Download I. Simple Resistor Circuit

In Unit 305 we turn our attention to the last of the major units on circuits. In this unit we
focus on resistors and current flow through a circuit. The lessons in this unit will parallel
the approach of the unit on capacitors. At the end of this unit however, you will consider
a circuit composed of both resistors and capacitors. This will leave you with a two more
units on magnetism. During the second of those units you will learn of one more type of
circuit component, an inductor. For this unit be prepared to answer test items over
material in from the outline below.
I. Simple Resistor Circuit
A.
B.
C.
D.
Current and current density
Resistance, resistivity and conductivity
Ohm’s Law
Electrical Power
II. Multiple Resistor Circuits
A.
B.
C.
D.
Resistors in series
Resistors in parallel
Reduction method
Kirchoff’s Laws
1. Loop Rule
2. Junction Rule
III. Batteries and Meters
A. Real Batteries
1. EMF
2. Terminal Voltage
3. Internal Resistance
B. Meter Resistance and Galvanometers
1. Ammeters
2. Voltmeters
3. Ohmmeters
IV. Resistor and Capacitor (RC) Circuits
A. Initial conditions of RC circuits
B. Steady-state conditions of RC circuits
C. Functions of time during charging
1. charge
2. current
3. potential differences
D. Functions of time for discharging
1. charge
2. current
3. potential differences
Lesson 3-24
Read Chapter 27:1-9
Simple Resistor Circuits
When different points of a wire are at different potential values an electric field will be
established within a wire via, E = -V/x. The electric field will move charged particles
along the wire. We call the flow of charge, electrical current. In real currents, electrons
flow against the electric field; we will use conventional current notation that considers
the flow of positive charge in the opposite direction. The symbol for electric current is
“i” and has units of Amperes, Amps or C/s. In this lesson we consider the use of a single,
ideal battery to create a constant potential difference in order to develop a current flow
through a single resistor component.
Electric Current i
Electrical current is the rate of flow of charge through a wire
i = dq/dt
as defined in equation 27-1. Since current is measured in
amperes or amps the meter used to determine the current flow
through a wire is known as an ammeter. Checkpoint 1 in the text checks your
comprehension of Kirchoff’s Junction rule- the sum of currents flowing into a point equal
the sum of the currents flowing out of the point. This is necessary in order to conserve
charge at any point in a circuit.
Current Density J
i =  J  dA
Current actually tells us how much charge flows past a point each
second but does not tell us how closely packed the charges are as
they flow through a wire. Current density, J, gives an indication
of how close the charges are together or how far apart they are spread. Current density,
J, is measured in Amps/m2. Study Sample Problem 27-2 concerning current density.
Drift Speed vd
The drift speed is the rate at which charges move through the wire. It is usually a
surprise to most students to find that charge moves very slowly through a wire. Drift
velocities on the order of mm/sec are typical. One wonders, if drift velocities are so slow
then why does an electrical circuit turn on so fast? The key is that the electric fields are
established in a closed circuit at near the speed of light. As soon as the electric field is
established in part of a circuit the charge will flow in that part of the circuit.
Drift velocity depends on the charge density or charge per unit volume of the
conductor and can be found directly from this density and the current density. In order to
avoid confusion of too many densities the charge per unit volume is noted as the product
of “ne”. The charge per unit volume is calculated from the
J = (ne)vd
number of atoms per unit volume and assuming that each
atom contributes one free electron. Study Sample Problems 27-3 and 27-4.
Resistance
Different materials have the ability to resist electrical current more or less. Consider that
identical objects made of silver and glass will allow less current to flow for a given
electric field strength. The ability of a material to resist electrical current is known as
resistance, R. Resistance depends on the length of the material, L, the cross-sectional
area of the object, A and the composition of the material, .
R = L/A
The resistivity of different materials is shown in Table 27-1.
Resistance is measure in ohms or .
Ohm’s Law
J = E
Ohm’s law can be stated in two contrasting ways. In one respect,
a given electric field, E, can produce a particular current density,
J in a material with resistivity, . In this equation the direction of the electric field and
the current density are assumed to be in the same directions.
Consider a uniform resistor as shown in figure 27-7 and as diagrammed in figure
27-8. Suppose that a battery is used to create a potential difference across the leads of the
resistor. The electric field, E, could be determined from E = V/L where L is the length
of the resistor. The resulting current density can be represented
V = iR
as J = i/A. Subbing both expressions into the above box will give
us the more familiar form of Ohm’s Law that is shown in the box
to the right. The delta symbol is assumed for the rest of the unit.
Electric Power
In circuits, a resistor usually converts electrical energy to heat. The amount of heat
developed depends upon the power and time that current flows through the resistor.
Electrical power can be determined according to any of the equations shown below.
P = iV = i2R = V2/R
Also, Heat = Pt
Recall that power is measured in Watts or J/sec.
Homework Problems Ch 27: 1, 2, 15, 17, 28, 29, 43, 45.
Lesson 3-25
Read Chapter 28:1-6
Multiple Resistor Circuits
The previous lesson considered a single resistor connected to a single battery. Suppose
there are multiple resistors within the circuit. How do you determine the potential
difference across each individual resistor? How do you find the current flow through
each resistor? To analyze these circuits you need to know the rules for pure series
circuits and the rules for pure parallel circuits stated below
Pure Series Circuits

Resistors in series are shown in figure
28.5. The current flowing out of one
RTOT = R1 + R2 + R3 
must flow directly into the other. This
causes current to be the same for all
With equal currents and summing
resistors in series.
resistances the potential differences will
distribute. This is also a consequence of
iTOT = i1 = i2 = i3 
Kirchhoff’s Loop rule where  V’s = 0.
Resistances in series merely add. As
you combine more and more resistors in
VTOT = V1 + V2 + V3 
series you are increasing the overall path
length for collisions. Since the length of
the resistance is in the numerator in R
Study example problem 28.1. Add all of
the potential differences at the end of the
=L/A the relation for series resistors is
simple.
example to verify Kirchhoff’s loop rule.
Pure Parallel Circuits
Resistors in parallel are shown in figure
28.8. When connected in parallel each
resistor forms its own loop with the
battery. This causes all objects in parallel
to have the same potential difference.
VTOT = V1 = V2 = V3 
By placing resistors in parallel junctions
are created. At junctions, currents are
allowed to split creating a distributive
property for currents in parallel.
Consider example circuits shown below:
Example #1
2
3

RTOT-1 = R1-1 + R2-1 + R3-1 
Study example problem 28.8. Add all of
the currents at the junction to check
Kirchhoff’s junction rule. The sum of
currents entering a junction must equal the
sum of currents leaving a junction.
12 
4
12 
Example #2
3
iTOT = i1 + i 2 + i3 
As you add more and more resistor loops
in parallel you are increasing the net
cross-sectional area through which the
current will flow. Since area is in the
denominator of the R=L/A equation you
wind up with a reciprocal relation for
combining resistors in parallel.
6
EMF =
38 V
EMF =
48 V
2
1
Example #3
5
8
10 
EMF =
30 V
3
6
Simple Combination Circuits
If you consider figure 28-1 you see a single resistor with a single battery. What if you
have multiple resistors in a circuit as in figure 28-9? To analyze these circuits requires a
two-step process. The first step is to determine the equivalent resistance and net current
flowing out of the battery. Sample Problem 28-3 demonstrates the first part of the
analysis. To reduce the circuit to its equivalent resistance you need merely use’d
equations from the previous page. Every time you combine parallel branches or if you
combine a single branch with resistors in series redraw the circuit. In the long run this is
safer. The second half of the analysis can be done in one of two approaches.
You can calculate at each step using V=IR. Then you carry-back a value to the
previous step. If you are going back to a series step then the current is what is carried
back. Going back to a parallel step calls for taking voltage back. You start with the last
circuit you drew in the reduction of a circuit to its equivalent resistance.
1. In Example 28.3d for example you have a 12 V battery and equivalent resistance of
40 ohms. Using V = iR gives 12V=i*40 or i = 0.30 Amps.
2. Since the previous step is figure 28.9c with resistors in series you take back the
current of i=0.30 Amps to each resistor.
3. Now using V=IR on each resistor gives V4 = 2.4 Volts, V23 = 3.6 Volts and V1 = 6
Volts. It is a good sign that the potential differences add up to a total of 12V.
4. Since the 12 resistor in figure 28.9c represents two resistors in parallel in figure
28.9a you take the 3.6 V back to those resistors. Using V=iR or I = V/R will give the
individual current through each resistor.
5. Finally in circuit 18.8a you use V=iR on each resistor to find remaining unknowns.
V1 = 0.3A*20 = 6V ; V4 = 0.3A*8 = 2.4V ; i2 = 3.6V/20 = 0.18A ; i3 =
3.6V/30 = 0.12A
The “calculate and carry-back” method is slow and takes up a lot of paper space. The
good news is that the steps are consistent, methodical and will always work as long as
there is a single battery and no cross branching. Figure 28.10 for example will not work
with this method.
As an alternate approach, you can use Kirchhoff’s rules after you have the equivalent
resistance and total current. This method is super-fast but does require some creativity on
occasions. You can analyze Figure 28.3a when connected to a 12V battery with
Kirchhoff’s rules in only three steps after finding the equivalent resistance.
1. In Example 18.3d for example you have a 12 V battery and equivalent resistance of
40 ohms. Using V = iR in figure 18.8d gives 12V=I*40 or i = 0.3 Amps.
2. The 8 and 20 resistors are in series with the battery and thus have the same
current of 0.3A. Using V=iR on each device gives potential differences of 2.4V and
6.0V respectively.
3. A closed loop would include the battery, 8, 20 and either one of the other
resistors. The sum of the voltages around the closed loops would give the following
equations for V’s: +12V – 2.4V – V2 – 6.0V = 0; V’s: +12V – 2.4V – V3 – 6.0V =
0. In either case you find a potential difference of V2 = V3 = 3.6 Volts.
4. Now use i = V/R to get the individual currents.
Homework Problems For each of the following determine equivalent resistance and total
current flowing out of the battery. Then use either method you prefer to find the current
through each remaining resistor and potential difference across each remaining resistor.
Homework Problems Ch 28: 29, 31, 38, 47, 49, 50.
Lesson 3-26
Read 28. 6
Combination Resistor Circuits
Part II
As mentioned in the previous lesson you cannot use a “calculate and carry-back method”
if there are multiple power supplies or cross branching. If either of these problems shows
up you must use Kirchhoff’s Laws as outlined on page 676 & 681. You must also write
N equations for finding N unknowns. What the textbook and authors fail to realize is that
the Matrix buttons on your graphing calculators come in handy here. We demonstrate
these buttons using the equations of example 28.4 at the bottom of page 683.
i1 + i2 = i3
rewrite as 
i1 + i2 – i3 = 0
3 + 4i1 – 4i2 – 6 = 0
rewrite as 
4i1  4i2 + 0I3 = 3
6  4i3 – 4i2  6 = 0
rewrite as 
0i1  4i2  4i3 = 0
Write Matrix [A]
Write Matrix [B]
Now do [A]-1 [B]
Use Kirchhoff’s Rules to write N equations for N unknown currents. Note that resistors
on the same branch have the same current and do not need two different unknowns. You
may as well combine the resistors into a single resistance. A single resistor of 6, for
example, can replace the left branch of figure 28.10.
Sample #1
10 
R?
EMF?
EMF=
34 V
2
8A
3A
Homework Problems Ch 28: 32, 43, 48, 56, 80, 81
Lesson 3-27
Read 28. 4&7
Actual Meters and Real Batteries
Internal Resistance of a Battery
All batteries have some small, internal resistance. When a real battery is connected to an
external resistance some of the battery’s EMF is lost inside of the battery itself. The
circuit of a simple resistor connected to a battery with an internal resistance is shown in
figure 28.4. You can perform an experiment to measure internal resistance (r) by placing
known external resistors across the terminal of a battery and measure either the terminal
voltage (VT) or by measuring the resulting current.
Experiment #1
Place different external resistors across the terminal of the battery and use an ammeter to
measure resulting current and a voltmeter to measure resulting terminal voltage. Place
the current in the X list and terminal voltage in the Y list. The data should be a linear
plot with the internal resistance as the slope and the battery EMF as the y-intercept. The
best-fit line should be y = -rX + E.
Experiment #2
Place different, known external resistors(R) across the battery terminals. Measure the
resulting current (I). A plot of the external resistance on the X-axis and resulting current
on the Y-axis should give a curve. A voltmeter can be used to measure the open circuit
voltage or E. You can guess at different values of internal resistance until you have a
best-fit line using y = E (X + r)-1 that matches your scatter-plot data. See equation 28-4.
Ammeters
1. An ammeter is used to measure amperage of resulting current flowing through any
device.
2. In order to measure all of the current flowing through a device an ammeter must be
connected in series with the component.
3. If the ammeter has any internal resistance, the meter resistance will decrease the value
of the current that the meter is trying to measure. An ideal ammeter has an internal
resistance that is approaching 0.
Voltmeters
1. A voltmeter is used to measure the potential difference across any device.
2. In order to measure potential difference one lead is connected before and another lead
connected after in the scheme of current flow through the device. In other word,
voltmeters are always connected in parallel to the device they are measuring.
3. If the voltmeter has a low value of resistance then it will draw off some of the current
flowing through the device. The result is a lowering of the voltage difference that the
meter is attempting to measure. To prevent the voltmeter from affecting the voltage
measurement the internal resistance of a voltmeter should approach a value of  .
The high resistance of the voltmeter protects it from most student mistakes. 
The low resistance of the ammeter usually causes them to be prone to student destruction
or at least serious damage. 
Homework Problems Ch 28: 11, 16, 17, 21, 23, 24
Lesson 3-28
Read 28. 8
RC Circuits
What happens when a resistor and capacitor are placed on the same branch of a circuit
such as in figure 28.13? Is there a current flowing through the resistor? Does charge
collect on the plates of the capacitor? What about charge, current and voltages as
functions of time?
Initial Values
An empty capacitor acts like a closed switch or like bare wire with zero resistance. To
determine initial values of current and voltage in an RC circuit redraw the circuit without
the capacitor present. Find the currents and potential differences in the redrawn circuits.
The charge in the capacitor is zero and the potential difference across a capacitor is zero.
Steady-State Conditions
A fully charged capacitor will shut off current flow to the branch it is on. For that
particular branch there will be no current flow and all of the potential difference across
the branch is found across the capacitor. You can use q=CV to get the maximum charge.
Example
Find the current flowing through each resistor just after the switch is closed in the circuit
below.
S
4 f
EMF =
24 Volts
3
12 
The current flowing down the 3 ohms branch is 8 A since that resistor forms a closed
loop with the battery. The current flowing down the 12 ohms resistor is 2 A. This is also
the rate at which charge is entering the positive plate of the capacitor and rate at which
charge is leaving the negative plate of the capacitor. The initial current flowing through
the battery is 10 A.
After a long time what is the current flow through each resistor and the charge stored
within the capacitor?
The current flow through the 12 ohms resistor is zero since the charged capacitor will
switch off current flow to that branch. This branch does form a closed loop with the
battery however. Since V12 = 0 volts then all of the battery’s EMF shows up across the
capacitor. Using q = CV the final charge on the capacitor is 96 C. The current flowing
through the 3 ohms resistor is still 8 A. The total current flowing through the battery is
now just 8A.
Discharging Capacitors
A discharging capacitor acts like a battery with an exponentially decaying terminal
voltage. The initial discharge voltage depends on the potential difference across the
capacitor just before it starts to discharge.
Suppose after S has been closed for a long time it is suddenly thrown to the open
position. What will be the new current readings through each resistor as well as the
battery?
The capacitor acts like a battery with an EMF of 24 Volts. This is set up over a closed
loop with a resistance of 15 . The resulting, initial current through the resistors is 1.6A
down the 3 ohms resistor and up the 12 ohms resistor. The battery is on an isolated
branch and experiences no current.
Functions of Time
All of the RC equations for voltages, currents, charge etc. are either growth or decay
functions. These functions have a time constant in them that is  = R*C. It should be
noted that the product of ohms times farads comes out in seconds.
Growth Function
Decay Functions
If a value is increasing in time look for an
If a value is decreasing with time look for
equation of the form:
an equation of the form:
F(t) = FMAX ( 1 – e -
( t / )
)
FMAX is found from the final values on the
previous page.
F(t) = FMAX e -
( t / )
FMAX is found from steady-state value.
Finding functions of time
In order to find the functions of time you can do the simple steps outlined below:
1. Use Kirchoff’s rules to sum the voltages around a loop. For potential differences for
capacitors use VC = q/C and for resistors use VR = iR.
2. Replace the currents with their differential equivalent, i = dq/dt.
3. Separate the variables and integrate to get charge as a function of time in the capacitor.
4. Multiply q(t) by C in order to get VC(t).
5. Differentiate q(t) in order to get the current through the branch.
6. Multiply i(t) by R in order to get VR(t).
Sometimes it is easier to merely guess the function of time using the above forms. For
more complicated circuits with multiple branches the time constant for a capacitor is not
necessarily the product of RC for that branch. It can be shown that the time constants for
complex RC circuits with multiple branches is found by
 = qMAX/ iMAX
dividing the maximum charge in a capacitor by the
maximum current flow into that capacitor.
Homework Problems Ch 28: 65, 66, 67, 69, 71, 73, 77, 82