Introduction to

CurrentIn 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)

In any typical wire

e-e-e-

e-+++ +

E

I

vd

J

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

V= JρL and V/L = JρSo does V/L = E inside a wire?

Well E = -dV/dL, so

E inside wire = ρJ

resistivity

E inside wire = ρJFind 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

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.

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

(1.6 x 10-19 Coul/electron)(1x10-5m/s)(π {1x10-3m}2)

= 5.9 E 26 electrons/cubic meterWhat 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 conductivityLow resisitivityLoose electrons (for most electrical circuits)

Insulators

High resistivityLow conductivityTightly 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 geometryR = L/AOhms ()

Conductivity,

= 1/The inverse of resisitivity R = L/A =L/σA

Electrical PowerP = IVP: Power in WattsI: Current in AmperesV: Potential Drop in VoltsP = i2RP = 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 ResistanceThe resistance that is an integral part of a cell.Tends to increase as a cell ages. (refrigeration helps slow this aging down)

re

Internal ResistanceWhen voltage is measured with no current flowing it gives e.

re

V

Internal ResistanceWhen voltage is measured with current flowing, it gives VT, equal to e – iR.

re

i

V

Resistors in series

R1 R2 R3

Req = R1 + R2 + R3

Resistors in parallelR1

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 AnalysisUnlike 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 RuleJunction rule.The sum of the currents entering a junction equals the sum of the currents leaving the junction.Conservation of…charge.

Kirchoff’s 2nd RuleLoop rule.The net change in electrical potential in going around one complete loop in a circuit is equal to zero.Conservation ofenergy.

Using Conventional

Current& the Loop Rule

Internal resistance of real batteries

Through resistors: Going with the current is like going downhill, negative V, Going against current is like going uphill, +V.

Voltage DropsWhen 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.

Water analogy

Applying Kirchhoff’s Laws

13

13

4624

0)4()6(Loop Red

II

IIV

32

32

6212

0)6()2(Loop Blue

II

IIV

Goal: Find the three unknown currents.

321 III

Using Kirchhoff’s Voltage Law

First decide which way you think the current is traveling around the loop. It is OK to be incorrect.

Using Kirchhoff’s Current Law

Applying Kirchhoff’s Laws

213

32

13

6212

4624

III

II

II

21212212

21121121

86662)(6212

6104664)(624

IIIIIIII

IIIIIIII

2

2

2121

2121

2121

4424

80601203660144

)8612(108612

)61024(661024

I

I

IIII

IIII

IIII

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

1

113

3

332

4(?)6244624

6)545.0(2126212

I

III

I

III

545.018.273.2

:been have shouldIt

:of Instead

321

213

III

III

2.18 A

2.73 A

From top, looping clockwise:

+4V –I2 3Ω – I3 5Ω = 0

From middle section, looping clockwise:

+ I2 3Ω - I1 5Ω +8V= 0

Subtract these two equations to eliminate I1;4V + I2 11Ω = 0

So I2 = 4/11 = - 0.36 Amperes

Junction rule I1 + I2 = I3

+4V –I2 3Ω – (I1 + I2 )5Ω = 0

+4V –I2 3Ω – I15Ω - I25Ω = 0

+4V –I2 8Ω – I15Ω = 0

- 0.36 x 3Ω - I1 5Ω +8V= 0

So I1 = +1.38 Amps and I3 = 1.02 Amps

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

Variable Resistors

BQ

V

12 Volts

330

330

For the drawing shown, calculatea) the current at Ab) the total power dissipated by the resistor pair.

A

A current of 4.82 A exists in a 12.4- resistor for 4.60 minutes.

a)How much charge andb)How many electronspass through a cross section of

the resistor in this time?