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.)