week 6 chapter 31 Current and Resistance

Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 2) 3)

Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 2) 3) Current can only flow if there is a continuous connection from the negative terminal through the bulb to the positive terminal. This is only the case for Fig. (3).

Why does the light in a room come on instantly when you flip a switch several meters away? 1. Electrons travel at the speed of light through the wire. 2. Because the wire between the switch and the bulb is already full of electrons, a flow of electrons from the switch into the wire immediately causes electrons to flow from the other end of the wire into the lightbulb. 3. The switch sends a radio signal which is received by a receiver in the light which tells it to turn on. 4. Optical fibers connect the switch with the light, so the signal travels from switch to the light at the speed of light in an optical fiber.

1. i d > i a > i b > i c These four wires are made of the same metal. Rank in order, from largest to smallest, the electron currents i a to i d. 2. i b = i d > i a = i c 3. i c > i b > i a > i d 4. i c > i a = i b > i d 5. i b = i c > i a = i d

The two charged rings are a model of the surface charge distribution along a wire. Rank in order, from largest to smallest, the electron currents ia to ie at the midpoint between the rings. (Note: electron current is number of electrons moving across the wire section per unit time) 1. i c > i e > i a > i b = i d 2. i d > i b > i e > i a = i c 3. i c = i d > i e > i a = i b 4. i b = i d > i a = i c = i e 5. i a = i b > i e > i c = i d

Electrons flow through a 1.6-mm-diameter aluminum wire at 2.0 x 10-4 m/s. How many electrons move through a cross section of the wire each day? Ans: 2.1 x 10 24 electrons (used n= 6.0 x 10 28 m -3 for aluminum, from table 31.1)

The current in an electric dryer is 10.0 A. How much charge and how many electrons flow through the hair dryer in 5.0 min Ans: 3.0 x 10 3 C, 1.88 x 10 22

1. J c > J b > J a > J d Rank in order, from largest to smallest, the current densities J a to J d in these four wires. 2. J b > J a = J d > J c 3. J b > J a > J c > J d 4. J c > J b > J a = J d 5. J b = J d > J a > J c

1. J c > J b > J a > J d Rank in order, from largest to smallest, the current densities J a to J d in these four wires. 2. J b > J a = J d > J c 3. J b > J a > J c > J d 4. J c > J b > J a = J d 5. J b = J d > J a > J c J = I/A, does not depend on conductivity σ. Ja = I/πr 2 (2I)/πr 2 = 2Ja (2I)/π(2r) 2 = 1/2 Ja I/πr 2 = Ja

Two copper wires of different diameter are joined end-to-end, and a current flows in the wire combination. When electrons move from the larger-diameter wire into the smaller-diameter wire, 1. their drift speed increases 2. their drift speed decreases 3. their drift speed stays the same 4. not enough information given to decide Recall J = ne e vd, and J = I/A Moving from larger to smaller diameter implies moving from smaller to larger J so J increases and hence drift velocity increases too

Electrons in an electric circuit pass through a resistor. The wire has the same diameter on each side of the resistor. Compared to the drift speed of the electrons before entering the resistor, the drift speed of the electrons after leaving the resistor is 1. faster 2. slower 3. the same 4. not enough information given to decide Recall, again, J = ne e vd, and J = I/A Throughout the resistor and in the wires on either end the current is I and since the diameter of the wire is the same before entering and after leaving the resistor the current density J is the same in both wires, and hence the electrons have the same drift velocity

Electrons in an electric circuit pass through a resistor. The wire has the same diameter on each side of the resistor. Compared to the potential energy of an electron before entering the resistor, the potential energy of an electron after leaving the resistor is 1. greater 2. less 3. the same 4. not enough information given to decide On general grounds: energy is dissipated in the resistor so the charge carriers (electrons) must loose energy. In equations, the electric potential decreases from one end of the resistor to the other in the direction of the current. Electrons move opposite the current, so they move towards higher electric potential V. The potential energy of electrons is U = ev, it decreases when V increases.

A 3.0-mm-diameter wire carries a 12 A current when the electric field is 0.085 V/m. What is the wire resistivity? Ans: 5.0 x 10-8 Ω m

The two segments of the wire in the figure have equal diameters but different conductivities σ1 and σ2. Current I passes through this wire. If the conductivities have the ratio σ2/σ1 =2, what is the ratio E2/E1 of the electric field strengths in the two segments of the wire? Ans: 1/2

1. 15 A into the junction What are the magnitude and the direction of the current in the fifth wire? 2. 15 A out of the junction 3. 1 A into the junction 4. 1 A out of the junction 5. Not enough data to determine

A wire connects the positive and negative terminals of a battery. Two identical wires connect the positive and negative terminals of an identical battery. Rank in order, from largest to smallest, the currents I a to I d at points a to d. 1. I c = I d > I a > I b 2. I a = I b > I c = I d 3. I c = I d > I a = I b 4. I a = I b = I c = I d 5. I a > I b > I c = I d

You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? 1) Ohm s law is obeyed since the current still increases when V increases 2) Ohm s law is not obeyed 3) This has nothing to do with Ohm s law Ohm s law, V = I R, states that the relationship between voltage and current is linear. Thus for a conductor that obeys Ohm s Law, the current must double when you double the voltage. Follow-up: Where could this situation occur?

Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? 1) d A = 4 d B 2) d A = 2 d B 3) d A = d B 4) d A = 1/2 d B 5) d A = 1/4 d B The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than B.

A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? 1) it decreases by a factor 4 2) it decreases by a factor 2 3) it stays the same 4) it increases by a factor 2 5) it increases by a factor 4 Keeping the volume (= area x length) constant means that if the length is doubled, the area is halved. Since, this increases the resistance by four.

1. R a > R c > R b > R d Conductors a to d are all made of the same material. Rank in order, from largest to smallest, the resistances R a to R d. 2. R b > R d > R a > R c 3. R c > R a > R d > R b 4. R c > R a = R d > R b 5. R d > R b > R c > R a

An aluminum wire consists of the three segments shown in the figure. The current in the top segment is 10 A. For each of these segments find the: a. current I b. current density J c. electric field E d. drift velocity vd e. mean time between collisions τ f. electron current i σ = 3.5 x 10 7 Ω -1 m -1 n = 6.0 x 10 28 m -3 Ans: Itop = Imid = Ibot = 10 A Jtop = Jbot = 3.18x10 6 A/m 2, Jmid = 1.27x10 7 A/m 2 Etop = Ebot = 0.0909 V/m, Emid = 0.364 V/m τ = 2.07x10-14 s itop = ibot = imid = 6.25x10 19 s -1

When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3)

When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3) The voltage is provided at 120 V from the outside. The light dimmer increases the resistance and therefore decreases the current that flows through the lightbulb. Follow-up: Why does the voltage not change?

Electrons in an electric circuit pass through a source of emf. The wire has the same diameter on each side of the source of emf. Compared to the drift speed of the electrons before entering the source of emf, the drift speed of the electrons after leaving the source of emf is 1. faster 2. slower 3. the same 4. not enough information given to decide Recall, again, J = ne e vd, and J = I/A. This is just as in the case with a resistor, done earlier. In the wires on either end of the source of emf the current is I and since the diameter of the wire is the same before entering and after leaving the source of emf the current density J is the same in both wires, and hence the electrons have the same drift velocity

Electrons in an electric circuit pass through a source of emf. The wire has the same diameter on each side of the source of emf. Compared to the potential energy of an electron before entering the source of emf, the potential energy of an electron after leaving the source of emf is 1. greater 2. less 3. the same 4. not enough information given to decide On general grounds: source of emf does work on the charge carriers (electrons) which therefore must gain energy. In equations, the electric potential increases across the source of emf in the direction of the current. Electrons move opposite the current, so they move towards lower electric potential V. The potential energy of electrons is U = ev, it increases when V decreases.

The figure shows a wire that is made of two equal diameter segments with conductivities σ1 and σ2. When a current I passes through the wire, a thin layer of charge appears at the boundary between the segments. Find an expression for the surface charge density η on the boundary. Give your result in terms of I, σ1, σ2, and the wire s cross-section area A. Ans: η = ɛ 0 ( 1 σ 2 1 σ 1 ) I A

Example: What should be the diameter of an aluminum wire that carries 15 A when the voltage across 1.0 m of the wire is 0.25 V? Ans: 1.4 mm (need resistivity of aluminum: 2.65 x 10-8 Ω.m)