Resistance
Current is somewhat like fluid flow. In fluid
flow it takes a pressure difference to make
the fluid flow due to the viscosity of the
fluid and the size (area and length) of the
pipe. So to in electricity, it takes a voltage
difference to make electric current flow due
to the resistance in the circuit.
Resistance
By experiment we find that if we increase the
voltage, we increase the current: V is
proportional to I. The constant of
proportionality we call the resistance, R:
V = I*R
Ohm’s Law
UNITS: R = V/I so Ohm = Volt / Amp.
(We will do this experiment in lab!)
Resistance
Just as with fluid flow, the amount of resistance
does not depend on the voltage (pressure) or
the current (volume flow). It does, however,
relate the two. The same idea applied to
capacitance: the capacitance did not depend
on the charge and voltage - the capacitance
related the two.
As with fluid flow and with capacitance, the
amount of resistance depends on the materials
and shapes of the resistor.
Resistance
The resistance depends on material and
geometry. For a wire, we have:
R=rL/A
where r is called the resistivity (in Ohm-m) and
measures how hard it is for current to flow
through the material, L is the length of the
wire, and A is the crossectional area of the
wire. One of our lab experiments will deal with
Ohm’s Law and the above equation.
Electrical Power
The electrical potential energy of a charge is:
PE = q*V .
Power is the change in energy with respect to
time:
Power = DPE / Dt .
Putting these two concepts together we have:
Power = D(qV) / Dt = V(Dq) / Dt = I*V.
Electrical Power
Besides this basic equation for power:
P = I*V
remember we also have Ohm’s Law:
V = I*R .
Thus we can write the following equations for
power: P = I2*R = V2/R = I*V .
To see which one gives the most insight, we
need to understand what is being held constant.
Example
When using batteries, the battery keeps the
voltage constant. Each D cell battery
supplies 1.5 volts, so four D cell batteries in
series (one after the other) will supply a
constant 6 volts.
When used with four D cell batteries, a light
bulb is designed to use 5 Watts of power.
What is the resistance of the light bulb?
Example
We know V = 6 volts, and P = 5 Watts; we’re
looking for R.
We have two equations:
P = I*V and V = I*R
which together have 4 quantities:
P, I, V & R..
We know two of these (P & V), so we should
be able to solve for the other two.
Example
Using the power equation we can solve for I:
P = I*V, so 5 Watts = I * (6 volts), or
I = 5 Watts / 6 volts = 0.833 amps.
Now we can use Ohm’s Law to solve for R:
V = I*R, so
R = V/I = 6 volts / 0.833 amps = 7.2 W .
Example extended
If we wanted a higher power light bulb,
should we have a bigger resistance or a
smaller resistance for the light bulb?
We have two relations for power that involves
resistance:
P = I2*R and P = V2 / R .
(The P=IV does not have R in it at all!)
Which one do we use to answer the above
question?
Example extended
Answer: In this case, the voltage is being held
constant due to the nature of the batteries. This
means that the current will change as we change the
resistance. Thus, the P = V2 / R would be the most
straight-forward equation to use. This means that as
R goes down, P goes up.
The answer: for more power, lower the
resistance. This will allow more current to
flow at the same voltage, and hence allow
more power!
Connecting Resistors
Instead of making and storing all sizes of
resistors, we can make and store just certain
values of resistors. When we need a nonstandard size resistor, we can make it by
connecting two or more standard size resistors
together to make an effective resistor of the
value we need.
Connecting Resistors
There are two basic ways of connecting two
resistors: series and parallel.
In series, we connect resistors together like
railroad cars:
+
-
+
high V
-
low V
R1
R2
Series
If we include a battery as the voltage source, the
series circuit would look like this:
R1
+
Vbat
R2
Note that there is only one way around the
circuit, and you have to flow through BOTH
resistors in making the circuit - no choice!
Formula for Series:
To see how resistors combine to give an
effective resistance when in series, we can
look either at V = I*R, or at R = rL/A .
Formula for Series
We see that in series the current must move
through both resistors. Thus Itotal = I1 = I2 .
Also, the voltage drop across the two resistors
add to give the total voltage drop:
Vtotal = V1 + V2 . Thus, using V = I*R, or
Reff = Vtotal / Itotal = (V1 + V2)/Itotal = V1/I1 + V2/I2
=
R1 + R2 = Reffective .
Note that connecting resistors in series gives a bigger effective
resistance than any of the individual resistors.
Formula for Series
Using R = rL/A , we see that we have to go
over both lengths, so the lengths should add.
The distances are in the numerator, and so
the values should add. This is just like in R
= V/I where the V’s add and are in the
numerator!
Parallel
In a parallel connection, there is a branch
point that allows you to complete the circuit
by going through either one resistor or the
other: you have a choice!
High V
R1
R2
Low V
Parallel Circuit
If we include a battery, the parallel circuit
would look like this:
+
Vbat
R1
R2
Formula for Parallel R’s
We see that in parallel the voltage across
each resistor is the same and is the same as
the voltage of the battery: Vtotal = V1 = V2 .
Also, the current through the two resistors add
to give the total current: Itotal = I1 + I2 .
Thus, using V = IR, or Reff = Vtotal / Itotal
Reff = Vtotal / (I1 + I2), or
[1/Reff] = (I1 + I2) / Vtotal = I1/V1 + I2/V2 =
1/R1 + 1/R2 = 1/Reffective .
Formula for Parallel Resistors
If we start from R = rL/A , we can see that
parallel resistors are equivalent to one
resistor with more Area. But A is in the
denominator (just like I was in the previous
slide), so we need to add the inverses:
1/Reff = 1/R1 + 1/R2 .
Note that connecting resistors in parallel gives a smaller
effective resistance than any of the individual resistors.
Computer Homework
The Computer Homework, Vol 3, #5 & 6,
give both an introduction and problems
dealing with resistors. Your homework
assignment is to do #6 on Resistors
Advanced.
Connecting Capacitors
Together
Instead of making and storing all sizes of
capacitors, we can make and store just
certain values of capacitors. When we need
a non-standard size capacitor, we can make
it by connecting two or more standard size
capacitors together to make an effective
capacitor of the value we need.
Two ways
As with resistors, there are two basic ways of
connecting two capacitors: series and parallel.
In series, we connect capacitors together like
railroad cars; this is just like we have for
resistors. Using parallel plate capacitors it
would look like this:
+
-
+
high V
-
low V
C1
C2
Series
If we include a battery as the voltage source,
the series circuit would look like this:
C1+ +
+
Vbat
C2
Note that there is only one way around the
circuit, and you have to jump BOTH
capacitors in making the circuit - no choice!
Parallel
In a parallel connection, there is a branch
point that allows you to complete the circuit
by jumping over either one capacitor or the
other: you have a choice!
High V
C1
C2
Low V
Parallel Circuit
If we include a battery, the parallel circuit
would look like this:
+
Vbat
+
C1
+
C2
Formula for Series:
To see how capacitors combine to give an
effective capacitance when in series, we can
look either at C = Q/V, or at
Cparallel plate = KA / [4pkd] .
Formula for Series
Using C = Q/V, we see that in series the
charge moved from capacitor 2’s negative
plate must be moved through the battery to
capacitor 1’s positive plate. Thus
Qtotal = Q1 = Q2 .
Also, the voltage drop across the two
capacitors add to give the total voltage drop:
Vtotal = V1 + V2 . Thus, Ceff = Qtotal / Vtotal =
Qtotal / (V1 + V2), or [1/Ceff] = (V1 + V2) / Qtotal =
V1/Q1 + V2/Q2 = 1/C1 + 1/C2 = 1/Ceffective .
Capacitors in Series
From our formula for series:
 1/Ci = 1/Ceffective
we see that the effective capacitance will be
less than the least of all the capacitors in
the series.
WHY?
Capacitors in Series
Note that the charge moved by the voltage source
(battery) is the same as the charged moved on
each of the capacitors, including the least (Qtotal
= Q1 = Q2 ). But the voltage across this least
capacitor is less than the voltage across the battery
(since that voltage is split among all the capacitors
Vtotal = V1 + V2 ). Thus for the effective
capacitance (Ceff = Qtotal / Vtotal) we have the
same charge that is moved with the largest
voltage, giving us the least capacitance!
Formula for Series
Using Cparallel plate = KA / [4pkd] , we see that
we have to go over both distances, so the
distances should add. But the distances are in
the denominator, and so the inverses should
add. This is just like in C = Q/V where the V’s
add and are in the denominator!
Note: this is the opposite of resistors when
connected in series!
Formula for
Parallel Capacitors
The result for the effective capacitance for a
parallel connection is different, but we can
start from the same two places:
C = Q/V, or Cparallel plate = KA / [4pkd] .
For parallel, both plates are across the same
voltage, so Vtotal = V1 = V2 .
The charge can accumulate on either plate, so:
Qtotal = Q1 + Q2 . Since the Q’s are in the
numerator, we have simply: Ceff = C1 + C2.
Formula for Parallel Capacitors
If we use the parallel plate capacitor formula,
Cparallel plate = KA / [4pkd] , we see that the
areas add, and the areas are in the
numerator, just as the Q’s were in the
numerator in the C = Q/V definition.
Review of Formulas
For capacitors in SERIES we have:
1/Ceff = 1/C1 + 1/C2 .
For capacitors in PARALLEL we have:
Ceff = C1 + C2 .
Note that adding in series gives Ceff being
smaller than the smallest, while adding in
parallel gives Ceff being larger than the
largest!
Computer Homework
The Computer Homework, Vol 3, #7 & 8,
give both an introduction and problems
dealing with capacitors. Your homework is
#8, Capacitors Advanced.
Review:
Capacitors:
Series: 1/Ceff = 1/C1 + 1/C2
Parallel: Ceff = C1 + C2
series gives smallest Ceff, parallel gives largest Ceff.
Resistors:
Series: Reff = R1 + R2
Parallel: 1/Reff = 1/R1 + 1/R2
series gives largest Reff, parallel gives smallest Reff.
The RC Circuit
What happens when we close the switch (S)
in the circuit below? That is, what is I(t) ?
C
+
V
R
S
Charging the Capacitor
Initially, the battery will “pump” charge onto
the positive plate of the capacitor, removing
that charge from the negative plate.
(Actually, negative electrons will be removed
from the positive plate and “pumped” by the
battery onto the negative plate.)
The resistor will provide a resistance to the
movement of this charge.
Qualitative Expectations
However, as the capacitor starts to fill up, the
charges already on the capacitor will repel
the new charges coming from the battery.
This should slow the rate of charge being
delivered (the current).
Q
I
t
t
Getting an equation
How do we mathematically model this
process? What fundamental equation do we
start from?
Conservation of Energy
From Conservation of Energy, and from PE =
qV, we have the following relation:
 Vi = 0 around a closed loop .
In our circuit we have three voltage sources: the
battery whereVB = constant, the capacitor
whereVC = (1/C)Q [from C = Q/V]), and the
resistor whereVR = IR [from Ohm’s Law]).
VB - (1/C)Q - IR = 0 where I = dQ/dt .
Getting a differential equation
VB - (1/C)Q - IR = 0 where I = dQ/dt .
This gives a differential equation for Q(t), and
then from I=dQ/dt, we can get I(t).
R (dQ/dt) + (1/C) Q = VB .
This is an inhomogeneous first order
differential equation.
The Solution: Q(t)
The homogeneous equation:
R (dQ/dt) + (1/C)Q = 0
has the solution: QH = Qo e-t/RC .
The inhomogeneous equation:
R (dQ/dt) + (1/C)Q = VB
has the solution: QI = CVB .
Therefore we have: Q(t) = CVB + Qoe-t/RC .
Using initial conditions
Q(t) = CVB + Qoe-t/RC .
To find what Qo is, we look at initial
conditions:
Q(t=0) = CVB + Qo = 0 (capacitor is initially
uncharged). Therefore we have: Qo = -CVB
Q(t) = CVB (1 - e-t/RC) .
For t very big, this says: Q(t=large) = CVB, or
that the capacitor will be fully charged!
The RC Circuit: charging!
Q(t) = CVB (1 - e-t/RC) .
This means the current is: I(t) = dQ/dt =
-CVB(-1/RC) e-t/RC = (VB/R)e-t/RC = Io e-t/RC .
The current starts out as if the capacitor were
not there (VB=VR=IR), but decays exponentially
as the capacitor becomes charged.
These two functions are plotted on the next
slide, and do indeed graph like we expected
earlier! The values used were:
V = 24 volts, R = 24,000 W, C = 2 mF .
Q(t) and I(t) for charging an RC
circuit
Q(t) vs t
I(t) vs t
1.200
I in microamps
40.000
30.000
20.000
10.000
1.000
0.800
0.600
0.400
0.200
t in milliseconds
55
49
43
37
31
25
19
13
7
55
49
43
37
31
25
19
7
13
time in milliseconds
1
0.000
0.000
1
Q in microCoulombs
50.000
Discharging the Capacitor
If we allow the capacitor to become
essentially fully charged, what will happen
when we discharge the capacitor through a
resistor?
In this case, we have the same differential
equation except we no longer have the
inhomogenous term, VB. Thus:
Q = Qo e-t/RC .
Solution: Discharging!
Q(t) = Qo e-t/RC . The capacitor discharges
as a dying (decaying) exponential.
Note when t = RC, Q = Qo e-1 = 0.368 Qo .
Since C = Q/V, V = (1/C)Q, and since Q
discharges exponentially, and C remains
constant, we find VC also decreases
exponentially: VC = VCoe-t/RC , and
VC(t=RC) = 0.368 VCo .
We use this fact in the lab experiment on
capacitors!
Units
From the expression for a discharging
capacitor: Q(t) = Qoe-t/RC , we see that the
quantity RC must have units of time. Let’s
check it out:
From Ohms law: V=IR, Ohm=W= Volt/Amp;
By definition: C = Q/V, Farad = Coul/Volt.
units of RC: W*Farads = [Volt/Amp]*(Coul/Volt)
= Coul/Amp = Coul/[Coul/sec] = sec.