PHYS-2020: General Physics IICourse Lecture NotesSection IIIDr. Donald G. LuttermoserEast Tennessee State UniversityEdition 3.3AbstractThese class notes are designed for use of the instructor and students of the course PHYS-2020:General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. Thesenotes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille.III. Current & Res istanceA. Electric Current.1. Current is defined as the rate at which charge flows through asurface.a) Mathematically:I =∆Q∆t, (III-1)where I is the current, ∆Q is the amount of charge passingthrough an area of wire, and ∆t is the time interval inwhich ∆Q is measured.b) Current is measured in amperes in the SI unit system:1 A ≡ 1 C/s . (III-2)2. The direct ion of current is defin ed t o be the direction at which apositive charge would flow through a wire.++--+++---A-Ia) In metals, it is electrons that flow and not positive charges=⇒ the electrons fl ow in the opposite direction of thecurrent!b) Moving charge ( whether positive or negative) throu gh aconductor is known as a mobile charge carrier.III–1III–2 PHYS-2020: General Physics I I3. Electrons flow in the opposite direction of the~E-field.a) As an electron (or any charged particle) moves througha conductor, it collides with atoms (and/or molecules) inthe conductor =⇒ causes a zigzag m otion through theconductor.b) The amount of charge passing t hrough a wire can be de-termined as follows:i) Let A be the cross-sectional area of a wire and ∆ xbe a small slice along the length of the wire.ii) The volume of this small segment of the wire isthen V = A ∆x (note that V here is volume notvoltage).iii) Let N be the number of charge carriers containedin this volume and q be the charge per carrier.Then,n =NV=NA ∆x,represents the number of carriers per unit volume.iv) The total charge contained in this volume is thus∆Q = Nq = (n A ∆x) q . (III-3)c) Although the electron makes a zigzag path thr ough thewire, o n average, it continues to move down the electricfield (remember in the opposite sense) at an average speedcalled the drift speed vd:vd=∆x∆t=⇒ ∆x = vd∆t .We can then substitute this into Eq. (III-3) giving∆Q = (n A vd∆t) q . (III-4)Donald G. Luttermoser, ETSU III–3d) Dividing both sides by ∆t gives∆Q∆t= I = n q vdA (III-5)orvd=In q A.Since we normally talk about electrons in a metal wire,we can rewrite this asvd=In |e| A. (III-6)e) If no current exists in a cond uctor, the electric field iszero inside the conductor. However, if current exists, anelectric field exists insid e th e conductor (due to Maxwell’slaws — see §IX of the notes).Example III–1. Problem 17.7 (Page 611) from the Serway& Vuille textbook: A 200-km long high-voltage transmission l ine2.0 cm in diameter car ries a steady current of 1000 A. If the con-ductor is copper with a free charge density of 8.5 × 1028electronsper cubic meter, how long (in years) does it take one electron totravel the full length of the cable? Added question: How long wouldit take a photon to travel the same di stance?Solution:The drift speed of electrons in t he line is (fr om Eq. III- 6)vd=In q A=In |e| (πD2/4),where I = 1000 A = 1000 C/s is the current, n = 8.5 × 1028electrons/m3is the charge carrier density, |e| = 1.60 × 10−19Cis the charge per electron, and D = 2.0 cm = 0.020 m is thediameter of the wire. Solving for the drift velocity givesvd=4(1000 C/s)(8.5 × 1028m−3)(1.60 × 10−19C)π(0.020 m)2= 2.3×10−4m/s .III–4 PHYS-2020: General Physics I IThe time to travel the length L of the 200 km = 2.00 × 105mwire is then∆t =Lvd=2.00 × 105m2.34 × 10−4m/s 1 y r3.156 × 107s!= 27.1 yr .For the additional q uestion, since light (i.e., photons) travels ata speed of c = 3.00×108m/s, the time it takes a photon to travelthis length is∆t =Lc=2.00 × 105m3.00 × 108m/s= 6.67 × 10−4s = 667µs .Note that even though an individual electron in the flow of elec-tricity takes a long time to travel through the wire, the E-fieldinside the wire propagates along the wire at the speed of light.B. Resistance and Ohm’s Law.1. Resistance is the ratio of voltage difference to current:R ≡∆VI. (III-7)a) Measured in ohms:Ω =VA=voltsamperes. (III-8)b) Resistance measures how hard it is for electrons (or anycharged particle) to flow through material =⇒ it essen-tially measures the number of internal collisions an elec-tron (or any charged particle) has in a circuit.c) The higher t he resistance, the more collisions with internalatoms/molecules th at make up the wire or resistor. Thelarger the number of collisions per second, t he larger theamount of h eat generated in the wire or resistor.Donald G. Luttermoser, ETSU III–52. Ohm’s Law: Resistance that remains constant over a wide rangeof applied voltage differences such that the voltage d ifference islinearly dependent on current:∆V = I R . (III-9)a) Materials that obey this law are called ohmic =⇒ con-ductors are oh mic.b) Materials that do not obey this law are called non-oh mic=⇒ semiconductors are non-ohmic.3. A resisto r is a simple circuit element t hat provides a specificresistance in an electric circuit. It’s symbol isAAAAAAAAAAAAAC. Resistivity.1. Since resistance is related to the number of collisions an electronhas with atoms/molecules of the wire, we can descr ibe resistancein terms of the geometric pr operties of the conductor and thecomposition of th e conductor (i.e., ohmic material):R = ρLA, (III-10)where L is the length of th e conductor, A is the cross-sectionalarea of the conductor (both of these are the geometric part),and ρ is the resistivity of the material (which is related to thecomposition of th e material — see Table 17.1 in the textbook).2. Electric conductors have low resistivity, insulators have high re-sistivity.3. Resistivities can change with temperature ( see §III.D).III–6 PHYS-2020: General Physics I IExample III–2. Problem 17.17 (Page 611) from the Ser-way & Vuille textbook: A wire 50.0 m l ong and 2.00 mm in diam-eter is connected to a source w ith a potential di fference of 9.11 V,and the current is found to be 36.0 A. Assume a temperature of20◦C and, using Table 17.1, identif y the metal of the wire.Solution:From Ohm’s law (Eq. III-9) we can calculate the resistance ofthe wire t o beR