AP Physics C: E&M
CURRENT AND CIRCUITS
INTRODUCTORY TERMS
DC: Direct current. A constantly applied voltage causes
charged particles to drift in one direction
CAPACITORS IN A DC CIRCUIT
Adding capacitors in series will lower the capacitance of the
circuit when compared to the possible capacitance of just one
capacitor in the circuit.
C1
Voltage
source
+
-
C2
C3
Only the first plate of the first capacitor and the last plate of the
last capacitor are actually connected to the voltage source, so
only these plates will gain or lose electrons due to the potential
difference of the battery.
CAPACITORS IN A DC CIRCUIT
C1
Voltage
source
+
-
C2
C3
The inner plates are induced with charge. All capacitors carry an
equivalent charge Q.
The voltage across all elements in the series will add up to that of
the battery. Each capacitor has a different capacitance and has
the same charge, so the individual voltages will differ.
V  V1  V2  ...  Vn
CAPACITORS IN A DC CIRCUIT
C1
Voltage
source
+
-
C2
Q
Q Q
Q
 
 ... 
Ceq C1 C 2
Cn
C3
Q is the same for all so the equivalent capacitance can be found
with:
1
1 1
1
 
 ... 
Ceq C1 C 2
Cn
This should not be surprising since you are basically just making
one big capacitor with a larger separation (d).
CAPACITORS IN A DC CIRCUIT
Adding capacitors in parallel will raise the capacitance of the
circuit when compared to the possible capacitance of just one
capacitor in the circuit.
Voltage
source
+
-
C1
C2
C3
All capacitors are directly connected to the same voltage source
so they will each reach the same potential difference when
charged.
CAPACITORS IN A DC CIRCUIT
Voltage
source
+
-
C1
C2
C3
Since each capacitor may have a different capacitance, each may
hold a different amount of charge, but the sum of the charge will
equal that of one capacitor to replace those in parallel.
Q  Q1  Q2  ...  Qn
CAPACITORS IN A DC CIRCUIT
Voltage
source
+
-
C1
C2
C3
CeqV  C1V  C2V  ...  CnV
V is the same for all so the equivalent capacitance can be found
with:
Ceq  C1  C2  ...  Cn
This should not be surprising since you are basically just making
one big capacitor with a larger surface area
(A) for charge to be stored.
PRACTICE PROBLEMS #’S 8-12
CIRCUIT COMPONENTS
+
-
+
-
A
B
D
+
-
+
-
+
-
E
C
+
-
+
-
ELECTRIC CURRENT
Electric current is the amount of charge passing
through a certain area per second. It is measured
in amperes.
1 C of charge through any
cross section of wire per
second is one AMP!
It takes over 6.24 billion
billion electrons to add up to
one coulomb!
ELECTRIC CURRENT
Q
I av 
t
If the charge flow rate varies,
we define the instantaneous
current as:
dQ
I
dt
The direction of current is the direction that positive
charges would flow if free to do so.

Q  nAxq
n=number of charge carriers per unit volume
A=cross-sectional area of wire
Δx=length of section of wire
ΔQ=charge in a section of wire
q=charge on each particle
ELECTRIC CURRENT
If charge carriers move with a velocity vd,
then they move a distance Δx=vdΔt
Q  nAv dtq

Q
I
 nAv dq
t
DRIFT VELOCITY
With no voltage, charges in a metal bounce
around randomly similar to gas molecules. With
a voltage they still bounce around but slowly
drift in one direction.
DRIFT VELOCITY
A copper wire with cross-sectional area3x10-6m2
carries a current of 10.0A. Find the drift speed of the
electrons. The density of copper is 8.95g/cm3.
from the periodic table
atomic mass of copper:
m=63.5g/mol
m 63.5g /mol
3
V 
 7.09cm /mol
3
 8.95g /cm
DRIFT VELOCITY
A copper wire with cross-sectional area3x10-6m2
carries a current of 10.0A. Find the drift speed of the
electrons. The density of copper is 8.95g/cm3.
n A 6.0210 23 electrons /mol
22
3
n


8.48x10
electrons
/cm
V
7.09cm3 /mol
I
10A
vd 

nqA 8.48x1028 electons /m3 1.6x1019 C3x106 m2 
4
v d  2.46x10 m/s
THEN HOW DO THE LIGHTS COME ON SO FAST?
CURRENT DENSITY

We will define current density as:
I
J   nqv d
A
A current density J and an electric field E are
established in a conductor when a potential
 the conductor.
difference is maintained across
J  E
The proportionality constant is called the
conductivity of the conductor.
OHM’S LAW

Named after Georg Simon Ohm (1787-1854)
For many materials, the ratio of the current density to the
electric field is a constant, (sigma), that is independent of
the electric field producing the current.
This is not a law of nature, but an empirical relationship
found to be valid for certain materials (most metals)
If the potential difference is constant, the current is constant.
OHM’S LAW
J  E
For a segment of wire of length L:
V  EL
V
JL IL
 J  
V

L
 A

V L
R 
Resistance!
I A

RESISTANCE

The unit is the Ohm (Ω)
1V
1 
1A
The inverse of conductivity is resistivity!
1 


L
R
A
RESISTANCE AND TEMPERATURE:
For all metals, resistivity increases
with temperature increase.
  0 1 T  T0 
some reference value
usually at 20°C
Temperature coefficient
of resistivity
R  R 0 1 T  T0 
ELECTRICAL ENERGY AND POWER
U  qV
Divide both sides by time.
U qV

t
t
P  IV
ELECTRICAL ENERGY AND POWER
P  IV
PI R
2
V  IR
V
I
R
2

V 
P    R
R 

2
V
P
R
ELECTROMOTIVE “FORCE” – (EMF)
An emf is any device (generator/battery) that
produces an electric field and thus may cause
charges to move around in a circuit.
Is an emf (ε) any different than a voltage source (V)?
Any real emf has a certain amount of its own internal
resistance, so the voltage that it will supply to a
circuit between terminals is slightly different than its
own potential difference.
Both are measured in Volts.
ELECTROMOTIVE “FORCE” – (EMF)
An emf can be thought of as a charge pump.
V    Ir
V  IR
IR    Ir
V is the terminal voltage
Epsilon is the potential difference of the emf
I is the circuit’s current
r is the internal resistance of the emf
R is the equivalent resistance of the circuit
P is the power dissipated in circuit and emf device
  IR  Ir
P  I  I R  I r
2
2
KIRCHOFF’S RULES FOR COMPLEX CIRCUITS:
The sum of the currents
entering any junction must
equal the sum of the
currents leaving that
junction.
The algebraic sum of the
changes in potential
across all of the elements
around any closed loop
must be zero.
I am Bunsen. Have
you tried my burner?
KIRCHOFF’S RULES FOR COMPLEX CIRCUITS:
Of course Bunsen,
If charge is split between
two branches it must flow
down one path. it will not
build up in a location or
disappear.
Also, a charge must gain
as much energy as it loses
throughout the circuit
because it begins and
ends at the same point.
By the way, nice burner!
Do you mean that
Energy and Charge
are conserved?
RC CIRCUITS
What is different about a circuit with a resistor and a
capacitor than one with just a resistor?
The current does not flow at a constant rate!
Why is this?
+
-
The charge stops flowing when a
capacitor matches the battery voltage. It
drains charge through the resistor after
batter is disconnected.
At time t=0 the switch is closed and the full capacitor discharges.
No current
C
I
++++++
R
--------
ΔVR=0
C
+++
-- - -
ΔVC=Q0/C
ΔVC=Q/C
From the loop rule…

VC  VR  0
Q
 IR  0
C
R
ΔVR=-IR
Q and I are instantaneous values:
Q
 IR  0
C
dQ
I
dt


Q dQ

R0
C dt
Q
dQ
1
 Q   RC
Q0
t
 dt
0
 Q 
t
ln   
Q 0  RC
dQ
Q


dt
RC
dQ
dt
  
Q
RC

e
 Q 
ln 
Q 0 
e
Q
e
Q0


t
RC
t
RC
FIND THE CURRENT EXPRESSION FOR AN RC CIRCUIT
dQ
I
dt

  RC

Q  Q 0e

t
