Chapter XIX: Measurement of Resistance

=EXPERIMENT 125. To study the construction and use of a simple
"Wheatstone's Bridge."=

_Apparatus._ Fig. 102. A Wheatstone's bridge, W B (No. 80), (§
324); astatic galvanoscope, A G (No. 59); dry cell, D C (No.
51); key, K (No. 55); 7 wires with spring connectors, two of
which, R and X, are equal in length; metal plate, M P, for
connecting wires.

_Arrange_ as in Fig. 102. The carbon of D C is joined to K, and
this to the point, C, of the bridge. The zinc of D C connects
with the point Z on W B. The A G is placed between the branches
for clearness. Wire 3 is joined to the left-hand binding-post
of A G, and wire 4 joins M P with the right-hand one. When the
end of wire 3 does not touch G-s W, it is evident that as soon
as K is pressed, the current divides at C on its way to Z,
where the branches unite again. K is used so that D C will not
be polarized by steady use.

=324. The Simple Wheatstone's Bridge= (Fig. 103) consists of a
wooden base, W, at the ends of which are fastened two aluminum
conductors, 1 and 3. At one side of W is fastened another
conductor, 2. In Fig. 104 are side views of the conductors.
These are used merely for convenience in making connections,
and take the place of the metal plates used in previous
experiments. A German-silver wire, G-s W, is stretched between
1 and 3, and under this is a scale, S, divided into 100 small
parts, these being tenths of the larger divisions. The ends of
G-s W are held between eyelets, as shown at E, Fig. 104.

=Reading the Scale.= The value of part A can be read
directly from the scale, using the lower row of figures.
The point marked P, for example, would be read 3.7 (three
and seven-tenths large divisions); B would be 6.3, found by
subtracting 3.7 from 10. The sum of A and B must always equal
10. The 6.3 may also be read directly by using the upper row of
figures for the whole numbers, counting the tenths to the left.
Try to divide the smallest divisions into halves, at least;
that is, if A = 3.75, B = 6.25. Take the readings carefully.

=325. Directions.= (A) Touch the free end of wire, 3, to the
point, C, which has a higher potential than M P. Press down K
for an instant only. Some current should pass through A G, as
a shunt. Should it pass from C to M P or the reverse? Note in
which direction the right-hand end of the astatic needle is
deflected.

(B) Swing the end of 3 around and touch it to the point, Z,
which has a lower potential than M P. Press K for an instant,
watch the needle, and compare with the results in (A).

(C) Move the free end of 3 along on G-s W, touching K at
intervals, until a point is found at which the needle of A G
is not deflected. How does the potential of this point compare
with that of M P?

_=326. Discussion; Equipotential Points.=_ Since one end of the G-s W has a higher, and its other end has a lower potential than M P, there must be, somewhere on it, a point at which the potential is the same as at M P. This place is quickly found by sliding the free end of wire, 3, along, pressing K occasionally, until A G shows that no current tends to pass through it in either direction, when the current passes from C to Z through the two branches of the divided circuit. This point and M P are called _equipotential points_.

If the resistance of the part, X, be increased, it should be evident that the part of the bridge-wire, B, should be also increased to find a point having the same potential as M P; that is, the end of 3 should be moved towards C.

We have, in the bridge-wire, a simple means of varying the resistance of its parts, A and B.

=327. Use of Wheatstone's Bridge.= It will be found, upon
trial, if we put a resistance of 2 ohms in place of R, Fig.
102, and 2 ohms in place of X, that the free end of wire 3
will have to be at the center of the bridge-wire in order
to get a "balance"; that is, to find the place where A G is
not affected. No matter what the resistance of R and X are,
provided they are equal, this will be true. The value of both
A and B, on the scale, will be 5 whole spaces, no tenths. From
this we see that A: B:: R: X, which reads A _is to_ B _as_ R
_is to_ X; this means that A × X = B × R. Supplying the values
of the letters, we have 5 × 2 = 5 × 2. If we did not know the
value of X, that is, if we were measuring the resistance of
a coil of wire, using a 2-ohm coil as the standard, or R, we
could find the value of X, knowing the other 3 parts of the
proportion. 5 × X = 5 × 2, which means that 5 times the value
of X is 10; hence the value of X is 10 ÷ 5 = 2 ohms.

Suppose that we have R = 2 ohms, which is the standard
resistance coil (No. 79), and are trying to find the resistance
of a coil, X. We slide the end of wire, 3, along on the
bridge-wire until the correct place is found. (See Exp. 125,
126, for details.) Take the values of A and B (§ 324), supply
them in the equation given, and work out the value of X.

=328. EXAMPLE.= R = 2 ohms; A = 3.7; B = 6.3; to find the value
of X in ohms.

A: B:: R: X, which means that A × X = B × R, or 3.7 × X = 6.3 ×
2. X must equal, then (6.3 × 2) ÷ 3.7 = 3.405 ohms.

=Note.= In practice it is most convenient to make connections
as shown in Fig. 105 when measuring resistances (Exp. 126). The
arrangement given in Fig. 102 is simply for explanation. It
will be seen that the smaller A is, compared with B, the larger
the unknown resistance compared with your standard.

=EXPERIMENT 126. To measure the resistance of a wire by means
of Wheatstone's Bridge; the "bridge method."=

_Apparatus._ Same as in Exp. 125; the two-ohm resistance coil,
R C (No. 79); a coil of wire, X, as, for example, the 15-turn
coil on the galvanoscope, G V (No. 58).

_Arrange_ as in Fig. 105. You will observe that the central
conductor of the bridge (2, Fig. 104) takes the place of M P in
previous explanations. We still have the same kind of a divided
circuit as explained in Exp. 125, A G being connected with
points of equal potential. It will be found convenient to have
D C at the right, and A G facing you at the left, the key being
in front. (See Exp. 107 in regard to adjusting A G.)

Notice that you have a standard resistance (2 ohms) in place of
R, Fig. 102, and an unknown resistance (galvanoscope coil) in
place of X. (See § 330.)

=329. Directions.= (A) Touch the free end of wire, 3, to
the left-hand side of the bridge-wire, press the key for an
instant, only, and note the direction taken by the right-hand
end of the needle. Move the end of wire, 3, to the right-hand
side of the bridge-wire, touch key, watching needle. Does the
needle move more or less than before? In the same or opposite
direction? If the deflections are opposite, the point that has
the same potential as binding-post, 2, must be _between_ the
two points touched.

(B) Be sure that all connections are good. Find the point on
G-s W, at which there is no deflection, as directed in Exp. 125
(C). Note the readings on the scale, as explained in § 324.

(C) Make the proper calculation, § 327, 328, and find the
resistance of the coil of G V, the resistances of the wires
joining R C and G V to the bridge being neglected.

(D) Make proper allowances for the resistances of the wires
just mentioned (see § 330), and compare them with the results
found in part (C).

=330. Allowances for connections.= It should be remembered that
the wires joining R C and G V to the bridge also have some
resistance. Such connections, in regular instruments, are made
by heavy copper straps or by thick, short wires, so that their
resistances can be neglected. In case you use the ordinary No.
24 copper wire, as directed, the resistances of the pieces
can be measured by means of the bridge, or you can calculate
their resistances from the wire tables. The resistances should
be allowed for. It is evident that your standard resistance
is 2 ohms _plus_ the resistance of the connecting wires, and
that the resistance of the coil, X, is found by deducting the
resistance of its connecting wires from that found from the
proportion previously used.

_Example._ We see from the table that the resistance of about
39 ft. 1 in. of No. 24, B and S copper wire is 1 ohm. This
equals 469 in. If 469 in. have a R (resistance) of 1 ohm, 1 in.
will have a R of one-469th of an ohm; that is 1 divided by 469,
which equals a little over .002 ohm. For every inch of No. 24
wire used, then, for connections, we may allow .002 ohm. This
will be near enough for our purposes.

Suppose that each connection is 18 in. long, the regular wires
with connectors being used. The R of the 36 in. joined to R
C will then be 36 times .002 = .072 ohm. Our standard R must
then be considered as 2.072 ohm. If we substitute this in the
example, as stated in § 328, we have 3.7 × X = 6.3 × 2.072. X
must equal (6.3 × 2.072)/3.7 = 3.528 ohm, which includes the
unknown resistance and 36 in. of connections, the R of which
is .002 ohm; 3.528 - .072 = 3.456, the resistance of X alone.
Compare this with the answer to example, § 328. Make allowances
according to length of connectors used.

_=Note.=_--Carefully keep all the results of these experiments in a note book for future reference. Be sure that connections are good.

=EXPERIMENTS 127-137. To measure the resistances of various
wires, coils, etc., by the "bridge method."=

_Apparatus._ The coils of wire, etc., as stated in the
"Directions" of each experiment. The details of each piece of
apparatus may be found by referring, from the numbers given,
to the "Apparatus List," and to descriptions in the paragraphs
mentioned. Also all the apparatus of Exp. 126.

=Note.= Make proper allowances for connections (§ 330) in all
experiments in measuring resistances.

=EXPERIMENT 127.=

=331. Directions.= (A) As explained in Exp. 126, measure the
resistance of the 10-turn coil of G V, allowing for connections
(§ 330). Read the bridge-scale carefully.

(B) Use one-half of the 2-ohm coil as standard and repeat.

=EXPERIMENT 128.=

=332. Directions.= (A) Measure the resistance of the 5-turn
coil of G V (see Exp. 126, etc.), using 2 ohms as standard.

(B) Use 1 ohm as standard, repeat, and compare results.

(C) Add the resistances of the 5 and 10-turn coils, and compare
the sum with the resistance of the 15-turn coil, as found in
Exp. 126, D. The difference should be but a few hundredths of
an ohm.

=EXPERIMENT 129.=

=333. Directions.= (A) Measure the resistance of the coil
of No. 24 copper wire (No. 89). This coil is used for later
experiments. Spring connectors are fastened to the ends of this
coil, allowing it to be directly connected to the conductor on
the bridge, so no allowance should be made for its connecting
wires. (See Exp. 126 for details.) Mark the resistance upon the
coil for future use. (See Note.)

=Note.= The student will be surprised, perhaps, to find that
different results are obtained for the resistance of a given
wire in case he uses different standard resistances in the
various tests; that is, he will probably get a different result
in Exp. 127 (A) from the result of Exp. 127 (B). The difference
here, however, may not be large. The best results are obtained
by making the standard resistance as nearly equal as possible
to the resistance to be measured, so that a balance can be
found when the end of wire 3 (Fig. 105) is near the center of
the bridge-wire. If R, Fig. 105, is much larger or smaller than
X, the point desired on G-s W will be near one of its ends, and
large errors thereby produced. The approximate resistance of
X can be found by trial, then more or less resistance can be
used for R to suit. The student should make several coils as
explained in Apparatus Book, Chapter XVII. The resistance of
the different coils furnished should be measured and marked.
These can be used to vary the value of R.

=EXPERIMENT 130.=

=334. Directions.= (A) Measure the resistance of the coil of
No. 25 copper wire (No. 90). (See Exp. 126 for details and the
Note, Exp. 129.)

=EXPERIMENT 131.=

=335. Directions.= (A) Measure the combined resistance of the
two coils used in Exps. 129 and 130, when they are joined in
"series"; that is, when one end of one coil is joined to one
end of the other by means of a metal plate, the free ends being
connected to the bridge (Exp. 126). The current has to travel
through the entire length of both coils.

(B) Compare this result with the sum of their separate
resistances found in Exps. 129 and 130. (See Exp. 129, Note.)

=EXPERIMENT 132.=

=336. Directions.= (A) Measure the resistance of the two coils
(Exp. 131) when they are joined "in parallel." (See § 293.)
They may be joined in parallel by connecting them both to the
bridge at the same time, one end of each being slipped onto 2
(Fig. 103), the other end of each being joined to 3. In this
way the current has two paths, side-by-side, to get from 2 to
3. (See Exp. 129, Note.)

(B) Compare this resistance with that of Exp. 131.

=EXPERIMENT 133.=

=337. Directions.= (A) Measure the resistance of 1 meter of No.
28 German-silver wire. Use the wire as arranged on a board,
Exp. 122 (Figs. 96 and 98), making the connections with the
bridge from binding-posts, X and Z. (See Exp. 129, Note.) The
wires connecting the bridge with the ends of the G-s wire will
each have to be about 2 ft. long. In making deductions (§ 330)
figure according to the length used.

(B) Divide the total resistance by 100 to get the resistance of
1 cm. of the wire, and carefully mark off the board into cm.
This will give 100 parts between X and Z.

=EXPERIMENT 134.=

=338. Directions.= (A) Using the No. 28 G-s wire on the board,
as arranged for Exp. 122, measure the resistance of the 2
meters in series, the connections being made with the bridge
from X and Y, Fig. 98.

(B) Compare the result with that of Exp. 133. What is the
relation between the length of a wire and its resistance? See
Summary of Laws. (See Exp. 129, Note.)

=EXPERIMENT 135.=

=339. Directions.= (A) Measure the resistance of the above two
meters of No. 28 G-s wire when joined in parallel. (§ 293.)
The binding-posts, X and Y, can be joined by a short wire with
connectors on its ends, or by clamping a thin strip across by
means of spring connectors. Use the 2-ohm coil as the standard,
and make proper allowances. (§ 330.)

(B) From the results of Exps. 132 and 135 what can be said
about the resistances of parallel circuits as compared with the
resistances of the separate branches?

=EXPERIMENT 136.=

=340. Directions.= (A) Arrange the 2 meters of No. 30 G-s wire
on the table or board, again (Exp. 121, Fig. 96).

(B) Measure the resistance of one meter. Find the value of X
approximately, and use a resistance for R that will suit. (See
Exp. 129, Note.)

(C) Divide the result by 100 to get the resistance of 1 cm. of
the wire.

(D) Compare the resistance of one meter of No. 28 G-s wire,
found in Exp. 133, with the resistance of 1 meter of No. 30 G-s
wire. What is the relation, then, between the size (area of
cross-section) of a wire and its resistance? (See the results
of Exp. 122, and § 319, also Summary of Laws.)

=EXPERIMENT 137.=

=341. Directions.= (A) Measure the resistance of 2 meters of
No. 30 copper wire, arranged on a board as in Fig. 96. (See
Exp. 129, Note.) Get the resistance of 1 meter.

(B) Compare the conductivities of copper and German silver
by studying the results of Exps. 136 and 137. Which has the
greater resistance? To find out how many times greater one
resistance is than the other, divide the larger by the smaller.

=EXPERIMENT 138. To study the effect of heat upon the
resistance of metals.=

_Apparatus._ Same as for Exp. 126; the coil of No. 24 wire (No.
89); a lamp or other source of heat. Arrange as in Fig. 105.

=342. Directions.= (A) Measure the resistance of the coil as
before, Exp. 129. The result should nearly agree with that of
Exp. 129, provided connections, etc., are the same.

(B) Remove the coil from the bridge, hold it about a foot above
a lamp or stove, to warm it thoroughly, but do not heat it
enough to injure the covering. It will take a minute or so to
warm it so that the heat will get to the inside also.

(C) Replace the coil, measure its resistance, and compare the
result with its resistance when cold. Does heat increase or
decrease the resistance of a copper wire?

_=343. Effect of Heat upon Resistance.=_ Although there was but the fraction of an ohm difference in the resistances of the hot and cold coil, it was evident that changes of temperature affect the conducting power of copper. This is true of all metals; but German silver and other alloys are much less affected than pure metals, so they are used in making standard resistance coils. The resistance of liquids that can be decomposed by the electric current decreases as the temperature rises. Carbon acts like the liquids, while the resistance of metals _increases_ as their temperature rises.

=EXPERIMENT 139. To measure the resistance of a wire by the
method of "substitution."=

_Apparatus._ The coil of No. 24 wire (No. 89), the resistance
of which has been measured, but which will be considered an
unknown resistance, X; G V, 2-F C, M P, connecting wires, etc.,
previously used; rheostat (§ 344). Arrange as in Fig. 106
first, then as in Fig. 107.

_=344. Simple Rheostat.=_ The No. 28 and No. 30 G-s wires stretched upon the board (Fig. 96), make a convenient form of rheostat. The resistance per cm. being known from the results of Exp. 133 and 136, the resistance for any number of cm. is easily found. The 10-cm. divisions should be divided into centimeters. These spaces may be marked off from the rule (No. 88).

=345. Directions.= (A) Be sure that 2-F C gives a constant
current, shown by the uniform deflection at G V, when arranged
as in Fig. 106. Do not use a cell that quickly polarizes. The
coil, X, forms a part of the circuit; it is joined to wires,
1 and 2, by means of metal plates, so that it may be quickly
removed without disturbing either G V or 2-F C. Carefully read
the deflection at G V.

(B) Remove X from the circuit, and join the free end of
wire, 2, to binding-post, X, and the free end of wire, 1, to
a small piece of sheet copper, which can be firmly pressed
upon the G-s wire to make a contact. Move this along on the
G-s wire until the deflection produced equals that of part
(A), remembering that the longer the G-s wire in the circuit
the less the deflection. Make two or three trials, as one or
two cm. difference in length make but a little difference in
the deflection. Note the number of cm. of G-s wire used, the
resistance of which must equal that of the coil, X.

(C) Find the resistance of X by multiplying the length just
found by the resistance of each cm., and compare the result
with the value found by using the bridge method directly.

=EXPERIMENT 140. To measure the E. M. F. of a cell by
comparison with the two-fluid cell.=

_Apparatus._ Rheostat (§ 344); the two-fluid cell, 2-F C (Exp.
113), the E. M. F. of which may be taken as 1 volt; dry cell, D
C; galvanoscope, G V. Arrange first as in Fig. 107.

=346. Directions.= (A) Be sure that 2-F C gives a constant
current. Take the reading of G V without the rheostat in the
circuit; that is, with wires, 2 and 1, joined directly. The
deflection should be 50 or 60 degrees at least, and be constant.

(B) Attach a small piece of copper to the end of 1, and firmly
rub it along upon the G-s wire, thus introducing resistance
into the circuit, until the deflection is, say, 60° (50 or 55
degrees will do). Note the length of G-s wire used and call it
(B).

(C) Gradually add more resistance by moving the end of 1 along
until the deflection is 50°, 10 degrees less than before. (If
the original was 50° make the new 40°). Call the number of cm.
of wire used (C).

(D) Replace 2-F C with the dry cell D C. Add resistance, as
before, until G V indicates a deflection of 60°, being careful
not to keep the circuit closed long enough to partially
polarize D C. Make 2 or 3 trials, allowing D C to rest a few
minutes between each. Call the number of cm. of G-s wire used
(D).

(E) Again add more resistance, as in (C), until the deflection
is reduced to 50°. Call the length used (E).

=347. Calculation.= It is known that resistances that are able
to reduce the strength of the currents equally are proportional
to the electromotive forces; that is, the electromotive forces
of the two cells are to each other as the two resistances
necessary to produce equal changes in the deflections, which,
of course, indicate equal changes in the strength of the
currents. Since the resistances used in the two cases are
directly proportional to the lengths used, we have:

Length (C-B): Length (E-D):: E. M. F. of 2-F C: E. M. F. of D C.

Substitute the values found and find the E. M. F. of D C.

=EXPERIMENT 141. To measure the internal resistance of a cell
by the "method of opposition."=

_Apparatus._ All the apparatus of Exp. 126. Two simple cells
(§ 275), the plates of which should be of the same size, the
same distance apart, and immersed in acid to the same extent in
both. The acid in both should be of the same strength.

=348. Directions.= (A) Connect the two cells in opposition, so
that no current will be generated by them, and so that the two
can be treated as a dead resistance. Do this by joining the
two zinc plates by a wire with connectors, and use wires to
connect the copper plates to the bridge like any other unknown
resistance.

(B) Measure the resistance of the two by the regular bridge
method, allowing for wires used for connections. One-half of
the resistance found will give the internal resistance of one
cell. (See Note.)

=Note.=--The standard resistance will have to be arranged
to suit each particular case to make the calculations even
approximately correct. (See Exp. 129, Note.) The standard
resistance may be increased by adding the various coils and
rheostat wires, their values being known.

_=349. Summary of Laws of Resistance.=_ 1. _The resistance of a wire is directly proportional to its length, provided its cross-section, material, etc., are uniform._

=EXAMPLE.= If 39.1 ft. of No. 24 copper wire has a resistance
of 1 ohm, 78.2 ft. will have a resistance of 2 ohms, because
78.2 is twice 39.1; 70.38 ft. will have a resistance of 1.8
ohms, as (70.38 ÷ 39.1 = 1.8) it is 1.8 times 39.1.

2. _The resistance of a wire is inversely proportional to its area of cross-section._ The areas of cross-section of round wires vary as the squares of their diameters; so _the resistance of a wire is also inversely proportional to the square of its diameter, other things being equal_.

=EXAMPLE.= A No. 30 wire has a diameter of about .01 inch,
while the diameter of a No. 24 wire is about .02 in.; that is,
the No. 24 has _twice_ the diam. that the No. 30 has. The area
of cross-section of the No. 24, however, is four times that of
the No. 30, so its resistance is but 1/4 that of the No. 30,
the lengths, etc., being the same. (See Wire Tables.)

3. _The resistance of a wire depends upon its material, as well as upon its length, size, etc._

4. _The resistance of a wire depends upon its temperature._ (See Elementary Electrical Examples.)