PHYSICS
LAB
NOTES
FOR
ELECTRICITY, MAGNETISM
AND
GEOMETRIC OPTICS
EXPERIMENTS
PHYSICS 38
Los Angeles Harbor College
J. C. FU
R. F. WHITING
© 1991
Twelfth Edition (F)
August 2004
The set of lab experiments that you will be doing this semester will, hopefully elucidate for you
some abstract concepts, enable you to test a few hypotheses or theories using the scientific
method, realize the capabilities or limitations of certain equipment and procedures, and to think
analytically.
Each lab period will begin with a presentation / discussion of the experiment indicated in
your lab notes. Come prepared, having read the references given. An attempt has been made to
have the labs run in parallel with lecture. Share work with your partner so that each person will
have an opportunity to have hands-on experience.
The laboratory reports that you turn in should have the following format:
Cover
page
Exp # ____ Title ____
Name ____ Date ____
Successive
pages . . .
PURPOSE
APPARATUS
INTRODUCTION
PROCEDURE
ABSTRACT
DATA
GRAPHS
TABLE
TABLE
CALCULATIONS
and RESULTS
J. C. Fu, Ph.D.
R. F. Whiting, M.S.
All rights reserved. No part of this book may be reproduced, in any form or by any means, without
permission in writing from the authors.
ELECTRICAL SAFETY
Since we use electrical apparatus daily, we should understand the elements of electrical safety. Electricity can
kill a person in two ways: it can cause the muscles of the heart and lungs (or other vital organs) to malfunction or it
can cause fatal burns.
Even a small electric current can seriously disrupt cell functions in that portion of the body through which it
flows. When the electric current is 0.001A or higher, a person can feel the sensation of shock. At currents 10 times
larger, 0.01A, a person is unable to release the electric wire held in his hand because the current causes his hand
muscles to contract violently. Currents larger than 0.02A passing through the torso paralyze the respiratory muscle
and stop breathing. Unless artificial respiration is started at once, the victim will suffocate. Of course, the victim must
be freed from the voltage source before he can be touched safely; otherwise the rescuer, too, will be in great danger.
A current of about 0.1A passing through the region of the heart will shock the heart muscle into rapid erratic
contractions (ventricular fibrillation) so the heart can no longer function. Finally, currents of 1A and higher through the
body cause serious burns.
The important quantity to control in preventing injury is electric current. Voltage is important only because it
can cause current to flow. Even though your body can be charged to a potential thousands of volts higher than the
metal of an automobile by simply sliding across the car seat, you feel only a harmless shock as you touch the door
handle. Your body cannot hold much charge on itself, and the current flowing through your hand to the door handle is
short-lived and the effect on your body cells is negligible.
In some circumstances, the 120-V house circuit is almost certain to cause death. One of the two wires of the
circuit is almost always attached to the ground, so it is at the same potential as the water pipes in a house. Suppose a
person is soaking in a bathtub; his body is effectively connected to the ground through the water and piping. If his
hand accidentally touches the high-potential wire of the house circuit (by touching an exposed wire on a radio or heater
for example), current will flow through his body to the ground. Because of the large, efficient contact his body makes
with the ground, the resistance of his body circuit is low. Consequently the current flowing through his body is so large
that he will be electrocuted.
Similar situations exist elsewhere. For example, if you accidentally touch an exposed wire while standing on
the ground with wet feet you are in far greater danger than if you are on a dry, insulating surface. The electrical circuit
through your body to the ground has a much higher resistance if your feet are dry. Similarly, if you sustain an electrical
shock by touching a live wire or a faulty appliance, the shock is greater if your other hand is touching the faucet on the
sink or is in the dishwater.
As you can see from these examples, the danger from electrical shock can be eliminated by avoiding a
current path through the body. When the voltage is greater than about 50 V, avoid touching any exposed metal
portion of the circuit. If a high-voltage wire must be touched (for example, in case of a power line accident when help
is not immediately available) use a dry stick or some other substantial piece of insulating material to move it. When in
doubt about safety, avoid all contacts or close approaches to metal or to the wet earth. Above all do not let your body
become the connecting link between two objects that have widely different electric potentials.
Reference: "Physics for Scientists and Engineers" by F. Beuche
MICROSHOCK
Microshock is electrical shock caused by very small amounts of current. As is shown in table 1, currents of
less than 1 milliampere are usually of no consequence. If a shock is delivered directly to the heart, however, even 20
microamperes of current can be dangerous. Current can be delivered directly to the heart through a pacemaker wire.
Wires for use with external (temporary) pacemakers come out to the body through the chest wall or through veins that
lead to an arm, the neck, or elsewhere. If such a wire were touched by a person who was holding onto a light switch,
electric bed frame, television set, or other appliance, many microamperes could be conducted to the pacemaker wire.
Many appliances will supply a good fraction of a milliampere to someone who is grounded. To see that for
yourself, connect an ammeter between the metal parts of an appliance and ground. (Start on a high range to protect
the meter.) Unless there is a very good third wire ground, significant currents will be measured.
EFFECTS OF A 60 Hz ELECTRIC SHOCK
Current
(held one second)
20 mA
1 mA
5 mA
1-10 mA
10-20 mA
30 mA
75-300 mA
5A
Effect
(current applied to skin, unless otherwise noted)
Ventricular fibrillation if applied directly to the heart
Sensation
Maximum harmless current
Mild to moderate pain
May cause muscular contractions, preventing release from shock source
Breathing may stop
Ventricular fribillation may occur
Burns tissues
TABLE OF CONTENTS
1. Mapping the Electric Field .....................................................................1
2. The Oscilloscope ..................................................................................4
3. Ohm's Law .............................................................................................8
4. Joule Heat ........................................................................................... 11
5. Resistance & Resistivity....................................................................... 15
6. The Wheatstone Bridge ...................................................................... 19
7. The RC Time Constant ....................................................................... 22
8. Series-Parallel Circuit ......................................................................... 26
9. Circuit Analysis ................................................................................... 30
10. Mapping The Magnetic Field................................................................ 34
11. The Tangent Galvanometer ................................................................ 37
12. AC-RL, AC-RC Circuits ........................................................................ 40
13. AC-RLC Circuit ................................................................................... 45
14. The Visible Spectrum ........................................................................... 49
15. Reflection & Refraction ....................................................................... 52
16. The Thin Lens ..................................................................................... 55
Experiment 1: MAPPING THE ELECTRIC FIELD
PURPOSE:
To study the nature of electric fields by:
a) plotting equipotential points.
b) sketching equipotential lines and electric field lines.
c) calculating the field strength near charged surfaces.
d) calculating the charge on a dipole.
INTRODUCTION:
The visualization of an electric field by the use of lines of force is a useful concept
introduced by Michael Faraday. The electric field lines indicate the direction and magnitude of
electrical forces acting on a positive electrical test charge placed in this region of space. The
electric field strength can be represented by a vector tangent to the electric field lines. The work
done to bring a positive test charge from infinity to any point in an electric field is stored as
electrical potential energy of the test charge at that location.
The electrical potential energy of one Coulomb of test charge at a certain location in an
electric field is defined as the potential energy per Coulomb, or voltage at that point. Points of
equal voltage can be readily determined with a voltmeter. These points can be joined to produce
equipotential lines. A test charge placed at any point along an equipotential line will have the
same electric potential or voltage. Electric field lines can be determined by tracing the pathway of
a hypothetical positive test charge as it is pushed by the repulsive electrical force of the positive
electrode, and pulled by the attractive electrical force of the negative electrode. The electric field
at any point along an electric field line is the vector sum of electrical forces exerted on a positive
test charge. Electric field lines are perpendicular to equipotential lines. In this experiment the
electric field between three different sets of electrodes on resistive paper will be examined.
q
V
Equal Voltage Points
Equipotential Lines
Electric Field Lines
APPARATUS:
Plotting board
DigiTek multimeter
2 thin conducting leads
4 banana leads
Metric ruler
Power supply, 10 VDC
Carbon paper
2 metal push pins
French curve
-1-
Resistive paper
White paper
2 alligator clips
Colored pencils
PROCEDURE:
1. Set up the apparatus as shown in Figure 1.
10 V
White paper
Carbon paper
Electrode paper
- + Voltmeter
+
Fig. 1.
2. Set the multimeter to function as a voltmeter by pushing in the DCV- button and setting the
dial to 20 V (pointed straight up). Press lightly across the electrode paper with the positive
wire from the voltmeter to locate several points, one to two centimeters apart, at 2.0 volts.
When you join all 2.0-volt points, you will have an equipotential line of 2.0 volts. Make a slight
impression with the end of the positive wire about every centimeter. Avoid punching holes in
the resistive paper.
3. Repeat step two for 4.0 V, 6.0 V, and 8.0 V.
4. Trace the outline of the silver-painted shapes onto the white paper. Mark these electrodes as
plus and minus to indicate their polarity. Use a colored pencil to connect equipotential points
of the same voltage marked on the white paper with a smooth line, using a French curve or
flexible ruler.
5. Draw some electric field lines (about 12 of them) using a French curve and using a different
colored pencil. Note that these lines have to be at right angles to the equipotential lines.
Indicate the direction of the electric field lines with arrows. Shown below are representative
electric field lines emerging from the positive electrode and terminating on the surface of the
negative electrode in the dipole configuration:
6. Repeat the above procedure for two other electrode configurations: Parallel lines (plates) and
a sharp point facing a long line.
7. In the presence of a strong electric field, the voltage change V is large when you move a
short distance s perpendicularly from one equipotential line to another. Calculate the electric
field strength, E = V/s for the regions shown in Table 1 and Table 2.
8. Calculate the charge in units of Coulombs on the positive electrode of a dipole system,
calculated approximately as q = Vr/(9.0 X 109).
-2-
DATA AND CALCULATIONS:
Table 1: Electric field strength between parallel "plate" electrodes.
+++++++++++
Region
R2
R4
R1
R
Distance s
(m)
Difference in
Voltage V
(V)
E = V/s
(V/m)
Distance s
(m)
Difference in
Voltage V
(V)
E = V/s
(V/m)
Charge q
(C)
Average
value of q
(C)
R1
R2
3
R3
----------------
R4
Table 2: Electric field strength near sharp point
Region
R1
R2
R3
R4
Table 3. Charge on electrode (dipole system)
q (+)
Point
V
X
X
X
P1
P2
P3
P1
_________
P2
r
P3
where V =
Voltage V
(V)
Distance r
from (+)
electrode
(m)
kq
Vr
and q =
r
k
-3-
Experiment 2: THE OSCILLOSCOPE
PURPOSE:
a) Introduce the principles of operation
b) Measure AC voltages and frequencies
c) Observe Lissajous figures
INTRODUCTION:
In the study of AC (alternating-current) circuits, the oscilloscope is an essential piece of
equipment. It provides a means of measuring, comparing, and displaying the voltage and
frequency of signals. A schematic diagram of the oscilloscope’s CRT (cathode ray tube) is shown
below
Vertical Input
Amplifier
Filament
Electron Beam
Synchronizing
Voltage
Amplifier
Vo
Switch
Generator
Horizontal Input
APPARATUS:
Hitachi dual trace oscilloscope
(2) Function generators, Simpson # 420
Digital multimeter DigiTec, model # 2180
Power outlet strip
-4-
Test leads as needed
Multifunction counter
BNC to banana adaptor
PROCEDURE:
A. Adjustments to obtain trace:
Set the intensity to medium. Trigger is on Ext. Position is at center. Coupling is set at AC.
Focus is adjusted to sharp. The sweep rate is set to 1 ms/div. The deflection is set to 1 V/div.
HITACHI
OSCILLOSCOPE
A
B
D
E
C
F
H
G
J
I
L
O
K
N
M
P
Q
R
Figure A
A) Power on/off and brightness
J) Type of signal for channel 2.
B) Horizontal position of both traces,
K) Channel trigger settings.
(pull switch for 10X horizontal)
L) Channel 1, vertical amplitude
C) Trigger setting for channel 2.
M) Chooses type of display for channels 1 & 2
D) Beam focusing
N) Channel 2, vertical amplitude
E) Time per screen division = 1 cm
O) Signal input for channel 1.
F) Knob making time / division variable.P) Vertical positioning of trace of channel 1.
G) Screen scale illumination.
(Pull switch for 5X vertical).
H) x-y setting for Lissajous figures.
Q) Vertical positioning of trace of channel 2.
I) Type of signal for channel 1.
R) Signal input for channel 2.
-5-
B. MEASURING AN AC (SINE WAVE) VOLTAGE
1.
2.
3.
4.
Set up- the apparatus as shown in Figure B.
Adjust the function generator to 100 Hz at 6 V peak-to-peak voltage. Record in Table 1.
Calculate the peak voltage (Vo = ½ Vpp) and root mean square voltage (Vr m s = 0.707 Vo).
Read the r m s voltage from the DVM (digital multimeter) and compare to the r m s voltage
calculated from the oscilloscope trace.
5. Record the sweep rate in time/cm and convert this to centimeters per cycle (sine wave).
Calculate the period  and the frequency f. The period is the time for one cycle. If 1 cycle is 3
cm, and the sweep rate is 1 msec/cm, then the period will be (3 cm/cycle) (1 X 10 -3 s/cm) =
3 X 10-3 s/cycle. The frequency will be f = 1/T = 1/(3 X 10-3) = 300 Hz.
6. Repeat the procedure for an AC signal of 200 Hz and 1000 Hz.
7. Sketch a trace of the 200 Hz AC signal seen on the oscilloscope. Indicate V pp, Vo and Vr m s
on the reticule in the data sheet.
Voltage
(Volts)
Vo
Vrms = 0.707 Vo

Vpeak-to-peak
Time (sec)
-Vo
OSCILLOSCOPE
Frequency
Counter
~
AC Power
Supply
F.G. 1
1000 
Sweep Rate
E
L
Time/Div
Volts/Div
To Ch. 1 of scope
and to DVM
Ch.1
Fig. B
C. LISSAJOUS FIGURES
1. Set up the apparatus as in Figure C.
2. Adjust function generator # 2 (F.G. 2) to the same frequency and voltage as function generator
# 1 (F.G. 1).
3. Observe the lissajous figures when F.G. 2 frequency is 2, 3 and 4 times that of F.G. 1.
4. Observe the lissajous figures when F.G. 1 frequency is 2, 3 and 4 times that of F.G. 2.
F.G. 1
~
1000 
N
To Ch. 1 of scope
and to DVM
O
Ch.1
R
Ch.2
-6-
To Ch. 2 of scope
and to DVM
F.G. 2
1000 
~
DATA AND CALCULATIONS:
Data Table 1:
Frequency
(Hz)
Vpeak-to-peak
(Volts)
Vo
(Volts)
Calculated
Vrms
(Volts)
Vrms From
Voltmeter
(Volts)
Period From
Oscilloscope
(milliseconds)
Frequency
From
Counter
(KHz)
Period
From
Counter
(s)
Frequency
f
(Hz)
100
200
1000
Trace of AC signal.
Horizontal: 1 ms/cm
Vertical: 1 V/div.
Data Table 2: LISSAJOUS FIGURES (2 waves with the same amplitude and different frequency
whole multiples.)
Channel 1
(horizontal)
Channel 2
(vertical)
60 Hz
100 Hz
100 Hz
100 Hz
200 Hz
300 Hz
400 Hz
60 Hz
200 Hz
300 Hz
400 Hz
100 Hz
100 Hz
100 Hz
Sketch Trace
-7-
Experiment 3: OHM'S LAW
PURPOSE:
a) To learn about simple DC circuit elements.
b) To verify Ohm's Law (V = IR).
c) To calculate the power consumption by a resistor in a series circuit.
INTRODUCTION:
Ohm's law states the relation between the electrical potential V (in volts), the current I (in
amperes) and the resistance R (in ohms, symbolized by the Greek letter ) in a circuit.
V = IR, or
R
V
I
Circuit Elements:
R
I
V
V = electrical potential, provided by a DC (direct current) power supply. Use the black and
red outlets by your sink.
I = current flow of charges. The direction of current flow is denoted by an arrow and is the
direction a positive test charge would follow.
R = resistance or load. A resistor resists current flow, and consumes energy or dissipates
power. A rheostat is a resistor with a resistance that can be varied.
P = electrical power measured in watts. The power P = IV.
APPARATUS:
DC power supply, 5 V
Rheostat, 20
Decade resistance box (R)
Analog multimeter (VOM)
Digital multimeter
Leads & connectors
DigiTec voltmeter
-8-
PROCEDURE:
V
R
A
V
V = IR,
Rheostat
5V
Slope = R
I
1. Use wires and spade lugs to connect the negative (black) outlet to the lower-left port of the
rheostat (the device with the long black wire coiled around a cylinder) and the positive (red)
outlet to the lower-right port.
2. Connect a red wire to the upper-right port of the rheostat. Slide the tap to the left, and the
voltage on this wire becomes small; slide the tap to the right, and the voltage becomes large.
This device now functions as a variable voltage source.
3. Set the resistance decade box to 75. Connect the red wire from the variable voltage source
to a port on the bottom of the resistance box. Set the Hewlett-Packard multimeter dial to mA,
turn it on, run a wire from the other port on the bottom of the resistance box to the right-hand
port of the multimeter, and run another wire from the middle port of the multimeter to the
negative outlet. The multimeter now functions as an ammeter, measuring the current I through
the resistance box in milliamperes. Set the DigiTec multimeter's dial to 20V, and push in the
DCV- button. Run two wires from the HI and LO terminals to the resistance box terminals.
This device now functions as a voltmeter, measuring the voltage across the resistance box.
4. Slide the tap of the variable voltage source all the way to the right, to create the largest voltage
available. Record the electric potential V (voltmeter reading) and the current I (ammeter
reading). Calculate the power P being dissipated by the resistance box.
5. Decrease V by approximately 0.5 volts. Record the resulting V and I.
6. Repeat step 5 for six more sets of readings.
7. Plot V vs. I and compute R from the slope of the line.
8. Compare this experimental value to the known value of R set on the resistance box, by
calculating the percent difference.
9. Repeat steps 4 - 8 for R set at 150  on the resistance box.
-9-
DATA AND CALCULATIONS:
Data Table 1: R = 75 
Reading
Voltage V
(Volts)
Current I
(Amperes)
Power = IV
(Watts)
Current I
(Amperes)
Power = IV
(Watts)
1
2
3
4
5
6
7
R = ___________ (Measured from graph)
Percent difference = _______________
Data Table 2: R = 150 
Reading
Voltage V
(Volts)
1
2
3
4
5
6
7
R = ___________ (Measured from graph)
Percent difference = _______________
- 10 -
Experiment 4:
JOULE HEAT
PURPOSE:
a) To determine the electrical power consumption of a heating coil.
b) To determine the ratio between Joules and calories, which are two units of energy.
INTRODUCTION:
When an electrical potential difference exists between the two ends of a resistance coil,
current will flow through the coil. The energy dissipated by the current flowing through the coil
can be used to heat a known mass of water.
V
I
Heat lost by heating coil
Electrical energy provided
Power consumption, Pt
(IV) t
Energy in Joules
= Heat gained by water + calorimeter
= Heat gained by water + calorimeter
= Heat gained by water + calorimeter
= mwcwT + (mcccT + mLcLT)
= Energy in calories
APPARATUS:
Power supply
Electronic balance
Stop clock
Banana wires
Calorimeter and heating coil
Hewlett-Packard multimeter
Thermometer and glycerine
DigiTec multimeter
Ice
Dewar flask
Spade lugs
PROCEDURE:
1. Determine the mass of the inner calorimeter cup mc. Record the mass. Fill the cup ¾ full of
water and add ice to cool it to 10o below room temperature. Wait for all the ice to melt, then
weigh the cup and water. Determine the mass of water m w and record the results in Table 1.
V
2. Set up the apparatus as in Figure 1 to the
right. Use the Hewlett-Packard multimeter as
the ammeter, with the dial set at 10A, the leftside wire to the Fused terminal and the rightside wire to the Common terminal.
A
Heating coil of calorimeter
+
-
DC Power Supply
Fig. 1
- 11 -
3. Lubricate the stopper hole with glycerine. Insert the thermometer about half way into the
calorimeter. Mix the water with the stirrer built into the calorimeter lid assembly. Record the
water temperature, Ti, to a tenth of a degree.
4. Set the battery charger to 6V. Close the switch and start the timer simultaneously.
5. Record the voltage V and the current I, initially and for every five-degree increase in
temperature.
6. Allow the water temperature to increase by T = 15.0oC, while stirring occasionally to ensure
a uniform water temperature.
7. Open the switch, and stop the timer simultaneously. Record T f and the time t.
8. Discard the warm water in the calorimeter and repeat the procedure with the power supply
set to 12 V. Remember to weigh the calorimeter cup with and without the new sample of
water to ensure accuracy.
9. Calculate the instantaneous power P = IV at your four temperature readings, and average
them.
10. The electrical energy provided = average power X time. The heat gained by the water is its
mass X specific heat X temperature change. The specific heat of water is 1.00 cal/gmoC.
The aluminum cup also gains heat, and the specific heat of aluminum is 0.22. The lid
(including the heating coil and the stirrer) also gains heat, and the water equivalent (W.E.) in
grams is written on the lid. These items absorb heat as if they were water, with m L = W.E.
and cL = 1.00 cal/gmoC. Calculate the heat gained.
11. The theoretical value of the mechanical equivalent of heat is 4.186 J = 1.000 calories.
Calculate your values, and the percent error when compared to the theoretical value.
- 12 -
DATA AND CALCULATIONS:
Data Table 1:
Initial temperature: Ti= _________ oC, Final temperature: Tf = ___________ oC
mc = __________ gm,
Temperature oC
mw + mc = __________ gm,
Current Reading
(Amps)
Voltage Reading
(Volts)
mw = __________ gm
Power
(Watts)
Ti
Ti + 5.0 oC
Ti + 10.0 oC
Ti + 15.0 oC
Average Power: P = _______________ Watts,
Data Table 2:
Time: t = _______________ sec.
Initial temperature: Ti = _________ oC, Final temperature: Tf = ___________ oC
mc = __________ gm,
Temperature oC
mw + mc = __________ gm,
Current Reading
(Amps)
Voltage Reading
(Volts)
mw = __________ gm
Power
(Watts)
Ti
Ti + 5.0 oC
Ti + 10.0 oC
Ti + 15.0 oC
Average Power: P = _______________ Watts,
- 13 -
Time: t = _______________ sec.
CALCULATIONS:
Electrical energy provided = Heat gained by the water, calorimeter and lid
(IV) t = mwcwT + mcccDT + mLcLT
Trial # 1 _________ J = _________ calories
_________ J = 1.000 calorie
Trial # 2 _________ J = _________ calories
_________ J = 1.000 calorie
Data Table 3: Mechanical Equivalent of Heat
Trial
Experimental Value
(J/cal)
Theoretical Value
(J/cal)
1
2
- 14 -
% difference
Experiment 5: RESISTANCE & RESISTIVITY
PURPOSE: a) To determine the resistance R and the resistivity  of nichrome and copper wires.
b) To observe the effects of the physical dimensions of the wires and the material
characteristics on R and .
INTRODUCTION:
R: Ohm’s law states that V = IR, where
R is the resistance. The equation
R = V/I permits R to be determined
by measuring V and I.
L
A
The resistance of a metal wire is
proportional to its length L and
inversely proportional to its crosssectional area A. These factors
may be combined to give
I
V
L , where  is a proportionality constant called the resistivity.
R A
:
Resistivity is a property which depends only on the material composition of a wire.
For example:
Copper:  = 1.70 X 10-8 m
Nichrome:  = 1.13 X 10-6 m
APPARATUS:
Board with 1 and 2 meters wire mounting
Hewlett-Packard multimeter
Nichrome wire 26, 22 and 18 gauge
Wire gauge and micrometer
Copper wire, 26 gauge
- 15 -
Power supply 12 VDC
DigiTec multimeter
Leads and connectors
Rheostat, 22 
PROCEDURE:
1. Determine the diameter d and calculate the cross-sectional area A = d2/4 of the wires, and
record them in Table 1.
V
A
Switch
R
-
+
Rheostat
+
-
Vo = 12 VDC
Fig. 1
2. Set up the apparatus as in Figure 1, with the nichrome #18 wire as the resistor R. The
rheostat acts as a variable voltage source for the apparatus in the upper half of the figure, with
the wire from the switch connected to the tap. The current will have to pass through exactly
1.000 meters of wire when the wire is screwed down between the two terminals of the board.
The ammeter will be the Hewlett-Packard multimeter, with the dial set to 10A, the Fused
terminal as the positive side and the Common terminal as the negative side. The voltmeter
will be the DigiTec multimeter with the DCV- button pushed in and the dial set to 2V.
3. Set the current flow in the circuit to about 0.50 A by adjusting the rheostat.
4. Record the current through and the voltage drop across 1.000 m of the nichrome #18 wire
(Table 2). Calculate the resistance R = V/I and the resistivity  = RA/L.
5. Repeat steps 3 - 4 for nichrome #22, nichrome #26 and copper #26.
6. Repeat steps 3 - 5 using 2.000 m length of wire instead of 1.000 m. Record the data in
Table 3.
- 16 -
Wire Gauge
Place wire here
Back:
Diameter
In inches
Front:
Gauge #
1 inch = 2.54 cm
MICROMETER
5
Object to be measured
0
45
0 1 2 3 4
40
Each division is 1 mm
= 0. 001 m
Each division is 0.01 mm
= 0. 00001 m
An example of how to read the micrometer when making a measurement:
(
)
Read the line at the contact.
If there is no line there, read
line just before the contact.
the
0
4. 5
mm
45
4
+ 0. 4 8
mm
4. 9 8
mm
0
45
Read to the nearest hundredth of a millimeter.
- 17 -
48 div. X 0.01 mm/div. = 0.48 mm
DATA:
Data Table I
Gauge #
Wire
Diameter
(m)
Cross-sectional Area
(m2)
18
Nichrome
22
26
Copper
26
Data Table 2: Wire length of 1.000 meter
Wire
Gauge #
Current
(A)
Voltage
(V)
Resistance
()
Resistivity 
(m)
Average
Given
(m)
(m)
__________
__________
18
Nichrome
22
26
Copper
26
% difference for Nichrome ____________
% difference for Copper ____________
Data Table 3: Wire length of 2.000 meters
Wire
Gauge #
Current
(A)
Voltage
(V)
Resistance
()
Resistivity 
(m)
Average
Given
(m)
(m)
__________
__________
18
Nichrome
22
26
Copper
26
% difference for Nichrome ____________
% difference for Copper ____________
- 18 -
Experiment 6: THE WHEATSTONE BRIDGE
PURPOSE: To measure the resistance of a resistor by using a null (zeroing) method.
INTRODUCTION:
b
Rs
Rx
I1
G
RR
RL
a
d
c
I2
1.5 Volts
Fig. 1. Wheatstone Bridge
Suppose that four resistors are connected together as shown in Figure 1. Usually, a
different amount of current will flow through each resistor. However, suppose that a very sensitive
ammeter called a galvanometer is placed between points b and c, and the resistance of the lowerleft resistor RL and the lower-right resistor RR are varied until the galvanometer reads exactly
zero. The resistors Rs and Rx must both experience current I1, and resistors RL and RR must
both experience current I2, because no current passes from b to c.
Therefore,
Vab = Vac
and
Vbd = Vcd.
From Ohm’s Law,
I1Rs = I2RL
and
I1Rx = I2RR.
Rearranging,
I2 / I1 = Rs / RL
Equating these two equations gives
Rx = Rs
a
and
RR
RL
d
c
LL
LR
d
c
a
I2 / I1 = Rx / RR.
RR
RL
Fig. 2. Slide-Wire Apparatus
- 19 -
According to this equation, an unknown resistance Rx can be accurately measured if an
accurately-known standard resistance Rs is available, and if the ratio of resistances RR / RL can
be determined. A uniform wire 1.000 meters long can be tapped along its length to split the wire
into two resistors, as shown in Figure 2.
Since RR =  (LR / A) and RL=  (LL / A),
 L A 
Rx = Rs  R  , giving
  LL A 
Rx = Rs
LR
LL
Eq. 1
Notice that neither  nor A need to be known.
APPARATUS:
Galvanometer
Leads & connectors
22 rheostat
Power supply 1.5 VDC
Wheatstone bridge
Carbon resistor 47 
Decade resistance box
Standard resistor 50  or 100 
Slide-wire apparatus with tap
Spool of wire #30 copper
PROCEDURE:
1. Set up the apparatus as shown in Figure 1. Use the 47 carbon resistor (the tiny cylinder with
a wire sticking out of each end) as Rx. Both keys on the top of the galvanometer can be
pushed down and rotated into a depressed position. Try this, then rotate both keys into their
elevated positions.
2. Tap the key onto the wire, and watch the motion of the galvanometer needle. Slide the tap to
another position and tap again, to see if the deflection is smaller. Continue changing the tap
position until the galvanometer deflection is very small. Be careful not to drag the tap across
the wire, as you will scrape some metal from the wire and it will no longer be uniform.
3. Push down and rotate the left-side key of the galvanometer to lock it into a depressed position,
to increase the sensitivity. Locate the null position on the slide wire, which should be only a
few millimeters from the position you have already located. Rotate and elevate the
galvanometer key to its less-sensitive position.
4. Record LL and LR to the nearest millimeter. Calculate Rx from Equation 1, and record the
nominal resistance. The nominal (named) resistance is the resistance assigned to it by the
manufacturer. For the carbon resistor, the band colors yellow-violet-black-silver represent 4-70-10%, indicating a resistance of 47 X 100 = 47 with a typical scatter of 10% around this
value. Calculate the percent difference, assuming the nominal value is the theoretical value.
5. Repeat steps 2 - 4 with the other resistors. Set the decade box at 70. The nominal value of
the spool can be calculated from R = L/A, with copper = 1.72 X 10-8 m, A = 5.10 X 10-8 m2
for gauge #30 wire, and L = 20.000 meters.
- 20 -
DATA:
Resistor
Rs
()
LL
(m)
LR
Rx
Rx
From Eq. 1
Nominal Value
(m )
()
()
Carbon
Decade
Spool
Rheostat
- 21 -
% difference
Experiment 7: RC TIME CONSTANT ()
PURPOSE: To determine the time constant  of an RC circuit.
INTRODUCTION:
The time constant  = RC determines the rate at which a capacitor with capacitance C will be
charged or discharged through a resistor of resistance R. The circuit diagram in Figure 1 shows a
fully-discharged capacitor (Q = 0 on both plates) connected to an open switch. When the switch
is closed, the voltage across the resistor will initially be large as current flows through the circuit,
then gradually drops to zero as the capacitor plates become fully charged. The voltage across
the resistor should decrease according to VR = Voe-t/RC = Voe-t/ (Figure 2) when the charge Q on
the capacitance plate builds up exponentially.
VR
Qo
Vo
R
0.63 Qo
E
C
0.368 Vo
Switch
0
0

VC = Vo e-t/RC
Fig. 1

t (sec)
t (sec)
Q = Qoe-t/RC
Fig. 2. Charging a Capacitor
A circuit with a charged capacitor, as shown in Fig. 3 will behave as shown in Fig. 4 when the
switch is closed. The voltage across the resistor should decrease as the capacitor discharges,
according to VR = Voe-t/RC = Voe-t/.
VR
Vo
R
Qo
C
0.37 Qo
0.368 Vo
Switch
0
0

t (sec)
VC = Vo(1 - e-t/RC)

t (sec)
Q = Qo(1 - e-t/RC)
Fig. 3
Fig. 4. Discharging a Capacitor
APPARATUS:
Two 1 ½ -Volt batteries
Capacitor (10F)
Carbon resistor (20M
Double pole, double throw switch
Leads & connectors
Hewlett-Packard multimeter
- 22 -
French curve
Plastic triangle
Timer
PROCEDURE:
A. Charging a Capacitor
1. Construct the circuit shown in the middle of Figure 5. The two batteries in series are
equivalent to a single 3.0-Volt battery. Notice that when switch A is closed, the switch acts like
a 3.0-Volt battery that charges the capacitor. When switch B is closed, the switch acts like a
wire that discharges the capacitor.
2. Zero the timer, and momentarily touch the ends of a wire to the two terminals of the capacitor
to make sure it is completely discharged.
3. Start the timer and simultaneously close switch A. A couple of seconds later, once the
voltmeter has adjusted to the larger voltages, start the timer and at the same moment press
the Data Hold button on the voltmeter to keep the reading visible. Press the Data Hold button
again as soon as the reading is recorded. Continue taking readings every 10.0 seconds by
pressing the Data Hold button, then pressing it again in order to release it for the next reading.
B. Discharging a Capacitor
1. Zero the timer. Unplug both voltmeter wires from the resistor, then plug each wire back in, but
on the opposite side of the resistor. This reversal permits the voltage to remain positive, while
the current reverses direction.
2. Close switch B. A couple of seconds later, once the voltmeter has adjusted to the larger
voltages, start the timer and at the same time press the Data Hold button on the voltmeter to
keep the reading visible. Press the Data Hold button again as soon as the reading is
recorded. Continue taking readings every 10.0 seconds by pressing the Data Hold button,
then pressing it again.
Charging:
Discharging:
C
VR
R
3.0 Volts
C
VR
C
VR
R
....
..
1.5 Volts
R
Switch B
Switch A
1.5 Volts
Switch A closed
Switch B closed
Fig. 5.
- 23 -
C. Graphs and Calculations
1. Graph VR as a function of t for both the charging and discharging circuits. Use the French
curve to connect the data points as smoothly as possible. When t equals the time constant ,
VR = Voe-1 = 0.368 Vo, where Vo = VR at t = 0. Calculate VR = 0.368 Vo, find it on the graph,
and obtain a value of , for each graph.
2. If VR = Voe-t/, then ln(VR) = ln(Vo) – t / where ln is the natural logarithm function, to the base
e = 2.7183 A graph of ln(VR) vs. t will have a slope m = – 1/. Construct this graph for the
data obtained from the discharging capacitor. Draw the best-fit straight line through these
data points, calculate its slope, and calculate . This will be your best estimate of , as it uses
all the data obtained. This sort of graph is called semi-logarithmic, as one axis is a logarithm
of the data.
3. The theoretical value of the time constant is  = RC. If the voltmeter had infinite resistance,
the value of R would equal the resistance of the 20 M resistor. However, the voltmeter has a
resistance of 11M, so R is the equivalent resistance of these two resistors in parallel.
Calculate R, and calculate  = RC from this value of R and from the nominal value of the
capacitance.
- 24 -
DATA:
Charging a Capacitor
Time
(s)
VR
(Volts)
Discharging a Capacitor
Time
(s)
0.0
0.0
10.0
10.0
20.0
20.0
30.0
30.0
40.0
40.0
50.0
50.0
60.0
60.0
70.0
70.0
80.0
80.0
90.0
90.0
100.0
100.0
110.0
110.0
120.0
120.0
130.0
130.0
140.0
140.0
150.0
150.0
160.0
160.0
170.0
170.0
180.0
180.0
 from graph of charging capacitor:________________ s.
 from graph of discharging capacitor:________________ s.
 from semi-logarithmic graph:________________ s.
Equivalent resistance R:_______________ 
 from nominal values of R and C:________________ s.
- 25 -
VR
(Volts)
ln (VR)
Experiment 8: SERIES -- PARALLEL CIRCUITS
PURPOSE:
To learn to wire simple series and parallel circuits on a breadboard and to verify the rules
pertaining to these circuits.
INTRODUCTION:
Breadboards allow an experimenter to build and modify circuits very easily, without
soldering. The Protoboard.10 that you will be using contains sockets that are connected together
in the following arrangement:
G
GLOBAL SPECIALTIES
proto-board.10
V1
V2
GND
Each vertical line represents a set of five sockets that are permanently wired together
under the breadboard. Each of the eight horizontal lines represents 25 sockets wired together.
Banana wires can be attached to the three terminals in the upper-right of the breadboard, and
short wires can be run from the base of these terminals to the horizontal sockets to provide
sources of voltage and a ground. Short wires can then connect different sets of sockets together,
with circuit devices such as resistors and integrated circuits plugged in as well.
SERIES CIRCUIT GEOMETRY AND EQUATIONS:
R1
R2
I1
R3
I2
Io = I 1 = I 2 = I 3
I3
Vo = V 1 + V 2 + V 3
Req = R1 + R2 + R3
Io
Io
Vo = Io·Req
Vo
Fig. 1a
- 26 -
PARALLEL CIRCUIT GEOMETRY AND EQUATIONS:
Io = I 1 + I 2 + I 3
I1
R1
I2
R2
I3
R3
Io
Vo = V 1 = V 2 = V 3
Req =
Io
1
1
R1

1
R2

1
R3
Vo = Io·Req
Vo
Fig. 1b.
APPARATUS:
Hewlett-Packard multimeter
Carbon resistors (1000, 470, 270)
BK Precision multimeter
Breadboard (protoboard.10)
Leads and connectors
Power Supply 6 VDC
SPST switch
PROCEDURE:
1. Set the DigiTec multimeter to function as an ohmmeter, plug it into the V1 and GND terminals,
insert the exposed end of each of two short wires through the hole at the base of the two
terminals and screw them tightly in place. By inserting these two short wires into various
sockets, convince yourself that the sockets are connected to each other as described in the
Introduction. Unplug the ohmmeter.
Resistor Color Code
2. Determine
the
nominal
resistance of each carbon
resistor by using the color
code bands. For example,
yellow-violet-brown-silver
becomes 4-7-1-10%, so the
resistance of the resistor is
47 X 101 = 470 ± 47.
Tens Digit
Ones Digit
Exponent
Tolerance
Black = 0
Brown = 1
Red = 2
Orange = 3
Yellow = 4
Green = 5
Blue = 6
Violet = 7
Grey = 8
White = 9
Silver 10% tolerance
Gold 5% tolerance
3. Before you plug into the DC power supply, set up the circuit as in Figure 2a with the switch
open. Then plug into the DC power supply and close the switch. Measure the current through
and the voltage across the resistors with the multimeters. Use alligator clip adaptors over the
banana plugs of the patch cords.
4. Repeat step 3 for the other seven circuits.
5. Calculate R = V/I, from Ohm’s law. Disconnect the breadboard from the power supply, and set
one of the multimeters to act as an ohmmeter. Measure the resistance of each resistor and of
the two combinations directly with the ohmmeter, and calculate the percent difference of this
reading from the value calculated from Ohm’s law.
- 27 -
I1
A
I2
R1
V
R2
R3
A
R1
R2
V
Fig. 2a
Fig. 2b
I3
R1
A
R2
R3
V
R3
V
R1
R2
R3
Io
A
Fig. 2d
Fig. 2c
I1
A
R1
I2
V
R2
A
V
Fig. 3a
Fig. 3b
V
I3
A
R3
V
Io
A
Fig. 3d
Fig. 3c
- 28 -
DATA:
Data and Calculations Table 1: Series Circuit
Fig.
Across
2a
R1
2b
R2
2c
R3
2d
Req
Nominal
Resistance
()
Voltage
(Volts)
Current
(Amperes)
R = V/I
()
Resistance
From
Ohmmeter
()
R = V/I
()
Resistance
From
Ohmmeter
()
% difference
Data and Calculations Table 2: Parallel Circuit
Fig.
Across
3a
R1
3b
R2
3c
R3
3d
Req
Nominal
Resistance
()
Voltage
(Volts)
Current
(Amperes)
- 29 -
% difference
Experiment 9: CIRCUIT ANALYSIS
PURPOSE: To analyze the current and voltage distribution in an electrical circuit.
INTRODUCTION:
The rules governing the relationship between current, voltage and resistance for series and
parallel circuits are outlined below:
Kirchoff's Loop Rule:
R1
Kirchoff's Junction Rule:
R2
I1
R3
I2
b
I3
I1
R1
I2
R2
I3
R3
Io
Vo
Io
Vo
a
n
n
 Vi  0
 Ii  0
+ Vi = Voltage rise
- Vi = Voltage drop
+ Ii = Current into junction
- Ii = Current out of junction
i
i
Loop Equation:
Junction Equation:
Clockwise from a: +Vo - V1 - V2 - V3 = 0
Junction b: +Io - I1 - I2 - I3 = 0
So:
So:
Vo = V1 + V2 + V3, and
Vo = I1R1 + I2R2 + I3R3
Io = I 1 + I 2 + I 3
APPARATUS:
Power supply 3V
Breadboard
Battery, 1½V
Leads and connectors
Carbon resistors, 100, 330, 470, 1000
Hewlett-Packard multimeter
Single pole, single throw switch
- 30 -