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Transcript
Introduction to Current In AP C Current I = dq/dt I: current in Amperes (A) q: charge in Coulombs (C) t: time in seconds (s) Current Density J = I/A J: current density in A/m2 I: current in Amperes (A) A: area of cross section of wire (m2) I = JA Drift Speed of Charge Carriers J = N e vd = {electrons/m3}{Coul/electron}{m/s} = C/s/m2 = A/m2 J: current density in Amperes/m2 vd: drift velocity in m/s n: # charge carriers per unit volume (per m3) e: charge of individual charge carrier (Coulombs) e- + In any typical wire e+ + eeE I vd J + On the APC reference table, current density J is not defined, but you have these 2 formulas related to J: resistivity J = I/A and V = IR so V = JAR. Now add in resistivity R = ρL/A so V = JA ρL/A V= JρL and V/L = Jρ So does V/L = E inside a wire? Well E = -dV/dL, so E inside wire = ρJ E inside wire = ρJ Find the electric field inside a copper wire of diameter 2 mm carrying a current of 3 milliamps. How would E inside change if…. …….. the wire were half as thick? ……. .aluminum were used instead of copper Lets consider the number of electrons per unit volume going through a wire Note: N is # of charges / m3 while Current density J is # of Amperes / m2 Let’s consider the same a copper wire of diameter 2 mm carrying a current of 3 milliamps. If 4 x 10-3 electrons move through at 1 x10-5 m/s. Find N. Let’s consider the same a copper wire of diameter 2 mm carrying a current of 3 milliamps. If 4 x 10-3 electrons move through at 1 x10-5 m/s. Find N. we get Rearranging N = I/ evdA = 3 x 10-3 Coul/sec (1.6 x 10-19 Coul/electron)(1x10-5m/s)(π {1x10-3m}2) = 5.9 E 26 electrons/cubic meter What would that be in moles of electrons /m3? 991 moles/m3 or 0.000991 moles per cm3 Ohm’s Law V = IR V : potential drop between two points (Volts, V) I : current (Amps, A) R : resistance (Ohms, ) Conductors High conductivity Low resisitivity Loose electrons (for most electrical circuits) Insulators High resistivity Low conductivity Tightly held electrons (for most electrical circuits) Resisitivity, Property of a material which makes it resist the flow of current through it. Ohm-meters (m) Resisitance, R Depends on resistivity and on geometry R = L/A Ohms () Conductivity, = 1/ The inverse of resisitivity R = L/A =L/σA Electrical Power P = IV P: Power in Watts I: Current in Amperes V: Potential Drop in Volts 2 iR P= P = V2/R Electromotive Force Related to the energy change of charged particles supplied by a cell. Designated as EMF or as e. A misnomer: not a force at all! Internal Resistance The resistance that is an integral part of a cell. Tends to increase as a cell ages. (refrigeration helps slow this aging down) e r Internal Resistance When voltage is measured with no current flowing it gives e. e V r Internal Resistance When voltage is measured with current flowing, it gives VT, equal to e – iR. e V r i Resistors in series R1 R2 R3 Req = R1 + R2 + R3 Resistors in parallel R1 R2 R3 1/Req = 1/R1 + 1/R2 + 1/R3 Current in a circuit Defined to be opposite direction of the flow of electrons Current in a circuit I Electrons move in opposite direction AP C Circuit Analysis Unlike the Regents, the AP C exams and college textbooks… 1) define current as the flow of + charge 2) have mixed circuits that with series and parallel elements 3) have capacitors, charging and discharging 4) can have more than one battery. The batteries aren’t necessarily ideal; they can have internal resistance that reduces voltage output. 5) you may need to use a loop rule to figure out voltage drops (Kirchoff's Laws) through simultaneous equations. 6) have coils magnetizing and demagnetizing Kirchoff’s st 1 Rule Junction rule. The sum of the currents entering a junction equals the sum of the currents leaving the junction. Conservation of… charge. Kirchoff’s nd 2 Rule Loop rule. The net change in electrical potential in going around one complete loop in a circuit is equal to zero. Conservation of energy. Using Conventional Current & the Loop Rule Internal resistance of real batteries Voltage Drops When doing a loop analysis, V =0. Some V are +, some -. Through batteries, going from – to + is an increase, or + V. going from + to - is logically a loss of potential or -V. Through resistors: Going with the current is like going downhill, negative V, Going against current is like going uphill, +V. Water analogy Applying Kirchhoff’s Laws Goal: Find the three unknown currents. First decide which way you think the current is traveling around the loop. It is OK to be incorrect. Red Loop V ( I 3 6) ( I1 4) 0 24 6 I 3 4 I1 Using Kirchhoff’s Voltage Law Blue Loop V ( I 2 2) ( I 3 6) 0 12 2 I 2 6 I 3 I1 I 2 I 3 Using Kirchhoff’s Current Law Applying Kirchhoff’s Laws 24 6 I 3 4 I1 12 2 I 2 6 I 3 I 3 I1 I 2 24 6( I1 I 2 ) 4 I1 6 I1 6 I 2 4 I1 10 I1 6 I 2 12 2 I 2 6( I1 I 2 ) 2 I 2 6 I1 6 I 2 6 I1 8I 2 24 10 I1 6 I 2 6(24 10 I1 6 I 2 ) 12 6 I1 8I 2 10(12 6 I1 8I 2 ) 144 60 I1 36 I 2 24 44 I 2 I 2 -0.545 A 120 60 I1 80 I 2 A NEGATIVE current does NOT mean you are wrong. It means you chose your current to be in the wrong direction initially. Applying Kirchhoff’s Laws 12 2 I 2 6 I 3 12 2(0.545) 6 I 3 I 3 2.18 A 24 6 I 3 4 I1 24 6(?) 4 I1 I1 2.73 A Instead of : I 3 I1 I 2 It should have been : I1 I 2 I 3 2.73 2.18 0.545 From top, looping clockwise: +4V –I2 3Ω – I3 5Ω = 0 +4V –I2 3Ω – (I1 + I2 )5Ω = 0 +4V –I2 3Ω – I15Ω - I25Ω = 0 +4V –I2 8Ω – I15Ω =0 From middle section, looping clockwise: + I2 3Ω - I1 5Ω +8V= 0 Junction rule I1 + I2 = I3 Subtract these two equations to eliminate I1; 4V + I2 11Ω = 0 So I2 = 4/11 = - 0.36 Amperes - 0.36 x 3Ω - I1 5Ω +8V= 0 So I1 = +1.38 Amps and I3 = 1.02 Amps If opposing batteries are simply in series, you can predict the direction of I accurately by looking at what the net voltage is. 6V 4v Net voltage is 2V rightward Variable Resistors Terminology: galvanometers measure small currents (mA), while ammeters measure large currents (whole Amps) 330 BQ 330 A V 12 Volts For the drawing shown, calculate a) the current at A b) the total power dissipated by the resistor pair. A current of 4.82 A exists in a 12.4- resistor for 4.60 minutes. a)How much charge and b)How many electrons pass through a cross section of the resistor in this time?