# Physics 1302W.400 Lecture 21 Introductory Physics for Scientists and Engineering II

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1 Physics 1302W.400 Lecture 21 Introductory Physics for Scientists and Engineering II In today s lecture, we will learn to: Calculate the resistance of a conductor depending on the material and shape Apply Ohm s law to correlate the current, potential difference, and resistance for a resistor Slide 31-1

2 Resistance and Ohm s Law The resistance of any element in general is defined to be R V I The units of current and resistance are the ampère (A) and ohm (Ω): 1 Ω = 1 V/A Current in a conductor is proportional to the potential difference (often simply denoted as V) across the conductor, and inversely proportional to resistance: I = V R This equation is referred to as Ohm s law. Materials that obey this law are called ohmic. Slide 31-2

3 Ohm s Law Fuse with increasing load Description: An ammeter and voltmeter are used to measure the voltage and current in a simple circuit consisting of a power supply and resistor. The voltage and/or resistance can be changed and the change in current is observed. Description: A piece of stainless steel wire is used as a fuse between a car battery and three light bulbs in parallel (each bulb increasing in wattage). As each bulb is switched on, the increase in current causes the wire to glow and eventually melt. Slide 31-3

4 Junctions and multiple loops Suppose that instead of a bulb we connect a wire in parallel with a bulb, as shown in the figure. What will happen? Because the wire is made from conducting material, the potential difference between the two junctions is effectively zero. Consequently there is nearly zero current in the bulb. However, because of the wire s very small resistance, there is a very large current in the wire. A current branch with negligible resistance in parallel to an element is called a short or a short circuit. Slide 31-4

5 Junctions and multiple loops The left figure shows a circuit with three light bulbs. The circuit diagram is shown in the figure on the right. Slide 31-5

6 Clicker Question Which of the three bulbs will light up? 1. All of them. 2. F only. 3. G and H. 4. F and G. 5. F and H. H does not light up because the wire in parallel with H carries all of the current through the right branch. (The wire shorts out H.) Slide 31-6

7 Clicker Question What electrical quantity is the same for two light bulbs that are connected in series? For two light bulbs connected in parallel? 1. Potential difference, potential difference 2. Current, current 3. Potential difference, current 4. Current, potential difference 5. Some other combination of properties Slide 31-7

8 Electric fields in conductors A conductor through which a charge current flows is not in electrostatic equilibrium. Therefore, the electric field does not have to be zero inside! Consider connecting two charged spheres with a metal rod of length l that is much thinner than the radius of the spheres. Assume a power source keeps the charge on spheres and potential difference V 12 constant. There is a feedback effect (of about 10-9 s) until the electric field is uniform and along the conductor everywhere. In steady state, the electric field is uniform through the rod: E = V 12 l. Recall that: V AB = B A E d Slide 31-8

9 Electric fields in conductors The figure shows the source of the electric field inside a bent conductor. There is accumulation of charge until the total electric field due to surface charge accumulation and applied potential difference has same magnitude everywhere inside the conductor and points parallel to the sides of the conductor. Inside a conductor of uniform cross section carrying a steady current, the electric field has the same magnitude everywhere and is parallel to the walls of the conductor. Slide 31-9

10 Clicker Question A thick resistor and a thin resistor of the same length and material are connected in series. Which resistor has the greater potential difference V across it and the greater resistance R? 1. V thin > V thick, R thin > R thick 2. V thin < V thick, R thin > R thick 3. V thin > V thick, R thin < R thick 4. V thin < V thick, R thin < R thick V AB = B A E d V thin > V thick because the density of electric field lines through the thin resistor has to be greater, and if the lengths of the resistors are the same and E is greater, then V is greater. (Electric field lines must satisfy the continuity principle) R thin > R thick because resistance is inversely proportional to crosssectional area. Alternatively, you can see this from Ohm s law: V = R*I] Slide 31-10

11 Resistance and Ohm s law A metal consists of a lattice of positively charged ions through which electrons can move relatively freely. The figure shows how an applied electric field affects the motions of the electrons in a metal. Slide 31-11

12 Resistance and Ohm s law In the presence of a uniform electric field E, the electrons are subject to a force ee, and therefore an acceleration a = ee/m e. For an electron travelling in a straight path between collisions, the electron s final velocity before the next collision is υ f = υ i + aδt = υ i e E Δt m e Slide 31-12

13 Resistance and Ohm s law The average velocity of all the electrons is ( υ f ) av = ( υ i ) av e E (Δt) m av e Even though the magnitude of υ i can be very large, its average value for all electrons is zero because the collisions produce a random distribution of the direction of initial velocities. The resulting average velocity is called the drift velocity: υ d = e E m e τ τ (Δt) av is the average time interval between collisions. Slide 31-13

14 Clicker Question (i) Does the electric field do work on the electrons as they accelerate between collisions? (ii) On average, does the kinetic energy of the electrons increase as they drift through the lattice? 1. Yes & Yes. 2. Yes & No. 3. No & Yes. 4. No & No. (i) Electric field does work to accelerate the electrons. (ii) No. Drift velocity is constant. Note: This implies that electric energy is converted into thermal energy of lattice. Slide 31-14

15 Resistance and Ohm s law Consider a constant current I, caused by a flow of charge carriers of charge q and speed v d, through a wire of cross-sectional area A. In a time interval Δt = τ, each charge carrier moves by l = v d τ on average. All carriers in a volume V = Al will pass through a cross section of the wire within τ. If the wire contains n charge carriers per unit volume, then the charge flowing through the cross section is: Q = nvq = nav d τq. The current through the wire will be: I = Q/τ = nav d q We can define the current density: J I /A = nv d q Note: analogous to I, the direction of J is the same (opposite) as that of the drift velocity for positive (negative) charge carriers Units: [ J ] = A/m 2 Slide 31-15

16 Physics 1302W.400 Lecture 22 Introductory Physics for Scientists and Engineering II In today s lecture, we will learn to model the electric properties of direct current single-loop circuits using charge continuity (Kirchhoff s first rule, or junction rule) and conservation of energy (Kirchhoff s second rule, or loop rule). Slide 31-16

17 Announcements 1. This week s homework is due on Friday. 2. We have assigned a new homework. The due date is Wednesday, March We also have posted additional Chapter 31 practice problems. 3. There will be a make-up class on Friday, March The next quiz will be on March 29 and 30. Slide 31-17

18 A few comments on the second quiz: Approximate average scores: Group Problem: 18.4 Problem 1: 10.8 Problem 2: 12.2 Multiple Choice: 16.9 Average total score: 58.3 Slide 31-18

19 Resistance and Ohm s law Consider a constant current I, caused by a flow of charge carriers of charge q and speed v d, through a wire of cross-sectional area A. In a time interval Δt = τ, the charge carriers move by l = v d τ on average. All carriers in a volume V = Al will pass through a cross section of the wire within the average time τ between collisions. If the wire contains n charge carriers per unit volume, then the charge flowing through the cross section is: Q = nvq = nav d τq. The current through the wire will be: I = Q/τ = nav d q We can define the current density: J I /A = nv d q Note: analogous to I, the direction of J is the same (opposite) as that of the drift velocity for positive (negative) charge carriers Units: [ J ] = A/m 2 Slide 31-19

20 Resistance and Ohm s law With q = e & v d = eeτ/m e : J = I /A = nv d e = ne 2 Eτ/m e Since J is proportional to E, we can define the conductivity as a measure of a material s ability to conduct a current for a given applied field - Drude model: σ J E = ne2 τ m e Units: [σ] = (A/m 2 )/(V/m)=A/(Vm) = (Ωm) -1 σ is independent of material s shape Conductivity acquires temperature dependence through τ The resistivity of a material is simply: ρ = 1/σ. Units: [ρ] = Ωm Slide 31-20

21 Resistance and Ohm s law Recall that the potential difference across a wire of length l is: V = E With J = I/A and σ = J/E, we get: V = I σ A l = I l σ A So, for ohmic conductors (R = V / I), the resistance is: R = l σ A = ρ l A Slide 31-21

22 Resistance and Ohm s law Example Copper wire: l = 10 m; d = 1 mm; I = 2 A For copper, we have n electrons/m 3 and σ (Ωm ) -1 We get: R wire 0.21 Ω and v d = I / (nea) 0.2 mm/s Note: The drift speed is an average property of the conductor. The reason that a light bulb turns on almost instantaneously is that the current is the same throughout the circuit all of the electrons throughout the circuit are set in motion almost simultaneously when we flip the switch. From the Drude formula, we get: τ s Note: This average time between collisions is very short because of large density of atoms and electrons. Slide 31-22

23 Clicker Question If the temperature of a metal is raised, the amplitude of the vibrations of the metal-lattice ions increases. What effect, if any, do you expect these greater vibrations to have on the resistance of a piece of that metal? (a) No effect. (b) Resistance decreases. (c) Resistance increases. Each ion moves around within a greater volume, which increases the probability of collisions with electrons: the average time between collisions and hence the conductivity decreases. The resistance R = l/(σa) increases. Slide 31-23

24 Flame and Liquid Nitrogen Description: Two lamps are wired to separate coils, but to the same battery. The lamps are plugged into the battery and have similar brightness. One coil is then placed in liquid nitrogen while the other is placed over a burner. The brightness of each bulb is observed again. Slide 31-24

25 Clicker Question What effect does running a current in a metal have on the temperature of the metal? (a) No effect. (b) Metal heats up. (c) Metal cools down. The work done on the electrons by the applied electric field is converted into thermal energy of the lattice ions. This effect is generally negligible in good metals at low currents. At high currents, once the metal heats up, the resistance increases and Ohm s law breaks down. Slide 31-25

26 Single-loop circuits We begin our analysis of single-loop circuits with the simple circuit shown. The emf of the battery establishes an electrostatic field that drives a current in the circuit. In steady state, I battery = I wire = I resistor = I Slide 31-26

27 Single-loop circuits Consider the simple single-loop circuit shown. The energy conservation law tells us that electric potential energy gained by the electrons as they travel through the source is converted to other forms of energy in the load. This observation leads to or Slide 31-27

28 Single-loop circuits For circuits containing several elements, we can generalize the previous equation to This equation is called the loop rule. (steady state, around loop) When evaluating the loop rule we need to pay close attention to the signs of the emfs and potential differences. Choose same direction of travel for all elements of the loop. The junction rule (I in = I out ) and the loop rule are also referred to as Kirchhoff s laws. Slide 31-28

29 Applying the loop rule in single-loop circuits 1. Choose a reference direction for the current in the loop (this may or may not correspond to the current direction) 2. Chose a direction for travel around the loop (arbitrary) 3. Traverse the loop in the direction chosen in step 2 from some arbitrary point on the loop. As you encounter circuit elements, each circuit element contributes a term to: 4. Solve your equation for the desired quantity; if I < 0, this means that the current is opposite to your reference direction. Slide 31-29

30 Series resistors (Exercise 31.8) Consider a single-loop circuit, containing two resistors with resistances R 1 and R 2 and a battery with an emf. Determine the current in the circuit in terms of R 1, R 2, and. Let s choose clockwise reference direction. Start at point a: Solve for current (same throughout loop): For resistors in series, we can therefore define an equivalent resistance: R eq = R 1 + R 2 + R 3 (resistors in series) Slide 31-30

31 Multiloop circuits Consider the multiloop circuit diagram shown. Applying the junction rule to junction a gives I = I 1 + I 2 + I 3 Using Ohm s law, I = V/R, we can write I = V ba + V ba + V! ba 1 =V R 1 R 2 R ba # " R 1 R 2 R 3 \$ & % Slide 31-31

32 Multiloop circuits We can replace the three resistors in parallel by a single resistor with equivalent resistance (R eq ). Then, in order to get the same current in the circuit, we must have 1 R eq = 1 R R R 3 Generalizing this equation to any number of resistors in parallel, 1 R eq = 1 R R R 3 (resistors in parallel) Slide 31-32

33 Multiloop circuits Similarly, we can replace capacitors (C 1, C 2, C 3, ) in series (magnitude of charge the same on all capacitors) or in parallel (potential difference the same across all capacitors) by a single capacitor with equivalent resistance (C eq ): Series: V = V 1 + V 2 + V 3 + = Q/C 1 + Q/C 2 + Q/C 3 + = Q/C eq 1 C eq = 1 C C C 3! Parallel: Q = C eq V = Q 1 + Q 2 + Q 3 + = C 1 V + C 2 V + C 3 V + C eq = C 1 + C 2 + C 3 Opposite of the rules for resistors! C 1 C 2 C 3 Slide 31-33

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