Physics 11.5-11.7 - Electric Current and Resistance PDF

11.5 – Electric Current Electric Current In an electric conductor, current (I) is described as a quantity of...

11.5 – Electric Current Electric Current In an electric conductor, current (I) is described as a quantity of charge (q) passing a given point during an interval of time (Δt). The unit of the electric current is amper (A) q 1C I 1A= t 1s 11.5 – Electric Current PHYSICS 11 - M.LLUPO 2 11.5 – Electric Current 11.5 – Electric Current Electric Current Water Model for Electric Current Direct current (DC) is the flow of electrons in only one direction More water is flowing if the cross section is bigger for the same through a circuit. water pressure. The electrons flow from the negative terminal of the source of electrical energy and travel through the conducting wires toward the positive terminal. PHYSICS 11 - M.LLUPO 3 PHYSICS 11 - M.LLUPO 4  11.5 – Electric Current 11.5 – Electric Current Practice – Electric Current Practice – Electric Current I = q / ∆t Electrical current and charge I = q / ∆t Electrical current and charge 1. The electrical system in your home operates at a potential difference of 120.0 2. Calculate the amount of current through a wire that has 0.85 C of electrons volts. A toaster draws 9.60 A for a period of 2.50 min to toast two slices of bread. passing a point in 2.5 min. a) Find the amount of charge that passed through the toaster. Given: q 0.85 C; Δt 2.5 min 60s/min 150 s; Required: I ? Given: V 120.0 V; I 9.60 A; Δt 2.50 min 60s/min 150 s; Required: q ? q 0.85C I q I I t q  I  t q  9.60A 150 s q  1, 440 C t 2.5 min(60s/min) I  5.67 103 A  5.67mA b) Find the amount of energy the toaster converted into heat (and light) while it 3. If a current of 0.55 A flows past a point for 12 minutes, how much charge has toasted the bread. travelled past that point, in C? Given: V 120.0 V; q 1,440 C; Required: ΔEE ? Given: I 0.55 A; Δt 12 min 60s/min 720 s; Required: q ? EE V q EE  q  V EE  1, 440C 120V EE  172,800 J q  I t q  0.55A  720s  396C q  4.0  102 C PHYSICS 11 - M.LLUPO 5 PHYSICS 11 - M.LLUPO 6 11.5 – Electric Current 11.5 – Electric Current Electric Current vs. Electron Flow Electric Current vs. Electron Flow 1752 – Benjamin Franklin thought that excess electricity was positive and that a The term current (I) means the flow of positive charge (from anode deficit of electricity was negative. The wrong idea that positive charges are to cathode) in a circuit. moving. The flow of negative charge (from cathode to anode) is called 1800 - Alessandro Volta invented the electrochemical cell. 1876 - An experiment at Harvard University showed that negative charges were electron flow. moving Practice: Analyze the circuits and determine which illustrates 25 years later - J.J. Thomson (1856–1940) discovered the electron, so moving conventional current, electron flow, and neither. negative charges were electrons. Explain your reasoning. PHYSICS 11 - M.LLUPO 7 PHYSICS 11 - M.LLUPO 8  11.5 – Electric Current 11.5 – Electric Current Elementary Charge Oil drop experiment – Elementary charge U.S. physicist, R. Millikan in his “oil-drop experiment” discovered that the size of one elementary charge is 19 The elementary charge is known to be the magnitude of the charge on a proton (+1 e) or an electron (−1 e). e  1.602  10 19 C If an object has N more protons than electrons, it has a charge q given: q  N e PHYSICS 11 - M.LLUPO 9 PHYSICS 11 - M.LLUPO 10 11.5 – Electric Current 11.5 – Electric Current Practice – Elementary Charge Electric Circuit Review q=Ne e = 1.602 10 19 C Elementary Charges Battery, load and conductor. A switch is an optional device that 3. Calculate the charge on a small sphere with an excess of 3.2 1014 electrons. allows you to open and close the circuit. Given: N 3.2 1014electrons; e 1.602 10 19 C; Required: q ? An open circuit means there is a disconnection or a break somewhere in the circuit that prevents current from flowing. q  N e q  (3.2 1014 electrons)(1.602 1019 C/electron) q  5.1105 C A closed circuit means that all connections are complete. 4. A light bulb draws a current of 0.60 A. If the bulb is left on for 8.0 min, how many In the circuit symbol for a battery, the longer line represents the electrons (elementary charges) pass through the bulb? positive pole of the battery, and the shorter line is the negative Given: I 0.60 A; ∆t 8.0 min 60s/min 480 s; Required: N ? pole. q I  t 0.60A  480s N N N e q  I  t e 1.602 1019 C/electron N  1.798 1021 electrons PHYSICS 11 - M.LLUPO 11 PHYSICS 11 - M.LLUPO 12  11.5 – Electric Current 11.5 – Electric Current Practice – Drawing a circuit Measuring Current In your notebook, draw the circuit below and label the parts. Electric current is measured in amps ( ) using With pen draw arrows showing electron flow and with pencil (or red a device called an ammeter. pen) draw arrows showing electric current. Very small electric currents are measured in milliamps ( ), using a milliammeter or galvanometer. Ammeters are always connected in series within a circuit in order to measure the current flowing through a resistor. A multimeter, when set as an ammeter, measures the current at one location in a circuit. PHYSICS 11 - M.LLUPO 13 PHYSICS 11 - M.LLUPO 14 11.5 – Electric Current 11.5 – Electric Current Ammeters Ammeters and Voltmeters Ammeter is an electrical device that measures electric current. Symbol of ammeter is Ammeter is connected in series because need to measure the current that flow through circuit. Ammeter is part of a circuit. If you disconnect it, you disconnect the circuit (open circuit). PHYSICS 11 - M.LLUPO 15 PHYSICS 11 - M.LLUPO 16  11.5 – Electric Current 11.5 – Electric Current Voltmeters and Ammeters in an Electric Circuit Practice – Voltmeters in an Electric Circuit In Grade 9, you learned that circuits could have one complete path In which of the circuits the voltmeter and ammeter are connected (series circuits) or more than one complete path (parallel circuits). correctly? Voltmeters are connected in parallel in the circuit. Ammeters are connected in series. So, ammeter is part of the circuit. PHYSICS 11 - M.LLUPO 17 PHYSICS 11 - M.LLUPO 18 11.5 – Electric Current 11.5 – Electric Current Practice – Voltmeters in an Electric Circuit Electric Circuit Elements Draw a voltmeter measuring potential difference gain (rise) and two others to measure the potential difference drop at loads. Symbol for a voltmeter is Draw an ammeter for each circuit. Symbol for an ammeter is PHYSICS 11 - M.LLUPO 19 PHYSICS 11 - M.LLUPO 20  11.7 – Electrical Resistance Factors Affecting the Resistance Everything that moves experiences frictional forces that resist that motion. The frictional effects in conductors offer resistance to the current passing through them. Electrical resistance (R ) is a property of matter that describes how difficult it is for electric current to travel through a material. You can compare a metal conductor carrying electric current to a water pipe carrying water current. 11.7 – Electric Resistance PHYSICS 11 - M.LLUPO 22 11.7 – Electrical Resistance 11.7 – Electrical Resistance Water Model for Potential Difference and Electric Current – The Ohm’s Law Ohm’s Law The bigger the cross area, the smaller the resistance, the easier the The potential difference across a load equals the product of the flow, the bigger the current. current through the load and the resistance of the load. For the same pressure (Voltage) V  IR more water is flowing (Current) The unit of resistance, the ohm (Ω), is defined in accordance with if the Resistance is small. Ohm’s law. The current is directly proportional One ohm is defined as the amount of electric resistance that will with Potential Difference and allow one ampere of current to move through the resistor when a inversely proportional with resistance. potential difference of one volt is applied across the resistor. V Devices and materials that do not obey Ohm’s law are said to be non- I R linear or non-ohmic. PHYSICS 11 - M.LLUPO 23 PHYSICS 11 - M.LLUPO 24  11.7 – Electrical Resistance 11.7 – Electrical Resistance Practice – Ohm’s Law Power, Potential Difference, Current, and Resistance V=IR R=V/I I=V/R Ohm’s law Every appliance is rated for its Power output (P) - the 4. What is the resistance of a load if a battery with a 9.0 V potential difference causes rate at which it can transform electric energy. Example: a current of 0.45 A to pass through the load? Electric range – 12,000 W. Given: V 9.00 V; I 0.45 A; Required: R ? Electric clothes dryer – 5,000 W V 9.00 V Light Bulb – 4.5 - 15 W R R I 0.45 A R  20  Power is defined as work done per unit time or energy 5. If the current through a 1460  resistor is 3.09 mA, what is the voltage drop across transformed or transferred per unit time. the resistor, in volts? W Work done Given: R 1460 Ω; I 3.09 mA 3.09 10 3A; Required: V ? P Power = t time V EE Energy transformed R P Power = I V  IR V  IR  (3.09 103 A)(1460) V  4.51V t time PHYSICS 11 - M.LLUPO 25 PHYSICS 11 - M.LLUPO 26 11.7 – Electrical Resistance 11.7 – Electrical Resistance Power, Potential Difference, Current, and Resistance Practice – Power, Potential Difference, Current, and Resistance Relation between Power, Potential Difference, Current and P = IV Power, Potential Difference, Current and Resistance P = RI2 = V2 / R Resistance EE V q Practice: P EE  V  q P P V  I 5. What is the power rating of a segment of Nichrome wire that draws a current of t t 2.50 A when connected to a 12.0 V battery? Given: V 12.0 V; I 2.50 A; Required: P ? Using Ohm’s Law: P V  I P  12.0V  2.50A P  30.0W V  R I P  V  I  (R  I )I  R  I 2 6. A stove element has a power rating of 4.40 kW. If it is connected to a 120 V P  R I2 supply, what is its effective internal resistance? V V V 2 V 2 Given: P 4.40 kW 4400 W; V 120 V; Required: R ? I P V I V( )  P R R R R V2 V 2 (120V) 2 P R  R P 4400W R  3.27Ω PHYSICS 11 - M.LLUPO 27 PHYSICS 11 - M.LLUPO 28  11.7 – Electrical Resistance 11.7 – Electrical Resistance Factors Affecting the Resistance Water Model for Electric Resistance For an electric conductor with a fixed diameter, resistance increases For the same pressure, more proportionately with the length (R L). water is flowing if there is little For an electric conductor with a fixed length, resistance varies resistance. inversely as the cross-sectional area (R 1/A). Bigger the cross-sectional area, the bigger the flow because there is less resistance. Resistance varies inversely as the cross-sectional area. (R 1/A). PHYSICS 11 - M.LLUPO 29 PHYSICS 11 - M.LLUPO 30 11.7 – Electrical Resistance 11.7 – Electrical Resistance Factors Affecting the Resistance Calculating the Resistance Summary The equation for the resistance of a conductor can now be written: L R  A The value of the proportionality constant (ρ) is called the resistivity and is a property of the material from which the conductor is made. PHYSICS 11 - M.LLUPO 31 PHYSICS 11 - M.LLUPO 32  11.7 – Electrical Resistance 11.7 – Electrical Resistance Calculating the Resistance Practice – Resistance – Your turn R = ρL / A ρ is resistivity Ohm’s law 3. Calculate the resistance of a 15.0 m length of copper wire, at 20°C, that has a diameter of 0.0500 cm. Given: L 15 m; d 0.0500 cm 5.00 10 4 m; 𝝆 1.70 10 8 𝛀∙m; Required: R ? L R  A A   r2 A   (2.50  104 m) 2 A  1.96  107 m 2 15.0m R  1.70  108 Ωm 1.96 107 m 2 R  1.30 Ω PHYSICS 11 - M.LLUPO 33 PHYSICS 11 - M.LLUPO 34