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 I = JA J: current density in A/m2 I: current in Amperes (A) A: area of cross section of wire (m2) I = JA

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)

In any typical wire e- + E I vd J

So does V/L = E inside a wire? Well E = -dV/dL, so E inside wire = ρJ On the APC reference table, current density J is not defined, but you have these 2 formulas related to J: J = I/A and V = IR so V = JAR. Now add in resistivity R = ρL/A so V = JA ρL/A resistivity 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

Current density J is # of Amperes / m2 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.

Rearranging we get N = I/ evdA = 3 x 10-3 Coul/sec 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. Rearranging we get 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

V = IR V : potential drop between two points (Volts, V) 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

P = IV P = i2R P = V2/R Electrical Power P: Power in Watts I: Current in Amperes V: Potential Drop in Volts P = i2R 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!

e r 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

e r Internal Resistance When voltage is measured with no current flowing it gives e. V e r

e r Internal Resistance When voltage is measured with current flowing, it gives VT, equal to e – iR. V e 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 Electrons move in opposite direction I

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 1st 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 2nd 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. Using Kirchhoff’s Voltage Law Using Kirchhoff’s Current Law

Applying Kirchhoff’s Laws A NEGATIVE current does NOT mean you are wrong. It means you chose your current to be in the wrong direction initially. -0.545 A

Applying Kirchhoff’s Laws

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. Net voltage is 2V rightward 6V 4v

Variable Resistors Terminology: galvanometers measure small currents (mA), while ammeters measure large currents (whole Amps)

BQ For the drawing shown, calculate a) the current at A 330  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. How much charge and How many electrons pass through a cross section of the resistor in this time?