Download i 1 - mrdsample

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Day 1 slides 1-11, hw 8-1 webassign
There are 4 webassigns for unit
8-2 = circuits 8-3 = kirchoff 8-4 RC etc
Copy DC circuits wkshts 1, 2, 3, + bulb demo
(saved as word doc) for use on day 2 and 3
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):
Units: C/s=A (Ampere)
I = dQ/dt
Scalar quantity
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?
E = 0 inside conductor was only for electrostatic
conditions. No longer dealing with that…charges are
now free to move. Only when charges are at rest,
does E = 0.
Current Density (j)
Current per cross sectional area
I   j  dA
where dA is the element of surface & I is the current through
the surface over which integration takes place
Current density is a vector field within a wire. The
vector at each point points in the direction of the E-field
I   j  dA
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. (nonconventional).
The thermal motion of electrons is very
random but fast (106 m/s). When Efield 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
V  IR
Rwire
L

A
Units for resistance are
ohms (Ω)
ρ = resistivity… a
parameter that depends
on the properties of
material
Units are Ωm
Materials that obey Ohm’s Law are said to be
ohmic. Incandescent light bulbs are non-ohmic.
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
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.
The rate at which charge
loses PE in resistor is given by
U Q

V  IV
t
t
From this we get power lost in the resistor:
P  IV
Using Ohm’s Law we can also get
2
V
PI R
R
2
Day 2
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Get out circuit board, batteries
Wire for short
Potentiometer
rheostat
Use drycell for more stable current flow.
Series Circuit Characteristics
RT  REQ  R1  R2  ...
IT  I1  I 2  ...
VT  V1  V2  ...
Parallel Circuit Characteristics
VT  V1  V2  ...
IT  I1  I 2  ...
1
1
1
 
 ...
RT R1 R2
Short Circuit
EXAMPLE 1
a) Determine the equivalent resistance of the circuit
shown if R1 = 860Ω, R2 = 640Ω, and R3 = 470Ω.
REQ = 836.9 Ω
b) Determine the voltage drop (potential difference)
across R3 .
V =6.73V
3
c) Determine the current through R2 .
I2 =8.23mA
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.
EXAMPLE 2
a) If V = 10V, find the
potential difference
across R4.
4.08V
D
C
b) If the wire at ‘C’ is cut, what
happens to the total current in
circuit?
RT goes from 13.5Ω to 60Ω, so IT decreases. Less
pathways means more total resistance.
EXAMPLE 2
RESET problem
c) If wire is cut at ‘D’
between R3 & R4 what
happens to VA? To VX?
D
C
Vx = 0 . Originally VA = VR2 = 6.89V.
VA still = VR2 , however, that value is
now 7.5V.
d) RESET…If R2 is “shorted out” what happens to the
total current?
The short leaves out R2, R3, & R4 . Total R goes
down to 6Ω from 19.5Ω, total current goes up.
R1
S
V
R3
R4
R2
Describe what happens to the voltage across each
resistor when the switch S is closed?
V1 and V2 increase; V3 and V4 decrease
Work on DC circuits #2 worksheet
S1
+
1
-
S2
2
S3
3
4
5
S4
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
An ideal battery has no internal resistance (friction).
However, a real battery has some internal resistance
where there is a voltage drop(EPE loss/charge)
within the 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,
Vab =  – Ir
Vab = 
Open circuit
voltage (ideal)
The current provided to the circuit depends on both the
internal, r, & external resistance, R. When R >> r, most
of the power that is delivered by the battery is
transferred to the load resistor, R.
As a battery ages, its internal resistance increases (the
chemical processes slows) . This affects the battery’s
ability to supply energy.
Different sized batteries (AAA vs D) have different
amp-hour (charge) ratings or capacities. The larger the
battery, the higher the amp-hour rating for the same V.
Larger-sized batteries have more charge to supply.
More charge carriers can deliver more energy.
A 12V car battery is more dangerous than 9, 1.5V D
batteries.
Series and Parallel EMFs
Batteries in series in the
same direction where
signs alternate: total
voltage is the sum of the
separate voltages.
Battery Charging
Batteries in series,
opposite direction where
signs are like signs: total
voltage is the difference,
where the lower voltage
battery is being charged.
Hooking up car batteries in series to charge. Like signs
must be connected together.
Batteries in parallel are not used to increase voltage or
charge one another, but serve 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 (charge) 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.
Vab worksheet
Power delivered to Load (R)
When is the power
delivered to the
load (R) a
maximum when
battery is NOT
ideal?
Meaning, how is R related to r in order to achieve
maximum power output, assuming there is a load
resistor?
PI R
2
Determine I that
flows through R,
then plug into P.
P(

rR
I   /( r  R)
2

2
) R
r  2rR  R
2
Clearly, PMAX is achieved when
denominator is a minimum.
2
R
P

2
r  2rR  R
2
2
R
Turn R in numerator to 1/(1/R) and simplify
P

2
1 2 1
1 2
( )r  ( )2rR  ( ) R
R
R
R
P

2
2
r
 2r  R
R
We know P will be maxed out when denominator is a
minimum. So, take the derivative of denominator and
set equal to zero to see when R is a minimum.
2
d r
(  R  2r )  0
dR R
d 2 1
( r R  R  2r )  0
dR
2
 r R 1 0  0
2
 r / R  1
2
R r
2
2
2
R  r
Since resistance
is + , then
Rr
2nd derivative test to prove minimum or maximum for
denominator,
2
d
2 2
2
r
(r R  1) 
3
dR
R
Since answer is positive,
solution refers to a
minimum. If denominator
is minimum, then P is
maximum when R=r.
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 (Vj ) = 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
Arbitrarily assign
currents and their
directions in each
branch abiding by
I1
junction rule. If
you are wrong
about direction,
your answer will
I2
yield a minus
sign.
Calculate the current in
each branch of the circuit.
I3
Junction Rule: I3 = I1 + I2
Assign loops
(direction is arbitrary)
When moving against
current ,it is a positive
change in ΔV
Top Loop (CW): {ahdcba}
-I1 (30) + 45 – I3 (1 + 40) = 0
Outside Loop (CW) {ahdefga}
-I1 (30) + I2 (20 + 1) – 80 = 0
When moving from –
terminal to + terminal of
battery, it’s a positive
change in ΔV
3 equations, 3 unknowns, solve!
I3 = I1 + I2
-I1 (30) + 45 – I3 (1 + 40) = 0
-I1 (30) + I2 (20 + 1) – 80 = 0
I2 = 2.6A
I1 = -0.86A
I3 = 1.7A
Negative indicates current is opposite
direction and that battery was charging
instead of discharging. Don’t disregard
minus sign when figuring out other
currents.
If a voltmeter was
connected between
points ‘c’ and ‘f’, what
would be the reading
(Vcf)? Vcf = Vc – Vf.
If current was a negative,
don’t switch the direction.
Just keep as is.
Voltmeter completes
loop!
Use loop rule for left side moving CW (fgabcf)
-80 + 41(1.7) + Vcf = 0 -10.3V + Vcf = 0 Vcf = 10.3V
Right side moving CW (cdefc)
(2.6)(21) -45 + Vcf =0
9.6V + Vcf = 0
Vcf = -9.6V
+ ΔV moving from ‘c’ to ‘f’, - ΔV moving from ‘f’ to ‘c’
therefore, ‘f’ is higher potential than ‘c’
Diff is
due to
rounded
current
Example: 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
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
c) At what rate is energy being transferred at the 4.0V
battery AND is the battery supplying or absorbing energy?
P = I2V2 = (1.94A)(4V) = 7.76W
Supplying since current
is moving from – to +
through the battery using
conventional current as
default. We expect
current to move from +
plate to the circuit.
4.0V
Kirchoff worksheet
Get out RC demo board
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 zero across an
uncharged capacitor (it acts like a
wire…short circuit).
As time passes, the capacitor reaches a
maximum charge where I = 0 in capacitor (it acts
like an open circuit). Resistance is infinite along
line of R and C. 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
ξ
1) At instant switch is
connected to ‘a’, C starts to
charge. I is a maximum in
circuit at t = 0, given by ξ/R.
2) As C charges, VC
increases as VR
decreases, keeping
ΣV = 0 around loop
(KVL).
Deriving V(t) and q(t)
for RC circuit using
Kirchoff’s Loop Rule.
We will move CW (after closing
switch) around loop:
ξ – iR – q/C = 0
( at t = 0, q = 0, & i = ε / R )
ξ – (dq/dt)R – q/C = 0
(since i = dq/dt)
Rearranging and dividing by R yields
dq/dt = -q/RC + ξ / R
dq/dt = (-q + ξC ) / RC
dq/dt = - 1/RC(q - ξC )
Multiplying last term by C/C
and combining yields
Factoring out -1/RC
dq/dt = - 1/RC(q - ξC )
q  C
e
 C
dq
1

dt
q  C
RC
q

0
dq
1

q  C
RC
Using u –sub (u=q-ξC) for left
side yields
q  C
t
ln

 C
RC
Exponentiating both sides
we get
(
t
)
RC
t

dt
FINALLY!
0
q (t )  C (1  e
VC   (1  e
t
t
RC
RC
)
)
Current, i, as function of time
q (t )  C (1  e
t
RC
)
q (t )  C  Ce
t
RECALL
RC
dq
t
1
RC
 0  (Ce
)( 
)
dt
RC
dq
  t RC
 ( )e
dt
R

i  ( )e
R
t
RC
d at
e  ae at
dt
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. 1 τ = 1RC.
Note: R needs to be in series in some way with C for there to be an
effect on the time constant for charging. (explain later)
It can be shown that after 1 time constant (RC), VC is
63% of its maximum voltage, Vo.
If t = RC, then Vc = Vo (1 – e-(t/RC))
= Vo (1 - e-1)
= Vo (1- 0.37)
= 0.63Vo
0.63 Vmax
0.37 Imax
1 time constant, RC
Discharging
After a very long time, the
capacitor would be fully charged.
If the switch was then moved
to ‘b’,
ξ
i
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.
Using the loop rule from before where at t = 0, q = Qo
ε – iR – q/C = 0 , but ε = 0 (no battery)
– (dq/dt)R – q/C = 0
dq
q

dt
RC
q
t
dq
1
Q q   RC 0 dt
o
q
t
ln

Qo
RC
q
t
RC
e
Qo
q  Qo e
t
RC
dq
Qo  t RC
i
 (
)e
dt
RC
Neg indicates reverse direction for
discharge
When discharging, C is now the power
source, so the time constant is C multiplied
by the equivalent resistance of remainder of
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) Ib = ε /(2R) d) Ib = ε / R
What is the current through the
battery after switch 1 has been
closed a long time?
C
S1
R
S2
Because C acts like a short
a) Ib = 0
b) Ib = V/(3R)
c) Ib = V/(2R)
d) Ib = V/R
Because C acts like R =

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
b) Ib = ε / (3R)
2R
ε
c) Ib = ε /(2R) d) Ib = ε / R
C
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
V2R = 0, VR = 0.13(VC) = 0.13 (ε/3). Capacitor only had 1/3
voltage across it…in parallel with R.
R
S2
S
Example
a) Find VR2 & VR1 after S has
been closed for 1τ.
12V
R2
Initially, each branch gets 12V
since they are in parallel so….VR2 = 12V always,
unaffected by R1C
R1
C
Using loop rule on right moving CCW:
– VR2 + VC + VR1 = 0
C is neg on left plate
-12V + (0.63(12V)) = - VR1
VR1 = 4.4V
b) Find total current at this time (1 τ )
if R1 = 10Ω and R2 = 20Ω
12V
4.4V
iR1  (
)  0.44 A
10
S
R2
R1

C
I R 2  ( )  0.60 A
R
Itotal = 1.04A
Could also use
this eqn to get iR1

i  ( )e
R
t
RC
1
 (12 / 10)(e )  0.44 A
Example
Each circuit below has a 1.0F capacitor charged to
100V. When the switch is closed:
a) Which system will be brightest?
b) Which lights will stay on longest?
2…each gets full current
Time constant of 1 is 4x as
long…2RC vs RC/2
c) Which lights consumes more energy assuming we
wait until both can’t be seen?
E = Pt…
Circuit 2 is 4x
the power
(2V2/R vs
V2/2R) but
lasts ¼ as
long as 1
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.
1MΩ
3MΩ
18V
6MΩ
1µF
Initially, C acts as a
short but it is in
series with 3MΩ .
REQ = (6-1 + 3-1)-1 + 1 = 3MΩ
iT = 18V / 3MΩ = 6uA
i1 = 6uA
i3 = 4uA
i6 = 2uA
1MΩ
3MΩ
18V
b) After a long time, C has
infinite resistance…
6MΩ
1µF
i3Ω = 0
Use KVL on outside loop…
18 - i1 (1MΩ) – i1 (6MΩ) = 0
i1 = 2.57uA
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.
Circuit will just be right hand loop
Find V6Ω first before switch was opened.
i6Ω = 2.57uA, so V6Ω = (2.57)(6) = 15.4V
Since C and 6MΩ were in parallel, they have
same V, therefore, VC = 15.4V.
1MΩ
3MΩ
18V
6MΩ
1µF
d) Find the charge on the capacitor 18s after the
switch is opened.
9MΩ is in series with C, so τ = RC = 9s
qo = CVo = (1uF )(15.4V) = 15.4uC
q(t) = qoe-t/RC = 15.4e-18/9
q = 2.1uC
e) How much energy is consumed by the
6MΩ during the discharging process?
UC = ½ CVo2
UC = 119uJ
Since P = I2R, the energy dissipated is
proportional to R, therefore
E = 6/9 (119uJ) = 79.3uJ
QUESTION: For the circuit shown, ξ = 6V, R1= 7Ω,
R2 = 3 Ω, R3 = 6Ω, and R4 =12 Ω . After operating for a
long time, equilibrium is established. What is the voltage
across the capacitor?
Use Kirchoff loop rule. Current in left branch will be
6/10, current in right branch will be 6/18 which gives
voltage for R1 and R3 . Top loop is 4.2V -2V = 2.2V=VC