X ~ Messenger Classes Read chapter 12 - Current & Resi X *Homework Help - Q&A from Onl X + X C A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTYSNTY5MDMO/details POF chapter 12.pdfneers 1 Open with 00-10:30 TF) Chapter 12 CURRENT AND RESISTANCE Read ch Assigned review th Objectives: dd or create marisol anto . Yes At the end of the chapter, the students should be able to: 100 points rk as done define electric current, electric resistance, and voltage; state Ohm's law and use it to calculate current, voltage, and resistance in a circuit; use resistivity to calculate the resistance of specified configurations of a material; use thermal coefficients of resistivity to calculate changes of resistance with temperature; and mments be able to calculate power, energy, and cost of electrical energy. to marisol anto Class comme 12. 1 Electric Current Add a class comm A current is any motion of a charge from one region to another. In this section, we study the flow of electric charges through a piece of material. The amount of flow depends on both the material current I through which the charges are passing and the potential difference V across the material. Whenever there is a net flow of charge through some region, an electric current is said to exist. Fig. 12.1 Current through the cross-sectional area A To define current more precisely suppose charges are moving perpendicular to a surface of area A as shown in s area A could be a cross- ? sectional area of a wire). The curren Page s through this surface. If AQ is the amount of a charge that passes through this surface in - time intonal ENG 7:08 am a US 02/05/2022X ~ Messenger X Classes Read chapter 12 - Current & Resi X * Homework Help - Q&LA from Onl X + C A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details potential difference V across the materi POF chapter 12.pdfneers 1 00-10:30 TF) Whenever there is a net flow of charge throug Open with some region, an electric current is said to exist. Fig. 12.1 Current through the cross-sectional area A To define current more precisely suppose Read ch charges are moving perpendicular to a surface of area A as shown in Fig. 12.1 (this area A could be a cross- sectional area of a wire). The current is defined as the rate at which charge flows through this surface. If Assigned review t AQ is the amount of a charge that passes through this surface in a time interval At, the average current lave is equal to the charge that passes through A per unit time. That is, dd or create marisol anto . Yes 100 points Coulombs ave or Amperes k as done At second (Eq. 12.1) One ampere (1 A) of current is equivalent to 1 Coulomb of charge passing through a surface in 1 second (1 A = 1 C/s ). This unit is named in honor of the French scientist Andre Marie Ampere (1775-1836). mments Example 12.1 A current of 50 mA exist in a resistor for 2 minutes. How many (a) Coulombs and (b) electrons pass through any cross section of the resistor in this time? to marisol anto Class comme Solution: Add a class comm (a) The amount of charge in Coulombs is found by Eq. 12.1 where the initial charge Qo = 0 at the initial time to = 0. At time t = 2 min = 120 s , the charge Q is Q = It = (50 x 10-3A) (120 s) = 6 C (b) Let n be the number of electrons. The charge Q is caused by the total number of electrons that pass through the cross-section of the resistor. Thus, ? Page 1 / 8 + ENG 7:08 am O N () O US 02/05/2022~ Messenger X Classes Read chapter 12 - Current & Resi X * Homework Help - Q&A from Onl X + X C A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details POF chapter 12.pdfneers 1 Open with 30-10:30 TF) 10 6 C Read ch n = - e 1.602 x 10-19c/e = 3.745 x 10 ' electrons Assigned review t 12.2 Resistivity and Conductivity dd or create marisol anto . Yes A perfect 100 points Table 12.1 Resistivities at room temperature (20 C) conductor would have Material Resistivity, p[Q-m) |Temperature Coefficient, rk as done zero resistivity, and a a(c)- perfect insulator would Conductor have an infinite resistivity. Silver 1.59 x 10- 8 0.0061 The SI units of resistivity Copper 1.68 x 10- 8 0.0068 mments (denoted by p) are Gold 2.44 x 10- 0.0034 to marisol anto ohm-meter (Q.m). Table Aluminum 2.65 x 10- 8 0.00429 Class comme 12.1 lists some Tungsten 5.6 x 10- 8 0.0045 9.71 x 10- Add a class comm representative values of Iron 0.0065 1 resistivity. Platinum 10.6 x 10- 8 0.003927 Metals and alloys Mercury 98 x 10 0.0009 have the smallest Nichrome (Ni, Fe, Cr alloy) 100 x 10- 8 0.0004 resistivities and are the Semiconductor best conductors. The Carbon (Graphite (3 -60) x 10- 5 -0.0005 resistivity of insulators are Germanium (1 - 500) x 10 -0.05 greater than those of the Silicon 0.1 - 60 -0.07 metals by an enormous Insulator factor, in the order of 1022 Glass 109 - 1012 ? The reciprocal of Page d 2 0ber 8 1013 10 resistivity is conductivity N .a. ENG 7:09 am US 02/05/2022~ Messenger X Classes Read chapter 12 - Current & Resi X * Homework Help - Q&LA from Onl X + X C A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details best conductors. The Carbon (Graphite) (3 - bUJ X IU - -U.UUUD POF chapter 12.pdfneers 1 resistivity of insulators are Germani Open with (1 - 500) x 10 5 -0.05 00-10:30 TF) greater than those of the Silicon 0.1 - 60 -0.07 metals by an enormous Insulator factor, in the order of 1022. Glass 109 - 1012 Read ch The reciprocal of Hard rubber 1013 - 1015 Assigned review t resistivity is conductivity (denoted by o). Its unit is (Q.m)". Good conductors of electricity have larger conductivity than insulators. dd or create marisol anto . Yes Conductivity is the direct electrical analogy of thermal conductivity. Good electrical conductors, such as 100 points metals, are usually also good conductors of heat. Poor electrical conductors, such as ceramic and plastic materials, are also poor thermal conductors. Tk as done Semiconductors have resistivities intermediate between those of metals and those of insulators. Those materials are important because of the way their resistivities are affected by temperature and by small amount of impurities. mments 12.3 Resistance and Ohm's Law to marisol anto Class comme A = area Suppose a conductor is a wire with a uniform L = length cross-sectional area A and length L, shown in Fig. 12.2. Add a class comm Let V be the potential difference between the higher- and lower-potential ends of the conductor. The p = resistivity =M direction of the current is always from the higher- to R the lower-potential end. As the current flows through the potential R=P- A difference, electric potential energy is lost; this energy Fig. 12.2 A conductor with uniform cross-setion is transferred to the ions of the conducting material. We can also relate the value of the current / to ? Page 2 / 8 + a ENG 7:09 am ()) O US 02/05/2022~ Messenger X Classes Read chapter 12 - Current & Resi X * Homework Help - Q&LA from Onl X + X C A https://classroom.google.com/u/2/c/NDQ3NzYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details POF chapter 12.pdfneers 1 Open with ... 00-10:30 TF) Read ch the potential difference between the ends of the conductor Assigned review t V pl. add or create marisol anto . Yes I A 100 points V - PL (Eq. 12.2) rk as done This shows that when p is constant, the total current I is proportional to the potential difference V. The ratio V to I for a particular conductor is called its resistance R. Imments R = 7 (Eq. 12.3) to marisol anto Class comm Comparing this to Eq. 12.2, we see that the resistance R of a particular conductor is related to the resistivity p of its material by Add a class comn R = (Eq. 12.4) Equation 12.4 shows that the resistance of a wire or other conductor of uniform cross-section is directly proportional to the length and inversely proportional to its cross-sectional area. It is also proportional to the resistivity of the material of which the conductor is made. If p is constant, as is the case for ohmic materials, then ? Page 3 1 8 + Equation 12.5 is often cal athematician and Physicist 7:09 am a ENG A () O US 02/05/2022~ Messenger X Classes Read chapter 12 - Current & Resi X *Homework Help - Q&LA from Onl X + X A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details Physics for Engineers 1 V = IR CEIT-06-201A (9:00-10:30 TF) (Eq. 125) Equation 12.5 is often called Ohm's Law, formulated by German Mathematician and Physicist Read ch Georg Simon Ohm (1787-1854). Ohm's Law states that the current is directly proportional to the voltage and inversely proportional to the resistance, provided physical conditions of the conductor such as length, Assigned review t area of cross section, temperature and material remain constant. The SI unit of resistance is ohm, equal to Volt per Ampere (1 0 = 1V/A) dd or create marisol anto . Yes A circuit device made to have a specific value of resistance between its end is called the resistor. 100 points Resistor has the property to absorb or tends to prevent the current to flow. The potential difference k as done (voltage) :'ss a sample of material that obeys Ohm's law is proportional to the current / through the sample. Example 12.2 in household wiring, a copper wire 2.05 mm in diameter is often used. Find the resistance of a 24-meter length of this wire at "C. nments . Solution: We use Eq. 12.4 to calculate the resistance R. The length 1 = 34 m and the diameter of to marisol anto Class comme the cross-section d = 2.05 x 10" m. From Table 12.1, Pew = 1.68 x 10 0-m. The cross-sectional area is given by Add a class comm md' "(2.05 x 10-3 m) A =- = 3.301 x 10 m R PL (1.68 x 10 8 0 m)(24 m) 3.301 x 10m? - = 0. 122 0 Example 12.3 A 14-gauge aluminum with a radius of 0.814 mm carries a current of 12.5 mA. (a) What is the potential difference across a 2.0-meter length of wire? (b) What could the potential difference in part (a) be if the wire is silver instead of aluminum? ? O N a. ENG 7:09 am A () O US 02/05/2022~ Messenger Classes Read chapter 12 - Current & Resi X * Homework Help - Q&A from Onl X + X A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details Physics for Engineers 1 Solution: (a) The given are the length of the aluminum wire L = 2.0 m; the radius of the cross- CEIT-06-201A (9:00-10:30 TF) section 7 = 0.814 mm and the current that passes through the wire / = 12.5 mA. From Table 12.1, the resistivity of aluminum is PAL = 2.65 x 10-80 . m. Solving first for the resistance RAI using Eq. 12.4, we Read ch get Assigned review t RAL = PALL _PALL (2.65 x 10-80 . m) (2.0 m) 0.0255 0 marisol anto . Yes A 17- 2 7: (0.814 x 10)2 d or create 100 points We substitute these values to Eq. 12.5 to find the value of the potential difference VAL. That is, k as done VAL = IRAI = (12.5 mA) (0.0255 0) = 0. 128 mV (b) From Table 12.1, the resistivity of silver is PAG = 1.59 x 10-8 0-m. Having the same dimensions of the wire as in (a), the resistance Rag is ments to marisol anto Class comm Page Pagl (1.59 x 10-80 . m) (2.0 m) RAg - = 0.0153 0 A RT (0.814 X 10-3)2 Add a class comm VAg = IRAg = (12.5 mA) (0.0153 0) = 0. 191 mV 12.4 Resistance and Temperature Because the resistivity of the material varies with temperature, the resistance of a specific conductor also varies with temperature. For temperature ranges that are not too great, over a limited range, the resistance of a conductor varies approximately linearly with temperature according to the expression, ? Page LEq. 12.6) the now resistance at temperature T a ENG 7:10 am A () O US 02/05/2022~ Messenger Classes Read chapter 12 - Current & Resi X *Homework Help - Q&LA from Onl X + X A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTYSNTY5MDMO/details Physics for Engineers 1 CEIT-06-201A (9:00-10:30 TF) VAg = IRAg = (12.5 mA) (0.0153 0) = 0. 191 mV 12.4 Resistance and Temperature Read ch Because the resistivity of the material varies with temperature, the resistance of a specific Assigned review t conductor also varies with temperature. For temperature ranges that are not too great, over a limited add or create marisol anto . Yes range, the resistance of a conductor varies approximately linearly with temperature according to the expression, 100 points Rn = Ri[1 + a(Tn - 7{)] (Eq. 12.6) rk as done where: Rn = the new resistance at temperature In R; = the resistance at the initial temperature 7, often taken to be 0C or 20"C a = the temperature coefficient of resistance (see Table 12.1) nments Example 12.4 Suppose the resistance of an aluminum wire is 1.1 0 at 20C, find the resistance at (a) 0"C and (b) 190C. to marisol anto Class comm Solution: (a) Let Ro be the resistance of the aluminum wire at To = 0C. From Table 12.1, the Add a class comm temperature coefficient of resistance of aluminum at 20C is at = .00429 (C) 1. We assume that this coefficient is constant throughout the given range of temperature so that we can use Eq. 12.6. . Ro = Rzo[1 + a(To - T20)] = 1.1 0[1 + 0.00429/C(0 - 20.C)] = 1.0056 0 (b) Let R190 be the resistance of the aluminum wire at T, = 190 C. R190 = Rzo[1 + a(7190 - 720)] = 1.1 0[1 + 0.00429/C(190 - 20.C)] = 1.9022 0 ? Page 4 / 8 + O N a ENG 7:10 am () O US 02/05/2022'2 -: A Ftead chapter'll - Current St Fte X t 1 7 classroo ma; crogleaxim 12.5 Power, Energy and Cost of Electrical Energy In eiectric circuits we are interested in the rate at which energy is either delivered to or extracted from a circuit-eiementt If the current through the element is i, then in a time interval cit an amount of charge :19 : idt passes through the element. The potential energy for this amount of charge is W!) = Vidtt Dividing this expression by cit, we obtain the rate at which energy is transferred either in Or out of the circuit element. The time rate of energy transfer is cailed the lawyer denoted by P, so we write _ dig _thr& _y____ . . F dt dt w (E21112?) Substituting Eu. 12.5 (V = ii?) to Eq. 2.1, we obtain P = iRU) = FR (quls) Similarly. we can get another expression for the power P by substituting i = gto Eq. 12.7. That is, v v? rat/[y]? (5.1.123) A very practical aspect of the use of any electric device is the cost of its operation. It should be noted thatthe thing for which the consumer pays the utility company is energy not power. Energy E is the product of power P and time t. Th elassroo rm; croglenxrm A very practical aspect of the use of any electric device is the cost of its operation. It should be noted that the thing for which the consumer pairs the utility company is energy not power. Energy E is the product of power P and time t. That is E : Pt (11.12. 10} A power of 1 W used for 1 second requires 1 Joule of energy. The most common unit is the kilowatt-hour [kWh], which is the energy consumed when 1 kw of power is used for 1 hour. if we multiply Eq.12.10 by the unit cost of energy, we obtain the total cost of energy consumption. Letting C be the-total cost of energyr consumption than C is given by the equation: c : Pt(urrit cost) (1511.12.11) Example 12.5 The power rating of a light buib is the power it dissipates when connected across a IZU-V potential dierence. la] What is the resistance of a 100 W bulb? {b} How much current does each bulb draw in normal use? [cl What is the monthly energy cost of this bulb if it is used'ii hours a day, 30 days per month, and the price of energy is P11.00 per kWh? Solution: la] We use Eq. 12.9 to nd the resistance ofthe loo-w bulb since the given is the potential difference = 120 V. V2 _1201 v? 71449 ' P ' too w ' lb} We use Eq. 12.? to solve for the current i' that passes through the bulb. 100W ~ Messenger X Classes Read chapter 12 - Current & Resi X *Homework Help - Q&A from Onl X + X A https://classroom.google.com/u/2/c/NDQ3NZYwNDEwOTkw/a/NDg1NTY5NTY5MDMO/details Physics for Engineers 1 CEIT-06-201A (9:00-10:30 TF) Read ch 154 . CURRENT AND RESISTANCE Assigned review t d or create marisol anto . Yes 100 points (c) Monthly energy cost, C is given by: