we can determine the energy by multiplying the potential difference by the charge. SOLUTION
Chapter 0 Problems 057 a. According to Equation 0., the current I is the amount of charge q divided by the time t, or I = q/t. Therefore, the additional charge that passes through the machine in normal mode versus standby mode is q I t I t I I t additional normal standby normal standby q in normal mode q in standby mode 0.0 A 0.067 A 60.0 s.6 C b. According to Equation 9.4, the potential difference is the difference (EPE) in the electric potential energy divided by the charge q additional, or = (EPE)/q additional. As a result, the additional energy used in the normal mode as compared to the standby mode is q EPE.6 C 0 30 J additional 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. SOLUTION a. The current is 3 750 9.5 0 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 9.600 C 5. EASONING AND SOLUTION Ohm's law (Equation 0., I) gives the result directly I 40 A
058 ELECTIC CICUITS 6. EASONING oltage is a measure of energy per unit charge (joules per coulomb). Therefore, when an amount Δq of charge passes through the toaster and there is a potential difference across the toaster equal to the voltage of the outlet, the energy that the charge delivers to the toaster is given by Energy q () The charge Δq that flows through the toaster in a time Δt depends upon the magnitude I of q the electric current according to I (Equation 0.). Therefore, we have that t q I t () We will employ Ohm s law, I (Equation 0.), to calculate the current in the toaster in terms of the voltage of the outlet and the resistance of the toaster. SOLUTION Substituting Equation () into Equation () yields Energy q I t (3) Substituting I (Equation 0.) into equation (3), we find that 0 4 Energy It t t 60.0 s 6.0 J 4 7. EASONING According to Ohm s law, the resistance is the voltage of the battery divided by the current that the battery delivers. The current is the charge divided by the time during which it flows, as stated in Equation 0.. We know the time, but are not given the charge directly. However, we can determine the charge from the energy delivered to the resistor, because this energy comes from the battery, and the potential difference between the battery terminals is the difference in electric potential energy per unit charge, according to Equation 9.4. Thus, a 9.0- battery supplies 9.0 J of energy to each coulomb of charge passing through it. To calculate the charge, then, we need only divide the energy from the battery by the 9.0 potential difference. SOLUTION Ohm s law indicates that the resistance is the voltage of the battery divided by the current I, or = /I. According to Equation 0., the current I is the amount of charge q divided by the time t, or I = q/t. Using these two equations, we have t I q / t q
Chapter 0 Problems 059 According to Equation 9.4, the potential difference is the difference (EPE) in the EPE electric potential energy divided by the charge q, or. However, it is q customary to denote the potential difference across a battery by, rather than, so EPE EPE. Solving this expression for the charge gives q. Using this result q in the expression for the resistance, we find that 9.0 63600 s t t t 6 q EPE / EPE 5.0 J 8. EASONING AND SOLUTION a. The total charge that can be delivered is 3600 s 5 q (0 Ah) 7.90 C h b. The maximum current is 0 Ah I 350 A h (38 min) 60 min 9. SSM EASONING The number N of protons that strike the target is equal to the amount of electric charge q striking the target divided by the charge e of a proton, N = (q)/e. From Equation 0., the amount of charge is equal to the product of the current I and the time t. We can combine these two relations to find the number of protons that strike the target in 5 seconds. The heat Q that must be supplied to change the temperature of the aluminum sample of mass m by an amount T is given by Equation.4 as Q cmt, where c is the specific heat capacity of aluminum. The heat is provided by the kinetic energy of the protons and is equal to the number of protons that strike the target times the kinetic energy per proton. Using this reasoning, we can find the change in temperature of the block for the 5 secondtime interval. SOLUTION a. The number N of protons that strike the target is 6 q I t (0.50 0 A)(5 s) N 4. 7 0 e e 9.6 0 C 3
060 ELECTIC CICUITS b. The amount of heat Q provided by the kinetic energy of the protons is Since Q 3 Q (4.7 0 protons)(4.9 0 J/proton) 30 J cm T and since Table. gives the specific heat of aluminum as c = 9.00 0 J/(kg. C ), the change in temperature of the block is Q 30 J T 7 C cm J 3 9.00 0 (5 0 kg) kg 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. esistance ank a b c 4L0 L L L 0 0 0 L0 L 4L 8L 0 0 0 L0 L 4L L 0 0 0 3 Therefore, we expect that a has the largest resistance, followed by c, and then by b. 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. SOLUTION a. The resistances can be found by using the results from the EASONING:
Chapter 0 Problems 06.50 0 m 0.600 L 0 5.00 0 m.50 0 m 0.0375 8L 0 8 5.00 0 m a b.50 0 m 0.50 L 0 5.00 0 m 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 0.600 3.00 b I 80.0 A 0.0375 3.00 c I 0.0 A 0.50. EASONING The resistance of a wire that has a length L and a cross-sectional area A is L given by Equation 0.3 as. Both wires have the same length and cross-sectional A area. Only the resistivity ρ of the wire differs, and Table 0. gives the following values: ρ Aluminum =.8 0 8 Ω m and ρ Copper =.7 0 8 Ω m. Applying Equation 0.3 to both wires and dividing the two equations will allow us to eliminate the unknown length and cross-sectional area algebraically and solve for the resistance of the copper wire. SOLUTION Applying Equation 0.3 to both wires gives L and A Copper Copper Aluminum Aluminum L A Dividing these two equations, eliminating L and A algebraically, and solving the result for Copper give L / A L / A Copper Copper Copper Aluminum Aluminum Aluminum 0.0 0. 8 Copper.70 m Copper Aluminum 8 Aluminum.80 m
06 ELECTIC CICUITS. EASONING AND SOLUTION The resistance of the cable is Since A = r, the radius of the cable is L I A 8.7 0 m0.4 m00 A LI 3 r 9.9 0 m.6 0 3. EASONING AND SOLUTION From Equation 0.5 we have that = 0 [ + (T T 0 )]. Solving for T gives 99.6 0 5 T T0 0.0 C 34. 6 C 3 3.7 0 (C ) 4. EASONING AND SOLUTION Using Equation 0.3 and the resistivity of aluminum from Table 0., we find.8 0 8 m0.0 0 3 m L 0.58 A 4 4.9 0 m 5. SSM WWW EASONING The resistance of a metal wire of length L, cross-sectional area A and resistivity is given by Equation 0.3: L / A. Solving for A, we have A L /. We can use this expression to find the ratio of the cross-sectional area of the aluminum wire to that of the copper wire. SOLUTION Forming the ratio of the areas and using resistivity values from Table 0., we have A 8 aluminum aluminuml / aluminum.80 m 8.64 A L /.70 m copper copper copper
Chapter 0 Problems 063 6. EASONING The resistance of the spooled wire decreases as its length L decreases, L according to (Equation 0.3). The resistivity and cross-sectional area A of the A wire do not change. Because the same battery is used, the potential difference across the wire is the same in both cases. Therefore, (Equation 0.) explains why the current I I increases as the resistance of the wire decreases. L A SOLUTION Solving (Equation 0.3) for L, we obtain L. Thus, the initial A length L 0 and final length L f of the wire are given by L A A () 0 f 0 and Lf where 0 is the initial resistance, and f the final resistance, of the wire. Taking the ratio of Equations () eliminates the unknown quantities A and, allowing us to solve for L f in terms of L 0 and the initial and final resistances: Lf L 0 A f 0 A f f or Lf L 0 0 0 () From (Equation 0.), the initial resistance I 0 and final resistance f of the spooled wire are and (3) 0 f I0 If Substituting Equations (3) into Equation () yields I Lf f I 0.4 A L0 L0 75 m 58 m I f 3. A I 0
064 ELECTIC CICUITS 7. EASONING Assuming that the resistance is at a temperature T and 0 at a temperature T 0, we can write the percentage change p in resistance as 0 p 00 0 Equation 0.5, on the other hand, gives the resistance as a function of temperature as follows: T T 0 0 where α is the temperature coefficient of resistivity. Substituting this expression into the expression for the percentage change in resistance gives 0 0 0 0 0 0 0 00 T T p 00 T T0 00 () The change in temperature is unknown, but it is the same for both wires. Therefore, we will apply Equation () to each wire and divide the two expressions to eliminate the unknown change in temperature. From the result we will be able to calculate the percentage change in the resistance of the tungsten wire. SOLUTION Applying Equation () to both wires gives p T T 00 and p T T 00 Tungsten Tungsten 0 Gold Gold 0 Dividing these two expressions, eliminating (T T 0 ) algebraically, and solving for p Tungsten give ptungsten Tungsten T T0 00 Tungsten p T T 00 Gold Gold 0 Gold p Tungsten 0.0045 C Tungsten pgold Gold 0.0034 C 7.0% 9.3% 8. EASONING AND SOLUTION Suppose that when the initial temperature of the wire is T 0 the resistance is 0, and when the temperature rises to T the resistance is. The relation between temperature and resistance is given by Equation 0.5 as = 0 [ + (T T 0 )], where is the temperature coefficient of resistivity. The initial and final resistances are related to the voltage and current as 0 = /I 0 and = /I, where the voltage across the
Chapter 0 Problems 065 wire is the same in both cases. Substituting these values for 0 and into Equation 0.5 and solving for T, we arrive at I 0 I.50 A.30 A T T 0 0C 4.5 0 4 360 C (C) 9. SSM WWW EASONING The resistance of a metal wire of length L, cross-sectional area A and resistivity is given by Equation 0.3: L / A. The volume of the new wire will be the same as the original volume of the wire, where volume is the product of length and cross-sectional area. Thus, = or A L = A L. Since the new wire is three times longer than the first wire, we can write A L A L A (3L ) or A A / 3 We can form the ratio of the resistances, use this expression for the area A, and find the new resistance. SOLUTION The resistance of the new wire is determined as follows: Solving for, we find that L / A 3 L A L A 9 L / A L A L A /3 9 9(.0 ) 89 0. EASONING The length L of the wire is related to its resistance and cross-sectional area A by L A/ (see Equation 0.3), where is the resistivity of tungsten. The resistivity is known (see Table 0.), and the cross-sectional area can be determined since the radius of the wire is given. The resistance can be obtained from Ohm s law as the voltage divided by the current. SOLUTION The length L of the wire is A L ()
066 ELECTIC CICUITS Since the cross-section of the wire is circular, its area is A r, where r is the radius of the wire. According to Ohm s law (Equation 0.), the resistance is related to the voltage and current I by / I. Substituting the expressions for A and into Equation () gives r 0.00300 3 m 0 A L I.4 A 0.049 m 8 5.60 m. EASONING AND SOLUTION The mass m of the aluminum wire is equal to the density d of aluminum times the volume of the wire. The wire is cylindrical, so its volume is equal to the cross-sectional area A times the length L; m = dal. The cross-sectional area of the wire is related to its resistance and length L by Equation 0.3; = L/A, where is the resistivity of aluminum (see Table 0.). Therefore, the mass of the aluminum wire can be written as m dal d L L The resistance is given by Ohm s law as = /I, so the mass of the wire becomes L d L I m d L 3 8 (700 kg/m )(.8 0 m)(75 m) (5 A) m 9.7 0 kg 0.300. EASONING AND SOLUTION According to Equation 0.6c, the power delivered to the iron is P 0 4 6.0 0 W 3. EASONING AND SOLUTION a. According to Equation 0.6c, the resistance is 4.4 P 33 W
Chapter 0 Problems 067 b. According to Equation 0.6a, the current is P 33 W I.8 A 4. EASONING AND SOLUTION The power delivered is P = I, so that we have a. P bd = I bd = (0 )( A) = 300 W b. P vc = I vc = (0 )(4.0 A) = 480 W c. The energy is E = Pt, so that we have E E bd vc Pbd tbd (300 W)(5 min).4 P t (480 W)(30.0 min) vc vc 5. EASONING The total cost of keeping all the Ts turned on is equal to the number of Ts times the cost to keep each one on. The cost for one T is equal to the energy it consumes times the cost per unit of energy ($0. per kwh). The energy that a single set uses is, according to Equation 6.0b, the power it consumes times the time of use. SOLUTION The total cost is Cost per set Total cost = 0 million sets $0. 0 million setsenergy in kw h used per set kw h The energy (in kwh) used per set is the product of the power and the time, where the power is expressed in kilowatts and the time is in hours: kw Energy used per set = Pt 75 W 6.0 h 000 W (6.0b) The total cost of operating the T sets is kw $0. 6 Total cost = 0 million sets75 W6.0 h $5.9 0 000 W kw h
068 ELECTIC CICUITS 6. EASONING a. The power delivered to a resistor is given by Equation 0.6c as P /, where is the voltage and is the resistance. Because of the dependence of the power on, doubling the voltage has a greater effect in increasing the power than halving the resistance. The following table shows the power for each circuit, given in terms of these variables, and confirms this fact. The table also gives the expected ranking, in decreasing order, of the power. Power ank a b c P 3 P 4 4 P P d b. The current is given by Equation 0. as I = /. Note that the current, unlike the power, depends linearly on the voltage. Therefore, either doubling the voltage or halving the resistance has the same effect on the current. The following table shows the current for the four circuits and confirms this fact. The table also gives the expected ranking, in decreasing order, of the current. Current ank a b c I I I 3 d I
Chapter 0 Problems 069 SOLUTION a. Using the results from the EASONING and the values of =.0 and = 6.00, we find that the power dissipated in each resistor is a b c d Power.0 P 4.0 W 6.00.0 P.0 W 6.00 4 4.0 P 96.0 W 6.00.0 P 48.0 W 6.00 ank 3 4 b. Using the results from part (b) and the values of =.0 and = 6.00, we find that the current in each circuit is Current ank a b c d.0 I.00 A 6.00.0 I.00 A 6.00.0 I 4.00 A 6.00.0 I.00 A 6.00 7. SSM EASONING According to Equation 6.0b, the energy used is Energy = Pt, where P is the power and t is the time. According to Equation 0.6a, the power is P = 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. SOLUTION The energy used by the dryer is 3
070 ELECTIC CICUITS Energy Pt It (6 A)(40 )(45 min) For the computer, we have Solving for t we find Energy.040 7 J It.7 A 60 s.00 min Converts m inutes to seconds 7 4 4 0 t.040 7 J.04 0 J.00 h t 3.0 s 3.0 s 8.9 h.7 A 0 3600 s 8. EASONING AND SOLUTION We know that the resistance of the wire can be obtained from P = / or = /P We also know that = L/A. Solving for the length, noting that A = r, and using = 00 0 8. m from Table 0., we find / Pr 0 6.5 0 4 m P 8 00 0 m4.00 0 W A r L 50 m 9. EASONING A certain amount of time t is needed for the heater to deliver the heat Q required to raise the temperature of the water, and this time depends on the power produced by the heater. The power P is the energy (heat in this case) per unit time, so the time is the heat divided by the power or t = Q/P. The heat required to raise the temperature of a mass m of water by an amount T is given by Equation.4 as Q = cmt, where c is the specific heat capacity of water [486 J/(kg Cº), see Table.]. The power dissipated in a resistance is given by Equation 0.6c as P /, where is the voltage across the resistor. Using these expressions for Q and P will allow us to determine the time t. SOLUTION Substituting Equations.4 and 0.6c into the expression for the time and recognizing that the normal boiling point of water is 00.0 ºC, we find that
Chapter 0 Problems 07 Q cmt cmt t P / 5 486 J/ kg C 0.50 kg00.0 C 3 C 0 90 s 30. EASONING The total volume of ice melted is the product of the thickness h and the area A of the layer that melts, which permits us to determine the thickness in terms of the volume and the area: olume ha olume or h () A Given the density = 97 kg/m 3, we can find the volume of the ice from (Equation.), where m is the mass of the ice. m olume olume m () The mass m of the ice determines the heat Q required to melt it, according to Equation.5 4 Q ml (.5) In Equation.5, L 33.5 0 J/kg is the latent heat of fusion for water (see Table.3). In order to find the maximum thickness of ice the defroster can melt, we will assume that all the heat generated by the defroster goes into melting the ice. The power output P of the defroster is the rate at which it converts electrical energy to heat, so we have that Q P or Q Pt (6.0b) t where t is the elapsed time (3.0 minutes). The power output P is, in turn, found from the operating voltage and current I: P I (0.6a) SOLUTION Substituting Equation () into Equation () yields
07 ELECTIC CICUITS Solving Q olume m m h (3) A A A ml (Equation.5) for m, we find that Equation (3), we obtain Substituting Equation (6.0b) into Equation (4) yields Q m. Substituting this result into L m QL Q h A A L A (4) Q Pt h LA LA (5) We note that the elapsed time t is given in minutes, which must be converted to seconds. Substituting Equation 0.6a into Equation (5), we obtain the maximum thickness of the ice that the defroster can melt: Pt It h LA LA 60 s 3 A 3.0 min min 4 3 33.50 J/kg97 kg/m 0.5 m 4 3.0 m 3. SSM EASONING AND SOLUTION As a function of temperature, the resistance of the wire is given by Equation 0.5: ( T T ) 0 0, where is the temperature coefficient of resistivity. From Equation 0.6c, we have P /. Combining these two equations, we have P0 P T T T T0 0 0 where P 0 = / 0, since the voltage is constant. But P P 0, so we find Solving for T, we find P P 0 0 or T T T T 0 T T 8 50 C 0 0.0045 (C ) 0
Chapter 0 Problems 073 3. EASONING Substituting 0 into Equation 0.7 gives a result that can be solved directly for the desired time. SOLUTION 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 0.54 0.54 3 t.39 0 s f 60.0 Hz 33. EASONING a. The average power P delivered to the copy machine is equal to the square of the rms-current I rms times the resistance, or P I (Equation 0.5b). Both I rms and are known. b. According to the discussion in Section 0.5, the peak power Ppeak is twice the average power, or Ppeak P. rms SOLUTION a. The average power is P I rms 6.50 A 8.6 786 W (0.5b) b. The peak power is twice the average power, so P peak P 786 W 57 W
074 ELECTIC CICUITS 34. EASONING AND SOLUTION The expression relating the peak current, I 0, to the rmscurrent, I rms, is I0.50 A Irms.77 A 35. SSM EASONING The average power is given by Equation 0.5c as P rms /. In this expression the rms voltage rms appears. However, we seek the peak voltage 0. The relation between the two types of voltage is given by Equation 0.3 as rms 0 /, so we can obtain the peak voltage by using Equation 0.3 to substitute into Equation 0.5c. SOLUTION Substituting rms from Equation 0.3 into Equation 0.5c gives Solving for the peak voltage 0 gives / 0 rms 0 P P 4.0 55 W 0 36. EASONING Because we are ignoring the effects of temperature on the heater, the resistance of the heater is the same whether it is plugged into a 0- outlet or a 30- outlet. The average power output P of the heater is related to the rms outlet voltage rms and the heater s resistance by rms P (Equation 0.5c). SOLUTION When plugged into an outlet in the US that has an rms voltage rms = 0, the average power output P of the heater is, from Equation 0.5c, rms P (0.5c) When operated in Germany, the average power output P0 550 W of the heater is given rms,0 by P0 (Equation 0.5c), where rms,0 = 30. Solving this relation for, we obtain
Chapter 0 Problems 075 rms,0 Substituting Equation () into Equation 0.5c yields rms rms rms 0 P 0 rms,0 rms,0 30 () P P 550 W 50 W P 0 37. EASONING AND SOLUTION The power P dissipated in the extension cord is P = I (Equation 0.6b). The resistance is related to the length L of the wire and its crosssectional area A by Equation 0.3, = L/A, where is the resistivity of copper. The crosssectional area of the wire can be expressed as 0 L I A P L where the ratio P/L is the power per unit length of copper wire that the heater produces. The wire is cylindrical, so its cross-sectional area is A = r. Thus, the smallest radius of wire that can be used is.7 0 m 3 r I (8 A). 30 m P (.0 W / m) L Note that we have used.0 W/m as the power per unit length, rather than.0 W/m. This is because an extension cord is composed of two copper wires. If the maximum power per unit length that the extension cord itself can produce is.0 W/m, then each wire can produce only a maximum of.0 W/m. 38. EASONING The higher the average power output P of the resistance heater, the faster it Q can supply the heat Q needed to raise the water s temperature. This follows from P t Q (Equation 6.0b), which implies that the recovery time is t. The average power output P P of the resistance heater is given by rms 8 P (Equation 0.5c), where is the resistance
076 ELECTIC CICUITS of the heater and rms = 0 is the heater s rms operating voltage. The heater must supply an amount of heat Q cm T (Equation.4) to cause an increase ΔT in the temperature of a mass m of water, where c 486 J/ kg C is the specific heat capacity of water (see Table.). The mass m of the water can be found from the density =.000 0 3 kg/m 3 (see Table.) and the volume water of the water held in the heater, according to m water (Equation.). rms SOLUTION Substituting P (Equation 0.5c) and Q cm T (Equation.4) into Q the expression t for the recovery time, we obtain P Q cmt cmt t P rms rms () Substituting m (Equation.) into Equation () yields water cmt c T t () water rms rms Before using Equation () to calculate the recovery time, we must convert the volume water of the water in the unit from gallons to m 3, using the equivalence gal = 3.79 0 3 m 3 : water 3 3 3.790 m 5 gal 0.0 m.00 gal 3 We note that the recovery time is to be expressed in hours, where.00 h = 3600 s. Therefore, from Equation (), we find that t 3 3 3 c T 3.0 486 J/ kg C.0000 kg/m 0.0 m 53 C C water rms 0.00 h 7300 s 7300 s.0 h 3600 s
Chapter 0 Problems 077 39. SSM EASONING a. We can obtain the frequency of the alternating current by comparing this specific expression for the current with the more general one in Equation 0.8. b. The resistance of the light bulb is, according to Equation 0.4, equal to the rms-voltage divided by the rms-current. The rms-voltage is given, and we can obtain the rms-current by dividing the peak current by, as expressed by Equation 0.. c. The average power is given by Equation 0.5a as the product of the rms-current and the rms-voltage. SOLUTION a. By comparing I 0.707 Asin 34 Hz t with the general expression (see Equation 0.8) for the current in an ac circuit, I I 0 sin f t, we see that 34 Hz f t 34 Hz t or f = 50.0 Hz b. The resistance is equal to rms /I rms, where the rms-current is related to the peak current I 0 by I I /. Thus, the resistance of the light bulb is rms 0 rms rms 0.0 I I rms 0 0.707 A.40 0 (0.4) c. The average power is the product of the rms-current and rms-voltage: I0 0.707 A P Irmsrms rms 0.0 60.0 W (0.5a) 40. EASONING AND SOLUTION The energy Q that is released when the water cools from an initial temperature T to a final temperature of 0.0 C is given by Equation.4 as Q = cm(t 0.0 C). The energy Q released when the water turns into ice at 0.0 C is Q = ml f, where L f is the latent heat of fusion for water. Since power P is energy divided by time, the power produced is Q Q cm( T 0.0 C) ml P t t f
078 ELECTIC CICUITS The power produced by an electric heater is, according to Equation 0.6a, P = I. Substituting this expression for P into the equation above and solving for the current I, we get cm( T 0.0 C) mlf I t (486 J/kg C )(660 kg)(0.0 C ) (660 kg)(33.5 0 J/kg) I 3 A 3600 s (9.0 h) (40 ) h 4 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. SOLUTION Solving for 0, we have Therefore, we find that I I I or ( I I) I 0 0 0 0 0 0 I (.0 A)(8.00 ) 3 I I 5.0 A.0 A 0 4. EASONING According to Equation 0., the resistance of the resistor is equal to the voltage across it divided by the current I, or = /I. Since the resistor, the lamp, and the voltage source are in series, the voltage across the resistor is = 0.0 L, where L is the voltage across the lamp. Thus, the resistance is 0.0 L I Since L is known, we need only determine the current in the circuit. Since we know the voltage L across the lamp and the power P dissipated by it, we can use Equation 0.6a to find the current: I P /. The resistance can be written as L
Chapter 0 Problems 079 0.0 P SOLUTION Substituting the known values for L and P into the equation above, the resistance is 0.0 5 4.0 0 60.0 W 5 43. SSM EASONING The equivalent series resistance s is the sum of the resistances of L the three resistors. The potential difference can be determined from Ohm's law according to I s. SOLUTION a. The equivalent resistance is s 5 45 75 45 b. The potential difference across the three resistors is I s (0.5 A)(45 ) 74 44. EASONING Ohm s law relates the resistance of either resistor to the current I in it and the voltage across it: L (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. SOLUTION 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 ()
080 ELECTIC CICUITS Let = 67 be the resistance of the second resistor. Again employing Equation 0., the potential difference across this resistor is given by I () Since the current in both resistors is the same, substituting Equation () into Equation () yields 7 I 67 86 45. EASONING Since the two resistors are connected in series, they are equivalent to a single equivalent resistance that is the sum of the two resistances, according to Equation 0.6. Ohm s law (Equation 0.) can be applied with this equivalent resistance to give the battery voltage. SOLUTION According to Ohm s law, we find Is I 0. A 47 8 9.0 46. EASONING The circuit containing the light bulb and resistor is shown in the drawing. The resistance of the light bulb is related to the power delivered to it by P I / (Equation 0.6b), where I is the current in the circuit. The power is known, and the current can be obtained from Ohm s law as the voltage of the source divided by the equivalent resistance S of the series circuit: I /. Since the two resistors are wired in series, the equivalent resistance is the sum of the resistances, or S. S Light bulb = 44 Ω P = 3.4 W 0.0 SOLUTION The resistance of the light bulb is P (0.6b) I
Chapter 0 Problems 08 Substituting I / (Equation 0.) into Equation 0.6b gives S P P P S I / S () The equivalent resistance of the two resistors wired in series is S (Equation 0.6). Substituting this expression for S into Equation () yields P P S Algebraically rearranging this equation, we find that 0 P This is a quadratic equation in the variable. The solution can be found by using the quadratic formula (see Appendix C.4): 4 P P 0.0 0.0 44 44 444 3.4 W 3.4 W 85.9 and 4 47. SSM EASONING The resistance of one of the wires in the extension cord is given by Equation 0.3: L / A, where the resistivity of copper is =.7 0. 8 m, according to Table 0.. Since the two wires in the cord are in series with each other, their total resistance is cord wire wire L / A. Once we find the equivalent resistance of the entire circuit (extension cord + trimmer), Ohm's law can be used to find the voltage applied to the trimmer.
08 ELECTIC CICUITS SOLUTION a. The resistance of the extension cord is 8 L (.7 0 m)(46 m). A.3 0 m cord 6 b. The total resistance of the circuit (cord + trimmer) is, since the two are in series, s. 5.0 6. Therefore from Ohm's law (Equation 0.: I), the current in the circuit is I s 0 6. 7.4 A Finally, the voltage applied to the trimmer alone is (again using Ohm's law), trim mer (7.4 A)(5.0 ) 0 48. EASONING The answer is not 340 W + 40 W = 580 W. The reason is that each heater contributes resistance to the circuit when they are connected in series across the battery. For a series connection, the resistances add together to give the equivalent total resistance, according to Equation 0.6. Thus, the total resistance is greater than the resistance of either heater. The greater resistance means that the current from the battery is less than when either heater is present by itself. Since the power for each heater is P = I, according to Equation 0.6b, the smaller current means that the power delivered to an individual heater is less when both are connected than when that heater is connected alone. We approach this problem by remembering that the total power delivered to the series combination of the heaters is the power delivered to the equivalent series resistance. SOLUTION Let the resistances of the two heaters be and. Correspondingly, the powers delivered to the heaters when each is connected alone to the battery are P and P. For the series connection, the equivalent total resistance is +, according to Equation 0.6. Using Equation 0.6c, we can write the total power delivered to this equivalent resistance as P () But according to Equation 0.6c, as applied to the situations when each heater is connected by itself to the battery, we have
Chapter 0 Problems 083 P or P or Substituting Equations () and (3) into Equation () gives P () P (3) P P P 340 W40 W P P 340 W + 40 W 40 W P P P P 49. EASONING Ohm s law provides the basis for our solution. We will use it to express the current from the battery when both resistors are connected and when only one resistor at a time is connected. When both resistors are connected, we will use Ohm s law with the series equivalent resistance, which is +, according to Equation 0.6. The problem statement gives values for amounts by which the current increases when one or the other resistor is removed. Thus, we will focus attention on the difference between the currents given by Ohm s law. SOLUTION When is removed, leaving only connected, the current increases by 0.0 A. In this case, using Ohm s law to express the currents, we have Curren t given b y Ohm's law wh en on ly is present Curren t given b y Ohm's law wh en and are present 0.0A () When is removed, leaving only connected, the current increases by 0.0 A. In this case, using Ohm s law to express the currents, we have Curren t given b y Ohm's law wh en on ly is present Curren t given b y Ohm's law wh en and are present 0.0A () Multiplying Equation () and Equation (), we obtain 0.0 A0.0 A
084 ELECTIC CICUITS Simplifying this result algebraically shows that 0.0 A0.0 A or 0.0 A0.0 A 0.4 A (3) a. Using the result for /( + ) from Equation (3) to substitute into Equation () gives 0.4 A=0.0 A or 35 0.0 A+0.4 A 0.0 A+0.4 A b. Using the result for /( + ) from Equation (3) to substitute into Equation () gives 0.4 A=0.0 A or 5.0 0 0.0 A+0.4 A 0.0 A+0.4 A 50. EASONING AND SOLUTION The rule for combining parallel resistors is which gives P P 5 or 55 446 5. SSM EASONING AND SOLUTION Since the circuit elements are in parallel, the equivalent resistance can be obtained directly from Equation 0.7: p 6 8.0 or p 5.3 5. 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 ranking of the voltages to be 3. 3
Chapter 0 Problems 085 b. The three resistors are in parallel, so the same voltage exists across each:. 3 The current 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 the ranking of the currents to be I I I. 3 3 SOLUTION 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 = 50.0 + 5.0 + 0.0 85.0 The current through each resistor is 4.0 I I I 3 0.8 A 85.0 s The voltage across each resistor is 3 3 I 0.8 A 50.0 4. I 0.8 A 5.0 7.05 I 0.8 A 0.0.8 b. The resistors are in parallel, so the voltage across each is the same as the voltage of the battery: 4.0 3 The current through each resistor is equal to the voltage across each divided by the resistance: 4.0 I 0.480 A 50.0 I 4.0 0.960 A 5.0 4.0 I 3.40 A 0.0 3
086 ELECTIC CICUITS 53. EASONING When the switch is open, no current goes to the resistor. Current exists only in, so it is the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are wired in parallel, we can use Equation 0.7 to find the equivalent resistance. Whether the switch is open or closed, the power P delivered to the circuit can be found from the relation P / (Equation 0.6c), where is the battery voltage and is the equivalent resistance. SOLUTION a. When the switch is open, there is current only in resistor. Thus, the equivalent resistance is 65.0. b. When the switch is closed, there is current in both resistors and, furthermore, they are wired in parallel. The equivalent resistance is or P 38.8 65.0 96.0 P (0.7) c. When the switch is open, the power delivered to the circuit by the battery is given by P /, since the only resistance in the circuit is. Thus, the power is 9.00 P.5 W 65.0 d. When the switch is closed, the power delivered to the circuit is the equivalent resistance of the two resistors wired in parallel: P / P (0.6), where P is 9.00 P.09 W 38.8 P (0.6) 54. EASONING AND SOLUTION The power P dissipated in a resistance is given by Equation 0.6c as P = /. The resistance 50 of the 50.0-Wfilament is 50 P (0.0 ) 50.0 W 88 The resistance 00 of the 00.0-W filament is 0 0 (0.0 ) P 00.0 W 44
Chapter 0 Problems 087 55. SSM EASONING Since the resistors are connected in parallel, the voltage across each one is the same and can be calculated from Ohm's Law (Equation 0.: I). Once the voltage across each resistor is known, Ohm's law can again be used to find the current in the second resistor. The total power consumed by the parallel combination can be found calculating the power consumed by each resistor from Equation 0.6b: P = I. Then, the total power consumed is the sum of the power consumed by each resistor. SOLUTION Using data for the second resistor, the voltage across the resistors is equal to I (3.00 A)(64.0 )= 9 a. The current through the 4.0- resistor is I 9 4.0 4.57 A b. The power consumed by the 4.0- resistor is P I (4.57 A) (4.0 ) = 877 W while the power consumed by the 64.0- resistor is P I (3.00 A) (64.0 )= 576 W Therefore the total power consumed by the two resistors is 877 W + 576 W = 450 W. 56. EASONING a. The two identical resistors, each with a resistance, are connected in parallel across the battery, so the potential difference across each resistor is, the potential difference provided by the battery. The power P supplied to each resistor, then, is The total power supplied by the battery is twice this amount. P (Equation 0.6c). b. According to Equation 0.6c, the resistor which is heated until its resistance is consumes only half as much power as it did initially. Because the resistance of the other resistor does not change, it consumes the same power as before. The battery supplies less power, therefore, than it did initially. We will use Equation 0.6c to determine the final power supplied.
088 ELECTIC CICUITS SOLUTION a. Since the power supplied to each resistor is P (Equation 0.6c), the total power is twice as large: Ptot P. Solving for, we obtain tot 5 30 P 9.6 W b. The resistances of the resistors are now = and =. From Equation 0.6c, the total power output P tot = P + P of the resistors is 3 3 5 P P P 7. W 30 57. EASONING The total power is given by Equation 0.5c as P rms / p, where p is the equivalent parallel resistance of the heater and the lamp. Since the total power and the rms voltage are known, we can use this expression to obtain the equivalent parallel resistance. This equivalent resistance is related to the individual resistances of the heater and the lamp via Equation 0.7, which is p heater lamp. Since heater is given, lamp can be found once p is known. SOLUTION According to Equation 0.5c, the equivalent parallel resistance is Using this result in Equation 0.7 gives p rms earranging this expression shows that lamp p rms P / P heater rms heater 0 4.0 0 lamp P W 5. 0 3 Therefore, lamp 90 3 5. 0
Chapter 0 Problems 089 58. EASONING The total power P 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: P /. eq When two resistors are connected in series, the equivalent resistance S of the combination is greater than the resistance of either resistor alone. This can be seen directly from (Equation 0.6). S When two resistors are connected in parallel, the equivalent resistance P of the combination is smaller than the resistance of either resistor alone. This can be seen directly by substituting values in (Equation 0.7) or by reviewing the discussion P in Section 0.7 concerning the water flow analogy for electric current in a circuit. Since the total power delivered by the battery is P / and since the power eq and the battery voltage are the same in both cases, it follows that the equivalent resistances are also the same. But the parallel combination has an equivalent resistance P that is smaller than B, whereas the series combination has an equivalent resistance S that is greater than A. This means that B must be greater than A, as Diagram at the right shows. If A were greater than B, as in Diagram, the equivalent resistances S and P would not be equal. esistance Diagram S = P esistance S B A A P Diagram B SOLUTION As discussed in the EASONING, the equivalent resistances in circuits A and B are equal. According to Equations 0.6 and 0.7, the series and parallel equivalent resistances are S A A A or P P B B Setting the equivalent resistances equal gives B
090 ELECTIC CICUITS As expected, B is greater than A. B A B or 4 A 59. SSM EASONING The equivalent resistance of the three devices in parallel is p, and we can find the value of p by using our knowledge of the total power consumption of the circuit; the value of p can be found from Equation 0.6c, P / p. Ohm's law (Equation 0., I) can then be used to find the current through the circuit. SOLUTION a. The total power used by the circuit is P = 650 W + 090 W +50 W = 3990 W. The equivalent resistance of the circuit is p P (0 ) 3990 W 3.6 b. The total current through the circuit is I p 0 3.6 33 A This current is larger than the rating of the circuit breaker; therefore, the breaker will open. 60. EASONING AND SOLUTION The aluminum and copper portions may be viewed a being connected in parallel since the same voltage appears across them. Using a and b to denote the inner and outer radii, respectively, and using Equation 0.3 to express the resistance for each portion, we find for the equivalent resistance that p Al Cu Cu Al Cu Al A b a Cu AAl a L L L L 3.000 m.000 m 3 3 3.000 m 0.006 8 8.70 m.50 m.80 m.50 m We have taken resistivity values for copper and aluminum from Table 0.
Chapter 0 Problems 09 6. EASONING Since the defogger wires are connected in parallel, the total resistance of all thirteen wires can be obtained from Equation 0.7: 3 p or p 3 where is the individual resistance of one of the wires. The heat required to melt the ice is given by Q ml f, where m is the mass of the ice and L f is the latent heat of fusion of the ice (see Section.8). Therefore, using Equation 0.6c, we can see that the power or energy dissipated per unit time in the wires and used to melt the ice is P ml f p t Energy/tim e or /3 ml f t According to Equation 0.3, L / A, where the length of the wire is L, its crosssectional area is A and its resistivity is ; therefore, the last expression can be written This expression can be solved for the area A. 3 3 ml L / A t SOLUTION Solving the above expression for A, and substituting given data, and obtaining the latent heat of fusion of water from Table.3, we find that LmL A 3 t f 8 4 (88.00 m)(.30 m)(.0 0 kg)(33.5 0 J/kg) 8 3.580 m 3(.0 ) (0 s) 6. EASONING Between points a and b there is only one resistor, so the equivalent resistance is ab =. Between points b and c the two resistors are in parallel. The equivalent resistance can be found from Equation 0.7: so bc bc f The equivalent resistance between a and b is in series with the equivalent resistance between b and c, so the equivalent resistance between a and c is
09 ELECTIC CICUITS ac ab bc 3 Thus, we see that ac > ab > bc. SOLUTION Since the resistance is = 0.0, the equivalent resistances are: ab bc 0.0 5.00 ac 5.0 3 63. SSM EASONING To find the current, we will use Ohm s law, together with the proper equivalent resistance. The coffee maker and frying pan are in series, so their equivalent resistance is given by Equation 0.6 as coffee + pan. This total resistance is in parallel with the resistance of the bread maker, so the equivalent resistance of the parallel combination can be obtained from Equation 0.7 as p ( coffee pan ) bread. SOLUTION Using Ohm s law and the expression developed above for p, we find I p coffee pan bread 0 46 3 9. A 64. 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. SOLUTION The.00.00 and 3.00 resistors are in series with an equivalent resistance of s = 6.00..00 6.00 4.00 3.00 6.00
Chapter 0 Problems 093 This equivalent resistor of 6.00 Ω is in parallel with the 3.00- resistor, so 6.00 3.00 P.00 6.00 4.00.00 P.00 This new equivalent resistor of.00 Ω is in series with the 6.00- resistor, so s ' = 8.00..00 4.00 8.00 s ' is in parallel with the 4.00- resistor, so 8.00 4.00 P.00.67.67 P Finally, p ' is in series with the.00-, so the total equivalent resistance is 4.67 Ω. 65. SSM EASONING When two or more resistors are in series, the equivalent resistance is given by Equation 0.6: s 3.... Likewise, when resistors are in parallel, the expression to be solved to find the equivalent resistance is given by Equation 0.7:.... We will successively apply these to the individual resistors p 3 in the figure in the text beginning with the resistors on the right side of the figure. SOLUTION 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 5.98. Finally, the 5.98-
094 ELECTIC CICUITS combination and the 0.0- resistor are in parallel, so the equivalent resistance between the points A and B is 4. 6. 66. 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. = 6 = 8 Ω A 4 = 6 B 3 = 48 SOLUTION Since and are wired in series, the equivalent resistance is 6 8 4 (0.6) The resistor 3 is wired in parallel with the equivalent resistor, so the equivalent resistance 3 of this combination is or 3 6 48 4 3 3 (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 4 3 6 6 4
Chapter 0 Problems 095 67. EASONING The circuit diagram is shown at the 60.0 right. We can find the current in the 0.0- resistor by using Ohm s law, provided that we can obtain a value for AB, the voltage between points A and B in the 0.0 A B diagram. To find AB, we will also apply Ohm s law, 0.0 this time by multiplying the current from the battery times AB, the equivalent parallel resistance between A 5.0 and B. The current from the battery can be obtained by applying Ohm s law again, now to the entire circuit and using the total equivalent resistance of the series combination of the 0.0- resistor and AB. Once the current in the 0.0- resistor is found, the power delivered to it can be obtained from Equation 0.6b, P = I. SOLUTION a. According to Ohm s law, the current in the 0.0- resistor is I 0 = AB /(0.0 ). To find AB, we note that the equivalent parallel resistance between points A and B can be obtained from Equation 0.7 as follows: or 40.0 AB 60.0 0.0 AB This resistance of 40.0 is in series with the 0.0- resistance, so that according to Equation 0.6, the total equivalent resistance connected across the battery is 40.0 + 0.0 = 60.0. Applying Ohm s law to the entire circuit, we can see that the current from the battery is I 5.0 60.0 0.50 A Again applying Ohm s law, this time to the resistance AB, we find that AB 0.50 A 0.50 A 40.0 0.0 Finally, we can see that the current in the 0.0- resistor is AB 0.0 0 0 AB I 0 8.33 0 A b. The power delivered to the 0.0- resistor is given by Equation 0.6b as P I 8.33 0 A 0.0 0.833 W 0
096 ELECTIC CICUITS 68. EASONING The total power P 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: P /. 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. + A + B + Since the total power delivered by the battery is P / 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. SOLUTION The total power delivered by the battery is C P / 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
Chapter 0 Problems 097 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 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. According to Equation 0.6, the equivalent resistance of this combination is C A We can now use case: Circuit A P / to determine the total power delivered by the battery in each eq 6.0 P 4.0 W 9.0 Circuit B Circuit C 6.0 P 9.0 6.0 P 9.0 8.0 W.0 W These results are as expected. 69. EASONING When two or more resistors are in series, the equivalent resistance is given by Equation 0.6: s 3.... When resistors are in parallel, the expression to be solved to find the equivalent resistance is Equation 0.7:... p 3 We will use these relations to determine the eight different values of resistance that can be obtained by connecting together the three resistors:.00,.00, and 3.00.
098 ELECTIC CICUITS SOLUTION When all the resistors are connected in series, the equivalent resistance is the sum of all three resistors and the equivalent resistance is 6.00. When all three are in parallel, we have from Equation 0.7, the equivalent resistance is 0.545. We can also connect two of the resistors in parallel and connect the parallel combination in series with the third resistor. When the.00 and.00- resistors are connected in parallel and the 3.00- resistor is connected in series with the parallel combination, the equivalent resistance is 3.67. When the.00 and 3.00- resistors are connected in parallel and the.00- resistor is connected in series with the parallel combination, the equivalent resistance is.75. When the.00 and 3.00- resistors are connected in parallel and the.00- resistor is connected in series with the parallel combination, the equivalent resistance is.0. We can also connect two of the resistors in series and put the third resistor in parallel with the series combination. When the.00 and.00- resistors are connected in series and the 3.00- resistor is connected in parallel with the series combination, the equivalent resistance is.50. When the.00 and 3.00- resistors are connected in series and the.00- resistor is connected in parallel with the series combination, the equivalent resistance is.33. Finally, when the.00 and 3.00- resistors are connected in series and the.00- resistor is connected in parallel with the series combination, the equivalent resistance is 0.833. 70. EASONING The power P dissipated in each resistance is given by Equation 0.6b as P I, where I is the current. This means that we need to determine the current in each resistor in order to calculate the power. The current in is the same as the current in the equivalent resistance for the circuit. Since and 3 are in parallel and equal, the current in splits into two equal parts at the junction A in the circuit. SOLUTION To determine the equivalent resistance of the circuit, we note that the parallel combination of and 3 is in series with. The equivalent resistance of the parallel combination can be obtained from Equation 0.7 as follows: or P 88 P 576 576 This 88-Ω resistance is in series with, so that the equivalent resistance of the circuit is given by Equation 0.6 as eq 576 88 864
Chapter 0 Problems 099 To find the current from the battery we use Ohm s law: 0.0 I 0.39 A 864 eq Since this is the current in, Equation 0.6b gives the power dissipated in as P I 0.39 A 576. W and 3 are in parallel and equal, so that the current in splits into two equal parts at the junction A. As a result, there is a current of 0.39 A in and in 3. Again using Equation 0.6b, we find that the power dissipated in each of these two resistors is I 3 I3 3 P 0.39 A 576.78 W P 0.39 A 576.78 W 7. SSM WWW EASONING Since we know that the current in the 8.00- resistor is 0.500 A, we can use Ohm's law ( = I) to find the voltage across the 8.00- resistor. The 8.00- resistor and the 6.0- resistor are in parallel; therefore, the voltages across them are equal. Thus, we can also use Ohm's law to find the current through the 6.0- resistor. The currents that flow through the 8.00- and the 6.0- resistors combine to give the total current that flows through the 0.0- resistor. Similar reasoning can be used to find the current through the 9.00- resistor. SOLUTION a. The voltage across the 8.00- resistor is 8 (0.500 A)(8.00 ) 4.00. Since this is also the voltage that is across the 6.0- resistor, we find that the current through the 6.0- resistor is I 6 (4.00 )/(6.0 ) 0.50 A. Therefore, the total current that flows through the 0.0- resistor is I 0 0.500 A + 0.50 A= 0.750 A b. The 8.00- and the 6.0- resistors are in parallel, so their equivalent resistance can be obtained from Equation 0.7,..., and is equal to 5.33. Therefore, p 3 the equivalent resistance of the upper branch of the circuit is upper 5.33 +0.0 5.3, since the 5.33- resistance is in series with the 0.0-
00 ELECTIC CICUITS resistance. Using Ohm's law, we find that the voltage across the upper branch must be (0.750 A)(5.3 ) = 9.0. Since the lower branch is in parallel with the upper branch, the voltage across both branches must be the same. Therefore, the current through the 9.00- resistor is, from Ohm's law, I 9 lower 9.0 9.00 9. A 7. EASONING The resistance of each of the identical resistors determines the equivalent resistance eq of the entire circuit, both in its initial form (six resistors) and in its final form (five resistors). We will calculate the initial and final equivalent circuit resistances by replacing groups of resistors that are connected either in series or parallel. The resistances of the replacement resistors are found either from S 3 (Equation 0.6) for resistors connected in series, or from (Equation 0.7), for resistors connected in parallel. P 3 Once the equivalent resistance eq of the circuit is determined, the current I supplied by the battery (voltage = ) is found from I (Equation 0.). We are given the decrease ΔI eq in the battery current, which is equal to the final current I f minus the initial current I 0 : I If I0.9 A () The algebraic sign of ΔI is negative because the final current is smaller than the initial current. SOLUTION Beginning with the circuit in its initial form, we see that resistors, 3, and 5 are connected in series (see the drawing). These three resistors, according to Equation 0.6, may be replaced with a single resistor S that has three times the resistance of a single resistor: 3 5 4 6 S 3 The resistor S is connected in parallel with the resistor 4 (see the drawing), so these two resistors may be replaced by an equivalent resistor P found from Equation 0.7: S = 3 4 6 P 3 4 3 3 3 3 3 4 P S 4
Chapter 0 Problems 0 After making this replacement, the three remaining resistors (, P, and 6 ) are connected in series across the battery (see the drawing). The initial equivalent resistance eq,0 of the entire circuit, then, is found from Equation 0.6: P = 3/4 6 3 eq,0 P 4 () 4 4 Next, we consider the circuit after the resistor 4 has been removed. The remaining five resistors are connected in series (see the drawing). From Equation 0.6, then, the final equivalent resistance eq,f of the entire circuit is five times the resistance of a single resistor: 3 5 6 eq,f 5 (3) From I (Equation 0.), the initial and final battery currents are eq I and I (4) 0 f eq,0 Substituting Equations (), (3), and (4) into Equation (), we obtain eq,f 4 9 I If I0 eq,f eq,0 5 5 55 4 (5) Solving Equation (5) for yields 9 9 35 3.0 55I 55.9 A 73. SSM EASONING The terminal voltage of the battery is given by terminal Emf Ir, where r is the internal resistance of the battery. Since the terminal voltage is observed to be one-half of the emf of the battery, we have terminal Emf/ I Emf / r. From and Ohm's law, the equivalent resistance of the circuit is emf / I r. We can also find
0 ELECTIC CICUITS the equivalent resistance of the circuit by considering that the identical bulbs are in parallel across the battery terminals, so that the equivalent resistance of the N bulbs is found from p N or p bu lb bu lb N This equivalent resistance is in series with the battery, so we find that the equivalent resistance of the circuit is This expression can be solved for N. r bu lb N r SOLUTION Solving the above expression for N, we have N bu lb r r bu lb 5 r 0.50 30 74. EASONING AND SOLUTION Ohm s law gives.5 r 0.054 I 8 A 75. EASONING When the terminal voltage of the battery (emf = 9.00 ) is 8.90, the voltage drop across the internal resistance r is 9.00 8.90 = 0.0. According to Ohm s law, this voltage drop is the current I times the internal resistance. Thus, Ohm s law will allow us to calculate the current. SOLUTION Using Ohm s law we find that 0.0 I r 0.0 8.3 A 76. EASONING The following drawing shows the battery (emf = ), its internal resistance r, and the light bulb (represented as a resistor). The voltage between the terminals of the battery is the voltage AB between the points A and B in the drawing. This voltage is not equal to the emf of the battery, because part of the emf is needed to make the current I go through the internal resistance. Ohm s law states that this part of the emf is I r. The current can be determined from the relation P = I AB, since the power P delivered to the light bulb and the voltage AB across it are known.
Chapter 0 Problems 03 Light bulb A I + + r = 0.0 B SOLUTION The terminal voltage AB is equal to the emf of the battery minus the voltage across the internal resistance r, which is I r (Equation 0.): AB = I r. Solving this equation for the emf gives I r () AB The current also goes through the light bulb, and it is related to the power P delivered to the bulb and the voltage AB according to I = P/ AB (Equation 0.6a). Substituting this expression for the current into Equation () gives I r P r AB AB AB 4 W.8 0.0.0.8 77. EASONING The voltage across the terminals of a battery is equal to the emf of the battery minus the voltage r across the internal resistance of the battery: = Emf r. Therefore, we have that Emf r () The power P being dissipated by the internal resistance is equal to the product of the voltage r across the internal resistance and the current I: P I (Equation 0.6a). Therefore, we can express the voltage r across the internal resistance as r r P () I
04 ELECTIC CICUITS SOLUTION Substituting Equation () into Equation (), we obtain P 34.0 W Emf r 3.4 4.0 I 55.0 A 78. EASONING The power P delivered to a resistor such as the light bulb is found from P I (0.6b) where I is the current in the resistor and is the resistance. When either battery is connected to the bulb, the circuit can be modeled as a single-loop circuit with the battery emf in series with two resistors, namely, the internal resistance r and the resistance of the bulb. As the two resistors are in series, their equivalent resistance S is given by S 3 (Equation 0.6): S r () The current I supplied by a battery connected to a resistor S is, according to Ohm s law, I (0.) where is equal to the emf of the battery. Thus, is the same for both circuits, since both batteries have the same emf, even though their internal resistances (r wet, r dry ) are different. SOLUTION Applying Equation 0.6b separately to each circuit, we obtain wet wet and dry dry S P I P I () Taking the ratio of Equations () eliminates the bulb resistance, yielding dry wet dry Pwet I P I wet dry I (3) I Substituting Equation (0.) for both I wet and I dry into Equation (3), we obtain P P wet dry I wet dry I S,wet S,dry S,wet S,dry (4)
Chapter 0 Problems 05 eplacing the equivalent series resistances in Equation (4) by Equation (), we find that dry 0.33.50 P wet S,dry r.39 P dry S,wet rwet 0.050.50 79. 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. SOLUTION 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: (5.0 )I (7)I 0.0 ()I (8.0 )I Potential drops 30.0 Potential rises Solving for the current gives I 0.38 A. b. The voltage between points A and B is c. Point B is at the higher potential. AB 30.0 (0.38 A)(7 ).0 0 80. EASONING In part a of the drawing in the text, the current goes from left-to-right through the resistor. Since the current always goes from a higher to a lower potential, the left end of the resistor is + and the right end is. In part b, the current goes from right-toleft through the resistor. The right end of the resistor is + and the left end is. The potential drops and rises for the two cases are: Potential drops Potential rises Part a I Part b I SOLUTION Since the current I goes from left-to-right through the 3.0- and 4.0- resistors, the left end of each resistor is + and the right end is. The current goes through the 5.0- resistor from right-to-left, so the right end is + and the left end is. Starting at
06 ELECTIC CICUITS the upper left corner of the circuit, and proceeding clockwise around it, Kirchhoff s loop rule is written as 3.0 I + 4.0 I 5.0 I 36 Potential drops Solving this equation for the current gives I =.0 A Potential rises 8. 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. SOLUTION Applying the loop rule to the top loop (going clockwise around the loop), we have 3.00 A4.00 3.00 A8.00 4.0 + I 3.00 A6.00 Potential drops This equation can be solved directly for the current; Potential rises I 5.00 A. Applying the loop rule to the bottom loop (going counterclockwise around the loop), we have I 3.00 A 6.00 4.0 + I.00 Potential drops Substituting I = 5.00 A into this equation and solving for gives Potential rises 46.0. 8. EASONING Because all currents in the diagram flow from left to right (see the drawing), currents I and I both flow into junction A. Therefore, the current I in resistor, which flows out of junction A, is, by Kirchhoff s junction rule, equal to the sum of the other two currents: =.70 I = 3.00 A I = 4.40 A I
Chapter 0 Problems 07 I I I () The current I is given. To determine I, we note that the resistors and are connected in parallel, and therefore must have the same potential difference = = across them. We will use Ohm s law I (Equation 0.) to determine the potential difference and then the current I. SOLUTION Using I (Equation 0.), we express the voltage across the resistors and in terms of currents and resistances as Solving Equation () for I yields I I () Substituting Equation (3) into Equation (), we obtain I I (3) I.70 I I I I I 3.00 A 4.84 A 4.40 83. 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 + = 8.0 + 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
08 ELECTIC CICUITS 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. SOLUTION Applying Kirchhoff s loop rule to the top loop (ABCF) gives I I Potential Potential rises drop () Similarly, for the bottom loop (FCDE), I Potential Potential rise drop () Solving Equation () for I gives I.0 6.0 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 4.0.0 A 8.0 Since I is a positive number, the current in the resistor goes from left to right, as shown in the drawing. 84. EASONING In preparation for applying Kirchhoff s rules, we now choose the currents in each resistor. The directions of the currents are arbitrary, and should they be incorrect, the currents will turn out to be negative quantities. Having chosen the currents, we also mark the ends of the resistors with the plus and minus signs that indicate that the currents are directed from higher (+) toward lower () potential. These plus and minus signs will guide us when we apply Kirchhoff s loop rule.
Chapter 0 Problems 09 A +.00 Ω B + 8.00 Ω E 3.00 I I 6.00 + + + 4.00 Ω I 3 9.00 + D C SOLUTION Applying the junction rule to junction B, we find F I I I () 3 Into junction Out of junction Applying the loop rule to loop ABCD (going clockwise around the loop), we obtain I I.00 6.00 4.00 3.00 () 3 Potential drops Potential rises Applying the loop rule to loop BEFC (going clockwise around the loop), we obtain I I 8.00 9.00 4.00 6.00 0 (3) 3 Potential drops Potential rises Substituting I from Equation () into Equation (3) gives Solving Equation () for I gives I I I 8.00 9.00 4.00 6.00 0 3 3 I I 8.00.00 5.00 0 (4) 3 I 4.50 A I3.00 This result may be substituted into Equation (4) to show that
0 ELECTIC CICUITS I I 4.50 A I3.00 8.00 3.00 5.00 0 5.00 8.00 5.00 0 or I.8 A 8.00 3 3 The minus sign indicates that the current in the 4.00- resistor is directed downward, rather than upward as selected arbitrarily in the drawing. 85. SSM EASONING We begin by labeling the currents in the three resistors. The drawing below shows the directions chosen for these currents. The directions are arbitrary, and if any of them is incorrect, then the analysis will show that the corresponding value for the current is negative. We then mark the resistors with the 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, I, and I 3, the plus and minus signs must be those shown in the drawing. We can now use Kirchhoff's rules to find the voltage across the 5.0- resistor. SOLUTION Applying the loop rule to the left loop (and suppressing units for convenience) gives 5.0 I 0.0I 3.0 0.0 () Similarly, for the right loop, 0.0I 0.0I 3.0 5.0 () If we apply the junction rule to the upper junction, we obtain Subtracting Equation () from Equation () gives I I I 3 (3)