Download Experiment EM-4S for Physics 105

Experiment EM-4S
1
Ohm's Law
Purpose
The purpose of this experiment is to familiarize the student with Ohm's law and with the proper techniques for
using two basic electrical measuring instruments -- the current meter and the voltmeter.
Note:
BEFORE COMING TO THE LABORATORY TO PERFORM THIS EXPERIMENT:
In addition to reading this instruction, study the separate set of notes on "Interpretation of Schematics." Unless
directed otherwise by your laboratory instructor, have the exercise sheet from this set of notes filled out and ready to
turn in at the beginning of the laboratory period in which you are scheduled to perform EM-4S.
Community College of Philadelphia
Physics Department
Student Name:
__________________________________________________________________
Partners: __________________________________________________________________
__________________________________________________________________
__________________________________________________________________
Physics 105
Section #: ____________________
Date performed: ____________________
Date due: ____________________
Experiment EM-4S
2
Theory
A flow of charged particles is called an electric current. An electric current in a metallic conductor consists of
moving electrons, which are subatomic particles with a negative charge. Nevertheless, we still often speak of a
conventional current of fictitious moving positive charges. The conventional current is assumed to flow in a
direction opposite to the actual electron current. The magnitude of the current is the amount of charge passing a
point in a conductor per unit time. For instance, if two coulombs of charge pass a point in a wire each second, the
current is said to be two amperes. Since each electron has a very small charge, a two ampere current corresponds to
a very large number of electrons passing a point in the wire each second (1.25 x 10 19 electrons per second).
In order for a current to flow in a conductor, a potential difference (or voltage) must exist between the ends of
the conductor. In 1826 G. S. Ohm found (for many conductors) that the current through the conductor is directly
proportional to the potential difference across the ends of the conductor. This relationship is known as Ohm's law
and can be expressed mathematically as follows:
I 
V
R
or
R 
V
I
or
V
=
IR
(l)
I is the current in amperes and V is the potential difference in volts. R is a constant of proportionality known as the
resistance and is measured in units of ohms. Heavy copper wires are often used when a low-resistance conductor is
desired. When certain higher resistances are needed, a device known as a resistor may be used. Some types of
resistors are described later in this paper. Commercial resistors usually obey Ohm's law quite accurately when used
within their power ratings.
The electrical potential difference (V) between two points is equal to the work per unit charge required to move
a positive charge from the point of lower potential to the point of higher potential. For instance, if 100 joules of
work must be done to move one coulomb of charge from one point to another, the potential difference between the
two points is 100 volts. When a potential difference exists between two points in a conductor, the electrostatic
energy associated with the separation of charge becomes transformed into kinetic energy of the charges and results in
a motion of charge that tends to neutralize the potential difference. To maintain a continuous flow of charge, the
electrostatic energy must be continuously replenished by some source of energy.
One familiar source of electric energy is an electric battery. Batteries normally maintain a nearly constant
potential difference between their terminals when small currents are drawn from them. But the battery voltage may
drop somewhat as the current approaches the maximum value recommended by the battery manufacturer. Figure 1a
is a schematic diagram of a battery connected by a pair of wires to an electrical resistor. If wires of a large enough
diameter are used and if they are made of a good conductor, there is a negligible potential difference along each
wire. The voltage across the resistor is virtually equal to that across the battery. According to Ohm's law, if the
voltage is doubled, the current will be doubled -- i.e., the number of coulombs of charge per second flowing around
the circuit will be doubled.
Figure 1b is an interesting fluid-flow analogy of the electric circuit just described. A water pump pumps water
around the loop of pipe, just as the battery pumps charge around the electrical circuit. The rate of water flow in
ft3/sec is analogous to the current in amperes (equal to the rate of charge flow in coulombs/second). The pressure
difference in lb/ft2 is analogous to the potential difference in volts. There is a negligible drop in water pressure
along each of the short, fat pipes; just as there was a negligible potential drop along each of the heavy copper wires.
Almost the entire pressure drop occurs in the long-narrow pipe, just as most of the drop in electrical potential
occurred in the resistor. If the pressure difference is doubled at the pump, the rate of water flow will (in some cases)
double -- in a manner analogous to Ohm's law for the electrical circuit.
Not all substances capable of carrying an electric current obey Ohm's law. In some cases where certain materials
of a different nature come in contact, the device may carry current more easily in one direction than another. This in
itself is not in accordance with Ohm's law, which suggests that a reversal in the sign of V merely causes a reversal in
the sign of I. Furthermore, the current is not proportional to the voltage for either direction of current flow. Ionized
gases can conduct an electrical current but do not obey Ohm's law. Ohm's law assumes a constant temperature. If
the temperature changes by a large amount in a conductor as the voltage is changed, the current may not be
proportional to the voltage.
Figure 1
Experiment EM-4S
a.
3
Battery and Resistor
b.
Water-Flow Analogy
3
High
Pressure
ft
2
sec
2 ampere current
Battery
+
-
lb
100
100 volt
Potential
Difference
ft 2
Pressure
Difference
50 ohm
Resistor
Water
Pump
2 A current
(2 coulombs per sec)
Low Pres.
Long
Narrow
Pipe
Short Fat Pipe
An electrical resistor can be constructed using a long narrow wire (analogous to the long, narrow pipe of Figure
1b). The wire can be wound on a small cylinder of insulating material. The ends of the resistance wire are attached
to terminals or heavy wires anchored to the ends of the insulating cylinder. Various alloys are used for the resistance
wire, all of them being poorer conductors than copper. Typically, the entire resistor assembly, excepting the
terminals, is covered with an insulating and protective coating of vitreous enamel. In addition to resistance, resistors
of this design have a property known as inductance, which may be objectionable in some applications. Some other
types of resistors are constructed of non-metallic materials such as carbon, which are poor enough conductors to
provide a resistance suitable for many applications when fashioned into small cylinders. Typically, wire leads are
attached to the ends of the cylindrical resistance element, and the entire assembly is encapsulated in plastic. In other
resistors the resistance path may be a helical stripe of metallic or non-metallic film on a ceramic cylinder. Various
types of variable resistors are also available.
In this experiment you will measure the resistance of certain resistors and devices, using the circuit of Figure 2.
You will measure the current through the resistor and the potential difference across the resistor, using an ammeter
and voltmeter. You will then calculate the resistance, using equation (1). In some of your runs you will determine
the resistance of the device twice at two different voltages. If the resistance appears nearly the same at the two
voltages, you will have verified (at least partially) that the device obeys Ohm's law.
Figure 2
Ammeter-Voltmeter Method of Measuring Resistance
Ammeter
I
Voltaic
Cell
Resistor
R
Voltmeter
V
Experiment EM-4S
You will be using digital multimeters to measure both current and voltage with an accuracy of a few tenths of a
percent. A label on the bottom of the meter specifies the accuracy on various ranges. Some higher priced digital
meters can measure voltages and currents with an accuracy as good as  .01 %. Other digital multimeters with an
accuracy of about 1 % are as inexpensive as analog meters with a needle and scale. For this reason analog meters are
rarely used now for measurements of voltage, current, and resistance. Although it is possible to measure resistance
with the same digital multimeters, you will calculate resistance from the voltage and current readings. If you
calculate the same resistance for a given resistor at different applied voltages you are confirming Ohm's law.
4
Apparatus
D Cells, alkaline, 1.5 V nominal, on board with terminals (2) (A rectangular 1.5 V battery may be substituted.)
Digital multimeters (2)
Ohm's Law board with 30 ohm resistor, 50 ohm resistor, and a # 1490 lamp
Assorted test leads
Experiment EM-4S
5
Procedure, Data, and Calculations
General Notes: You will use wires with a spade terminal on one end and a banana plug on the other to connect the
multimeter used as a current meter into the circuit. You can use either banana-to-spade leads or banana-tobanana leads for connecting the multimeter serving as a voltmeter. Other connections should be made with
spade-to-spade leads. On the multimeter used as the current meter, the jack labeled "COMMON" is used as the
negative terminal, and the jack labeled "mA" is used as the positive terminal. On the multimeter used as the
voltmeter the jack labeled "COMMON" is used as the negative terminal, and the jack labeled "V/k/S" is used
as the positive terminal. The labels are red for positive and black for negative. Likewise the positive terminals
on the voltaic cell boards are red, and the negative terminals are black. Possibly, rectangular 1.5 V batteries
may be substituted for the D cells on mounting boards. The terminals on the 1.5 V batteries are marked + and -.
1.
On the multimeter being used as a voltmeter, push in the buttons marked V and 2 for the 2 volt range. Set the
AC-DC button in the out position for dc. On the multimeter being used as a current meter, push in the buttons
marked mA and 200 for the 200 mA range. (1mA = 0.001 A) You will be converting currents in milliamperes
to amperes before recording them on the data table. Again set the AC-DC button in the out position for dc.
Plug the power cords of the meters into 120 V outlets. Hereafter, the two meters will be called simply the
"voltmeter" and the "current meter".
2.
Connect resistor R1 to a single voltaic cell, the current meter, and the voltmeter in accordance with the
schematic diagram below. A pictorial diagram provides additional details if you need them, but please try to
learn how to read these simple schematics as you go along. (R1 has a nominal value of 30 ohms. But this value
is subject to a manufacturer's tolerance of a few percent and should not be used in any future calculations.)
1.5 V


I


R

30 
(nom.)
Multimeter as Current Meter
R1
V

Voltaic
Cell
red
+
mA C V
blk
3.
R2
Multimeter as Voltmeter
mA C V
-
The meters should still be set up as specified in step 1. If the power switches on the meters are not on, depress
the buttons to turn them on. Read the current meter in mA (milliamperes) and convert to amperes by moving
the decimal point 3 places to the left. Then read the voltmeter.
I = ____________________ mA = ____________________ A
V = ____________________ V
Experiment EM-4S
6
4.
Calculate the resistance of R1.
R 
Resistance measured with one cell:
V

(Substitute)
I
= ____________________ ohms
5.
(Answer)
Modify the circuit by inserting a second voltaic cell in series with the first -- in accordance with the schematic
diagram below. Press the 20 button on the voltmeter to change the range to 20 V. The current meter should still
be set to its 200 mA range.
1.5 V



1.5 V
red
+
blk
-
red
+
blk
-

I


R

30 
V

Voltaic Cell
R1
Multimeter as Current Meter
mA C V
R2
Multimeter as Voltmeter
mA C V
Voltaic Cell
6.
Read the current in milliamperes and convert the reading to amperes by moving the decimal point three places to
the left. Then read the voltmeter.
I = ____________________ mA = ____________________ A
V = ____________________ V
7.
Again calculate the resistance of R1.
Resistance measured with two cells:
R 
V

I
= ____________________ ohms
(Substitute)
(Answer)
Experiment EM-4S
8.
7
Calculate the average of your two values of R1. Then calculate the percent difference between the two values.
Average R1 
R with one cell + R with 2 cells
2
=
Average R1
=
% Difference 
(Substitute)
____________________ ohms
R with 2 cells
-
R with 1 cell
(Answer)
 100
Average R1
% Difference
9.
=
(Substitute)
= ___________________ %
(Answer)
Transfer your measured and calculated results from steps 3 through 8 to the data table on page 10.
10. Press the 2 button on the voltmeter to set the range to 2 V. (The 2 V range of the voltmeter should be used for
measuring voltages between 0.2 V and 1.999 V.) Repeat steps 2, 3, and 4, using resistor R 2 instead of R1. (This
resistor has a nominal resistance of 50 ohms. But this nominal value has a tolerance of a few percent and should
not be used in any future calculations.) A small schematic is included on the data table. Essentially, you are
measuring R2 with the aid of one voltaic cell, a current meter, and a voltmeter. Make your calculations on
scratch paper; record your measured and calculated results on the data table.
11. Repeat steps 5 through 9 using resistor R2 instead of R1. A small schematic is included on the data table.
Essentially, you are measuring R2 again with a higher voltage obtained from two voltaic cells in series. Make
your calculations on scratch paper; record your measured and calculated results on the data table. Be sure to
use the 20 V range of the voltmeter when reading a voltage between 2 V and 19.99 V.
12. Press the 2 button on the voltmeter to set the range to 2 V. Repeat steps 2 through 4, using the lamp in place of
the resistor. A small schematic is included on the data table. Essentially, you are measuring the resistance of the
lamp filament with the aid of a voltmeter, a current meter, and one voltaic cell. Make your calculations on
scratch paper; record your measured and calculated results on the data table. Be sure to use the appropriate
range of the voltmeter.
13. Repeat steps 5 through 9, using the lamp in place of the resistor. A small schematic is included on the data
table. Essentially you are measuring the resistance of the lamp filament again at a higher voltage obtained from
two voltaic cells in series. You then compare your two values of lamp resistance. (There will be a much larger
percent difference for the lamp than for the resistors. Later you will be asked to explain this large percent
difference.) Make your calculations on scratch paper; record your measured and calculated results on the data
table. Be sure to use the appropriate range of the voltmeter.
Experiment EM-4S
8
14. Sometimes, two resistors are connected in parallel and used as a single resistor. See the small schematic
immediately below. Use the equation below and calculate the predicted resistance of the parallel combination of
R1 and R2. Use the Average values of each from your previous measurements.
R1
R2
R ppred
Rppred

R1 R 2
R1  R 2
=
(Substitute)
= ____________________ ohms
(Final answer)
15. Connect R1 and R2 in parallel. Then connect the parallel combination to the current meter, the voltmeter, and a
single voltaic cell -- in accordance with the small schematic on the data table. The voltmeter should be on the 2
V range.
16. Read the current I on the current meter and convert the reading to amperes. Then read the potential difference V
on the other meter. Finally, determine the measured value of the parallel resistance.
I = ____________________ mA = __________________ A
R pmeas

V
V = __________________ V
=
(Substitute)
= ___________________ ohms
(Final answer)
I
17. Calculate the percent difference between the measured and predicted values of the resistance of the parallel
combination of R1 and R2.
% difference

R ppred  R pmeas
R ppred  R pmeas
 200
=
(Substitute)
= ___________________ %
(Final answer)
18. Transfer the measured and calculated results of steps 14 through 17 to the data table.
Experiment EM-4S
9
19. Sometimes, two resistors are connected in series and used as a single resistor. See the small schematic
immediately below. Use the formula below and calculate the predicted resistance of the series combination of
R1 and R2. Use the average value of each from your previous measurements.
R2
R1
Rspred
=
R1 + R2
=
=
___________________ ohms
(Substitute)
(Final answer)
20. Connect R1 and R2 in series. Then connect this series combination to the current meter, the voltmeter, and the
two voltaic cells -- in accordance with the small schematic on the data table. Set the range of the voltmeter to 20
V by pressing the 20 button.
21. Read the current I on the current meter and convert the reading to amperes. Then read the potential difference V
on the other meter. Finally, determine the measured value of the resistance of the series combination of
resistors.
I = ____________________ mA = __________________ A
R smeas

V
V = __________________ V
=
(Substitute)
= ___________________ ohms
(Final answer)
I
22. Calculate the percent difference between your measured and predicted values of the resistance of the series
combination of R1 and R2. The calculation is similar to that in step 17; but do your calculations on scratch
paper this time. Transfer your measured and calculated results from steps 19 through 22 to the data table.
Experiment EM-4S
10
Data Table
Schematic Diagram
Step #
(Values are nominal.)
1.5 V
2-4



1.5 V
1.5 V

1.5 V
10

I




I

1.5 V

12

I


1.5 V
R2
50 


11
R1
30 

1.5 V
R1
30 


5-9

R2
50 

I


13
1.5 V



I


1.5 V




R2
% Dif.
I
amperes
Rsmeas
ohms
Rspred
ohms
% Dif.
V
volts






R1
R1
Rppred
ohms


R2
I
Rpmeas
ohms
V
volts
%

V

19-22

I
amperes
amperes

V
Lamp


1.5 V

volts

V
V
Step #
1.5 V
I
R

V
V

I

V
Lamp

1.5 V
14-18
I

% Dif.
ohms
Average
R
ohms
V


V

Experiment EM-4S
11
Questions
1.
Suppose that you measure the resistance of a resistor using a voltmeter, a current meter, and a battery. Suppose
that the voltmeter reads 0.2 % high and the current meter reads 0.3 % low. What will be the percent error in
your measurement of resistance. (Hint: If you do not know how to work this problem with abstract
mathematics, try making up a numerical example that meets the given conditions.)
2.
Why did you have such a large percent difference in the resistance of the lamp filament with the two different
applied voltages? What physical condition of a conductor must remain constant for Ohm's law to apply?