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)
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, )
Resisitivity, Property of a material which makes it resist the flow of current through it.Ohm-meters (m)
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 current flowing, it gives VT, equal to e – iR.
re
i
V
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.
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.
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
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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