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APC – UNIT 8
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
+
-
The time rate of the charge flow through A defines the
current (=charges per time):
Current moving from + to – is
called conventional current flow.
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
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)
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…
dQ = nV(-e) = -n A vd (dt) e
Minus means dirn
of + current
opposes dirn of vd
Ohm’s Law
Materials that obey Ohm’s Law are said to be ohmic.
Incandescent light bulbs are non-ohmic.
Conductivity
+
Vb
E
-
Within a wire of length L, I = jA and V = EL, substituting
into Ohm’s Law we get
Va
High resistivity
produces less
current density
for same E-field
Conductivity is the
reciprocal of
resistivity.
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
A
From this we get power lost in the resistor:
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 (DV) have a very
large resistance to reduce the
amount of current drawn from the
circuit being measured (short).
Connected in PARALLEL.
Compound Circuit
a) V = 10V, find the potential
difference across R4
C
b) If the wire before R2 is cut
(inoperable) what happens to
the total current?
RESET problem
c) If wire is cut at C between R3
& R4 what happens to VA? VX?
d) RESET…If R2 is replaced by a wire what happens
to the total current?
If V = 45V, determine the power dissipated in R5.
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.
The voltage available for external
circuit is called the terminal
voltage, Vab . The internal
resistance is r. Therefore,
When battery is not
connected or ideal
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
EMFs in series in the
same direction: total
voltage is the sum of the
separate voltages…it is
increased.
Battery Charging
EMFs in series,
opposite direction: total
voltage is the difference,
but the lower voltage
battery is charged.
EMFs in parallel are not used to increase voltage but to
provide more energy when large currents are needed.
Each cell only produces a fraction of total current so
loss due to internal resistance is less than for single
cell. Batteries will last longer.
In this case, VR = 12V and if R=1,
IT = 12A with each battery
providing only 6A each.
When connecting in parallel you are
doubling the capacity (amp hours)
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.
Power delivered to Load (R)
When is the power delivered to the load a
maximum when battery is NOT ideal?
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)
(conservation of charge)
The sum of the currents entering
any junction must equal the sum
of the currents leaving that
junction.
2) Loop Rule S (DVj ) = 0
(conservation of energy)
The sum of the potential
differences across all the
elements around any closed
circuit loop must be zero.
(valid for any closed loop);
Kirchoff Example
Calculate the current in
each branch of the circuit.
If a voltmeter was
connected between
points c and f, what
would be the
reading (Vcf)? Vcf
means Vc – Vf.
Ideal batteries ξ1 = 11.5 V and ξ2 = 4.00V,
and the resistances are each 3.2Ω.
a) What is the size and direction of
current i1? (Take upward to be
positive.)
ξ1 i1
i2
ξ2
b) What is the size and direction of
current i2? (Take upward to be
positive.)
c) At what rate is energy being transferred at the 4.00V battery and is
the battery supplying or absorbing energy?
a) Focus on left hand loop
ΣV around loop = 0,
therefore V across
resistor must be 11.5V.
Moving CW around loop
yields…
i1
11.5V
i1 = -11.5V/3.2Ω = -3.59A
Negative since following direction of ξ1
i2
4.0V
b) What is the size and direction of current i2? (Take
upward to be positive.)
Focus loop on
center rectangle
Move around loop, CW
starting at battery.
i2
3.59A
½ i2
i2
4.0V
4V + 3.59(3.2) – i2 (3.2) – (i2/2)(3.2) – i2 (3.2) = 0
i2 =1.94A,
down
RC CIRCUITS
Often RC circuits are used to control timing.
Some examples include windshield wipers,
strobe lights, and flashbulbs in a camera, some
pacemakers.
A closer look at current during charging process
ξ
Applying Kirchoff’s Loop Rule to RC circuit (CW) to
find V & q as function of time (after closing switch):
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.
Note: R needs to be in series in some way with C for there to be an effect on
the time constant (explain later)
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’…
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
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
S
R2
R1
C
S
Example
Find total current at this
time (1 τ ) if R1 = 10Ω and
R2 = 20Ω
12V
R2
R1
C
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?
1MΩ
3MΩ
18V
6MΩ
1µF
A capacitor is initially uncharged and then
the switch is closed.
a) At t = 0, when switch is closed, find the current in
each resistor.
b) Find these currents after a long time later.
After a terribly long time, the switch is opened.
1MΩ
3MΩ
18V
6MΩ
1µF
c) Find the voltage drop on the capacitor just after the
switch is opened.
1MΩ
3MΩ
18V
6MΩ
1µF
d) Find the charge on the capacitor 18s after the
switch is opened.
e) How much energy is consumed by the 6MΩ
during the discharging process?