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APC – UNIT 9
DC Circuits
Whenever electric charges move, an electric current is
said to exist. The current is the rate at which the charge
flows through a certain cross-section A. We look at the
charges flowing perpendicularly to a surface of area A
+
-
Atomic View of Current
Consider a wire connected to a potential difference…
E
+
-
Existence of E inside wire (conductor) does not contradict
our previous results for E = 0 inside conductor. Why?
Current Density (j)
Current density is a vector field within a wire. The
vector at each point points in the direction of the E-field
The current density and the electric field are
established IN a conductor whereas a potential
difference is maintained ACROSS a conductor.
Drift Velocity
Current was originally thought to be positive charge
carriers (Franklin) and therefore that became conventional
current flow. However, it is the free electrons (valence)
that move but they encounter many collisions with atoms
in the wire. (non-conventional)
The thermal motion of
electrons is very random but
fast (106 m/s). When E-field is
established, the e- ‘drift’.
Overall speed of electrons is
VERY slow. It is called the drift
velocity, vd. 5.5hr to move 1m.
We can relate the current to the motion of the charges
In a time Δt, electrons travel a
distance Δx = vd Δt.
Volume of electrons in Δt pass
through area A is given as
V = A Δ x = A vd Δ t
If there are n free electrons per unit volume (n= N/V)
where N = # of electrons then the total charge through
area A in time Δt is given by
dQ = (# of charges)x(charge per e-) Also…
I neAvd
dQ = nV(-e) = -n A vd (dt) e
j 
 nevd
A
A
dQ
I 
 neAvd
dt
Minus means dirn
of + current
opposes dirn of vd
Ohm’s Law
Conductivity
+
E
Vb
-
Within a wire of length L, I = jA and V = EL, substituting
into Ohm’s Law we get
EL  ( jA)(
L
A
Conductivity is the
reciprocal of
resistivity.
)
E

j
1
j
 
 E
Va
High resistivity
produces less
current density
for same E-field
Electrical Power
Consider the simple circuit below. Imagine a positive quantity
of charge moving around the circuit from point A through an
ideal battery, through the resistor, and back to A again.
As charge moves from A to B
B
through battery, its electrical
energy increases by an amount
QΔV while the chemical PE of
battery decreases by that amount.
When the charge moves
through the resistor, it loses
EPE as it undergoes collisions
with atoms in R and produces
thermal energy.
A
The rate at which charge
loses PE in resistor is given by
B
U Q

V  IV
t
t
A
From this we get power lost in the resistor:
Using Ohm’s Law we can also get
Series Circuit Characteristics
Parallel Circuit Characteristics
Short Circuit
Ammeter and Voltmeter
AMMETERS have a very small
resistance to limit their effect on
introducing resistance into the
circuit being measured. Connected in
SERIES.
VOLTMETERS (V) have a very
large resistance to reduce the
amount of current drawn from the
circuit being measured (short).
Connected in PARALLEL.
Compound Circuit
a) Find the potential
difference across R4
b) If the wire before R2 is cut
(inoperable) what happens to
the total current?
c) If R2 is replaced by a wire
what happens to the total
current?
10V
Potentiometer or Variable Resistor
Device that allows for you to vary the
resistance by changing the effective
length of wire
symbol
EMF (electromotive force),ε
A ideal battery has no internal resistance (friction).
However, a real battery has some internal
resistance where there is a voltage drop within
battery leaving less ΔV for external circuit.
Different sized batteries (AAA vs D) have different
amp-hour ratings. The larger the battery, the higher
the amp-hour rating for the same V. Larger-sized
batteries have more charge to supply
The battery capacity that battery manufacturers print on a
battery is usually the product of 20 hours multiplied by the
maximum constant current that a new battery can supply for
20 hours at 68 F° (20 C°), down to a predetermined terminal
voltage per cell. A battery rated at 100 A·h will deliver 5 A over
a 20 hour period at room temperature.
Series and Parallel EMFs
Battery Charging
EMFs in parallel can produce more current than a
single emf while maintaining voltage.
When connecting in
parallel you are doubling
the capacity (amp hours)
(60amphrs means if the
load drew 10A, it would
last 6hrs) of the battery
while maintaining the
voltage of the individual
batteries
Batteries MUST be the same, If not,
there will be relatively large currents
circulating from one battery through
another, the higher-voltage batteries
overpowering the lower-voltage
batteries.
Kirchoff’s Rules
Circuits that are complex in that they
cannot be reduced to series or parallel
combinations require a different approach.
1) Junction Rule (S Ij = 0)
2) Loop Rule S (Vj ) = 0
(conservation of energy)
Kirchoff Example
Calculate the current in
each branch of the circuit.
I1
I3
I2
If a voltmeter was
connected between
points c and f, what
would be the
reading (Vcf)? Vcf
means Vc – Vf.
RC CIRCUITS
Often RC circuits are used to control timing.
Some examples include windshield wipers,
strobe lights, and flashbulbs in a camera, some
pacemakers.
Initially, at t=0, at the instant a switch
closes there is a potential difference of
0 across an uncharged capacitor (it
acts like a wire…short circuit).
Ultimately, the capacitor reaches its
maximum charge and there is no current
flow through the capacitor (it acts like an
open circuit as t goes to infinity (R=∞)).
At this point, ΔVC = ε. Note that C
does not charge instantaneously.
Current (i) decays over time and is not
steady.
A closer look at current
during charging process
1) At instant switch is
connected to ‘a’,
2) As C charges,
ξ
ξ
Applying Kirchoff’s Loop Rule to above circuit (CW):
Current, i, as function of time
Time Constant, τ
There is a quantity referred to as the time constant of the
RC circuit. This is the time required for the capacitor to
reach 63% of its charge capacity and maximum voltage.
It also represents the time needed for the current to drop
to 37% of its original value.
It can be shown that after 1 time constant (RC), VC is 63%
of its maximum voltage, Vo.
0.63 Vmax
0.37 Imax
1 time constant, RC
Discharging
ξ
i
After a very long time, the
capacitor would be fully
charged. If the switch was
then moved to ‘b’…
The capacitor would
discharge through the
resistor as a function of time
similar to the previous
derivation.
Voltage across
resistor equals
voltage across
capacitor at all
times in above
circuit.
Example
2R
Both switches are initially open, and
the capacitor is uncharged. What is
the current through the battery just
after switch S1 is closed?
a) Ib = 0
ε
b) Ib = ε / (3R)
C
S1
c) Ib = ε /(2R) d) Ib = ε / R
Both switches are initially open,
and the capacitor is uncharged.
What is the current through the
battery after switch 1 has been
closed a long time?
a) Ib = 0
b) Ib = V/(3R)
c) Ib = V/(2R)
d) Ib = V/R
R
S2
Both switches are initially open, and
the capacitor is uncharged. What is
the current through the battery just
after switch S1 & S2 are closed?
a) Ib = 0
2R
ε
C
R
b) Ib = ε / (3R)
c) Ib = ε /(2R) d) Ib = ε / R
S1
After a long time what is the
current through the battery?
a) Ib = 0
After a long time S1 is opened.
What is the voltage across R
and 2R after 2τ?
c) Ib = ε /(2R)
b) Ib = ε /(3R)
d) Ib = ε /R
S2
S
Example
Find VR2 & VR1 after S has
been closed for 1τ.
12V
R2
R1
C
Find total current at this time if R1 = 10Ω and R2 = 20Ω
Example
Each circuit below has a 1.0F capacitor charged to
100 Volts. When the switch is closed:
a) Which system will be brightest?
b) Which lights will stay on longest?
c) Which lights consumes more energy assuming we
wait until both can’t be seen?