Download Review - AdvancedPlacementPhysicsC

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 = 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)

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?