Chapter 22: Current and esistance Solutions Questions: 4, 7, 17, 21 Exercises & Problems: 1, 16, 22, 28, 36, 38, 51, 53, 55 Q22.4: A lightbulb is connected to a battery by two copper wires of equal lengths but different thicknesses. A thick wire connects one side of the lightbulb to the positive terminal of the battery and a thin wire connects the other side of the bulb to the negative terminal. a) Which wire carries a greater current? Or is the current the same in both? Explain. b) If the two wires are switched, will the bulb get brighter, dimmer, or stay the same? Explain. Q22.4. eason: (a) Because there are no junctions or branching of the wires, they must carry the same current. This is because of conservation of charge. Charges can t appear or disappear. The current is the same everywhere in a single-loop circuit, including in the filament itself. (b) Because of the answer to part (a) the brightness of the bulb will stay the same. The current will be the same if the wires are switched. Assess: This question is asked because many students have difficulty with this concept. Concentrate on this idea until it makes complete sense. Q22.7: Metal 1 and metal 2 are each formed into 1-mm-diameter wires. The electric field needed to cause a 1 A current in metal 1 is larger than the electric field needed to cause a 1 A current in metal 2. Which metal has the larger resistivity? Explain. Q22.7. eason: The wire with the greater resistivity will have a greater resistance ( = ρla / ). The wire with the greater resistance will require a greater electric potential difference to maintain the same current as the wire with less resistance ( V = I). This larger electric potential difference will result in a larger electric field ( E= Vd / ). This allows us to conclude that the wire with the greater resistivity will require a greater electric field in order to sustain the same current as the wire with the smaller resistivity. Since the field required for metal 1 is greater, the resistivity of metal 1 is greater. Assess: In the electricity chapters it is important to know the basic concepts and how they are related. Q22.17: When lightning strikes the ground, it generates a large electric field along the surface of the ground directed toward the point of the strike. People near a lightning strike are often injured not by the lightning itself but by a large current that flows up on leg and down the other due to this electric field. To minimize this possibility, you are advised to stand with your feet close together if you are trapped outside during a lightning storm. Explain why this is beneficial.
Hint: The current path through your body, up one leg and down the other, has a certain resistance. The larger the current along this path, the greater the damage Q22.17. eason: After the lightning strike, you are standing in the electric field along the ground. There is an electric potential at every point in an electric field. If you stand with your feet close together, the electric potential difference between your feet is decreased. Since your body has a fixed resistance (from one foot to the other) the smaller the electric potential difference across your feet, the smaller the current you will experience. Assess: This question requires us to combine our knowledge of electric fields and Ohm s law. Q22.21: We can model the rear window defroster in a car as a resistor that is connected to the car s 12V battery. The defroster is made of a material whose resistance increases rapidly as the temperature increases. When the defroster is cold, its resistance is low; when the defroster is warm, it resistance is high. Why is it better to make a defroster with a material like this than with a material whose resistance is independent of temperature? Think about how the resistance, the current, and the power will change as the window warms. Q22.21. eason: If the material has a positive temperature coefficient, then as the window gets warmer (and the frost disappears), the resistance increases, which limits the current. When the defroster is cold, more current would flow, just as it is needed. It is kind of self-regulating, giving more current when the window is cold and less when it is warm (already having done its job). Assess: If the resistance were independent of the temperature, then the same current would be delivered to the defroster whenever it is on, whether the window is cold or warm, and that would waste energy. P22.1: The current in an electric hair dryer is 10 A. How much charge and how many electrons flow through the hair dryer in 5.0 min? P22.1. Prepare: We will find the total charge that flows through the hair dryer and then divide it by the electron charge to find the number of electrons. Solve: Equation 22.2 is Q = I t. The amount of charge delivered is 60 s Q = (10.0 A) 5.0 min = 3000 C 1min The number of electrons that flow through the hair dryer is Q 3000 C N = = = 1.9 10 e 19 1.60 10 C Assess: This is an enormous amount of charge and is typical of such devices. 22 P22.16: A 9.0V battery supplies a 2.5 ma current to a circuit for 5.0 hr. a) How much charge has been transferred from the negative to the positive terminal? b) How much work has been done on the charges that passed through the battery?
P22.16. Prepare: Charge, electric current, and time are related by W = q V. electric potential difference, and charge are related by Solve: The charge transferred is The work done on these charges by the battery is Assess: These values are reasonable for this case. q= I t = = I Q/ t. 3 3 (2.5 10 A)(5.0 hr)(3.6 10 s/hr) 45 C W = q V = = = Work, 2 (45 C)(9.0 V) 4.1 10 J. P22.22: esistivity measurements on the leaves of corn plants are a good way to assess stress and overall health. The leaf of a corn plant has a resistance of 2.0 MΩ measured between two electrodes placed 20 cm apart along the leaf. The leaf has a width of 2.5 cm and is 0.20 mm thick. What is the resistivity of the leaf tissue? Is this greater than or less than the resistivity of muscle tissue in the human body? (Hint: ρ muscle = 13 Ω m) P22.22. Prepare: esistivity is related to resistance, length, and cross-sectional area by = ρl/ A, and the area is related to the width and thickness of the leaf by A = WT. Solve: Combining these two expressions and solving for the resistivity, we obtain ρ = A L = WT L = = 6 2 4 / / (2.0 10 Ω)(2.5 10 m)(2.0 10 m) / (0.20 m) 50 Ω m Assess: This value for the resistivity is the same order of magnitude as other organic materials listed in Table 22.1. P22.28: The relatively high resistivity of dry skin, about 1 10 6 Ω m, can safely limit the flow of current into deeper tissues of the body. Suppose an electrical worker places his palm on an instrument whose metal case is accidently connected to a high voltage. The skin of the palm is about 1.5 mm thick. (i) Estimate the area of skin on the work s palm that would contact a flat panel, then (ii) calculate the approximate resistance of the skin of the palm. P22.28. Prepare: esistance is related to resistivity, length, and cross-sectional area = ρl A An electrical worker could have palms that are 10 cm by 12 cm or 1.2 10-2 by /. m 2. 6 3 2 2 5 / (1.6 10 Ω m)(1.5 10 m) / (1.2 10 m ) 1.3 10 Ω = ρl A= = Solve: The resistance is Assess: This built-in safety factor is a good thing and a reasonable value. P22.36: a) What is the resistance of a 1500 W (120 V) hair dryer? b) What is the current in the hair dryer when it is used? P22.36. Prepare: The 1500 W rating is for operating at 120 V. That is, the hair dryer dissipates 1500 W at V = 120 V. We will use Equation 22.14 and Ohm s law. Solve: (a) The hair dryer s resistance using Equation 22.14 is ( V ) (120 V) P 1500 W 2 2 = = = Ω (b) The current in the hair dryer when it is used is given by Ohm s law: 9.6
V 120 V I = = = 12.5 A 13 A 9.6 Ω Assess: Devices such as a hair dryer that operate at 120 V usually have relatively small resistance but large current. P22.38: A 70 W electric blanket runs at 18 V. a) What is the resistance of the wire in the blanket? b) How much current does the wire carry? P22.38. Prepare: Electric power, potential, and resistance are related by Electric potential, current, and resistance are related by Ohm s Law = V / I. Solve: (a) The resistance of the wire in the blanket is the electric blanket is I = V / = 3.9A Assess: These are reasonable values for an electric blanket. P= V 2 /. = V 2 / P= 4.6 Ω (b) The current in P22.51: The total charge a battery can supply is rated in ma hr, the product of the current (in ma) and the time (in hr) that the battery can provide this current. A battery rated at 1000 ma hr can supply a current of 1000 ma for 1.0 hr, 500 ma current for 2.0 hr, and so on. A typical AA rechargeable battery has a voltage of 1.2 V and a rating of 1800 ma hr. For how long could this battery drive current through a long, thin wire of resistance 22Ω? P22.51. Prepare: The electric potential difference, current, and resistance are related by Ohm s law I = V /. The amount of charge, the current, and the time it takes the charge to Q= I t. flow are related by Solve: The current is obtained by I = V / = 1.2 V / 22 Ω = 54.5 ma The amount of time the t = Q/ I = 1800 ma hr / 54.5 ma = 33 hr. battery can deliver this current is Assess: Since the current is very small (54.5 ma) we expect the battery to able to deliver this current for a long time. P22.53: Variations in the resistivity of blood can give valuable clues to changes in the blood s viscosity and other properties. The resistivity is measure by applying a small potential difference and measuring the current. Suppose a medical device attaches electrodes into a 1.5-mm-diameter vein at two points 5.0 cm apart. What is the blood resistivity if a 9.0 V potential difference causes a 230 μa current through the blood in the vein? P22.53. Prepare: We assume that Ohm s law applies to the situation. V I = We also use Equation 22.8 which gives in terms of ρ, L, and A. We are given that 1. 77 6 2 10 m. ρl = A V = 9. 0 V, L = 0. 050 m, I 230 µ A, = and A r d 2 2 2 = π = π( /2) = π(1. 5 mm/2) =
Solve: Combine the two previous equations. 6 2 A V A (9. 0 V)(1. 77 10 m ) ρ = = = = 1.4 Ω m 6 L I L (230 10 A)(0. 050 m) Assess: This resistivity is reasonably close to the value for blood (1.6 Ω m) given in Table 22.1. One of the m s cancels and V/A = Ω. P22.55: Wires aren t really ideal. The voltage drop across a current-carrying wire can be significant unless the resistance of the wire is quite low. Suppose a 50 ft extension cord is being used to provide power to an electric lawn mower. The cord carries a 10 A current. The copper wire in a typical extension cord has a 1.3 mm diameter. What is the voltage drop across a 50 ft length of wire at this current? Hint: ρcu = 1.7 10-8 Ω m P22.55. Prepare: We assume the extension cord is ohmic so we can use V I. also use = ρ LA /. We are given L = 50 ft = 15. 2 m, I = 10 A, and also look up the resistivity of copper in Table 22.1: Solve: Combine the two equations. ρl A= πr = π d = π. =. = We 2 2 2 6 2 ( /2) (1 3 mm/2) 1 33 10 m. 8 ρ Cu 1 7 10 m. =. Ω 8 (1. 7 10 Ω m)(15. 2 m) V = I = I (10A) 19V 6 2 A = =. 1. 33 10 m Assess: 1.9 V is less than 2 percent of 120 V, but in some situations it can be important. The longer the extension cord the greater the resistances (if the diameter stays the same); the greater the voltage drop across the cord, the more power is dissipated as thermal energy in the cord. If your extension cord gets hot to the touch, then you are dissipating too much power in the cord. Get a shorter or fatter extension cord. If you buy a long extension cord, make sure it has a low-gauge wire (large diameter) so the resistance will be low. We