CHAPTER XX.
CURRENT STRENGTH.
_=350. Strength of Current.=_ The water in a certain tank may be under great pressure, but if it is obliged to pass through long tubes before it can turn a water-wheel, for example, it is evident that the work done will depend not only upon the pressure in the tank, but upon the resistance to be overcome before the water gets to the wheel. The work that the water can do depends upon its _rate of flow_, and may be used to measure the _strength_ of the current.
The strength of a current of electricity is measured also by the _work_ that it can do, and it depends upon its _rate of flow_ at the point measured. The strength may be determined from its magnetic, heating, or chemical effects.
_=351. Unit of Current Strength; The Ampere.=_ A current having the strength of 1 ampere, when passed through a solution of silver nitrate under proper conditions, will deposit 0.001118 gramme of silver in _one second_; if passed through a solution of copper sulphate, copper plates being used for the electrodes, in the solution, 0.0003277 gramme of copper will be deposited in _one second_. (See Chemical Effects of the Current.) The thousandth of an ampere is called the milliampere. The strength of a current is proportional to the amount of chemical work that it can do per second. (See § 357.)
_=352. Measurement of Current Strength.=_ The _galvanoscope_ previously described simply shows the presence of a current, or whether one current is larger or smaller than another. When the degree-card is used to get the relative deflections, the instrument may be called a _galvanometer_.
_=The Tangent Galvanometer=_ is made on the same general idea as our galvanoscope, the diameter of the coil being twenty times, or more, the length of the needle. In these the strengths of the two currents compared are proportional to the tangents of the angles of deflection produced. (See Elementary Electrical Examples.) There are several varieties of galvanometers, each designed for its special work. They are often calibrated or standardized so that the amperes of current passing through them can be read off directly from the scale.
_=353. The Ammeter=_ is really a galvanometer from which may be read directly the strength of a current. The coil has a low resistance so that it will not greatly reduce the strength of the current to be tested.
_=The Voltameter=_ measures the strength of a current by chemical means.
_=354. Unit of Quantity; The Coulomb.=_ A current having a strength of 1 ampere will do more chemical work by flowing one hour than it can do in 1 second. In speaking of the _quantity_ of electricity we introduce the element of _time_. The unit of quantity is called the _coulomb_, just as a cubic foot of water may be taken as a unit of quantity for water. A coulomb is the quantity of electricity given, in one second, by a current having a strength of 1 ampere. Coulombs are found by multiplying amperes by seconds; thus, a current of 5 amperes will give 20 coulombs in 4 seconds.
_=355. Electrical Horse-power; The Watt.=_ The electric current has power to do work, and we speak of the horse-power of an electric motor in the same way as for a steam-engine. A current with the strength of 1 ampere and an E. M. F. of 1 volt has a unit of power called the watt. 746 watts make an electrical horse-power.
Watts = amperes × volts.
Watts ÷ 746 = the number of horse-power.
(See Transformers, also Elementary Electrical Examples.)
_=356. Ohm's Law.=_ It was first shown by Ohm that the strength of a current is equal to its E. M. F. divided by the resistance in the circuit; that is,
Strength of current (amperes) = E. M. F. (volts). / resistance (ohms).
If we let C stand for the strength in amperes, E for the E. M. F. in volts, and R for the resistance in ohms, we have the short formula, easily remembered,
C = E/R
_=357. An Ampere=_ would be produced by a current of 1 volt pushing its way through a resistance of 1 ohm. Knowing any two of the three, C, E, or, R, the other may be found. The resistance, R, it must be remembered, is the total resistance in the circuit, and is composed of the total internal and external resistances.
(See Elementary Electrical Examples.)
_=358. Internal Resistance and Current Strength.=_ It is evident that the internal resistance of a cell varies with the position and size of the plates. We shall now study the effects of these changes upon the strength of the current.
=EXPERIMENT 142. Having a cell with LARGE PLATES, to find how the strength of the current is affected by changes in the position of the plates, the external resistance being small.=
_Apparatus._ Galvanoscope, G V; materials for simple cell (Exp. 110); connecting wires. Arrange as in figure 108, omitting the wooden cross-piece.
=359. Directions.= (A) Connect the wires with the 5-turn coil of G V, which has but little resistance. Have the tumbler nearly full of dilute acid to get the effect of large plates; that is, the current has a large liquid conductor to pass through in the cell, and the _internal_ resistance will be small. G V should be properly placed N and S.
(B) Place the copper and zinc plates as far apart as possible in the acid, and press them against the bottom of the tumbler. Note the reading of G V. It is not necessary to take readings with reversed current.
(C) Still pressing them against the bottom of the glass, to keep the same amount of surface under acid, slowly bring them near each other and watch the needle.
(D) Hold the plates about an inch apart, and against the bottom, and note the reading of G V. Slowly raise the plates, keeping them the same distance apart until they are out of the acid. Watch the action of the needle.
Make a note of your readings in degrees and write your conclusions. Does a change in internal resistance affect the strength of the current?
=EXPERIMENT 143. Having a cell with SMALL PLATES to find how the strength of the current is affected by changes in the position of the plates, the external resistance being small.=
_Apparatus._ Same as in Exp. 142, the acid, however, being but 1 in. deep in the tumbler; that is, we have the effect of a cell with small plates, each being about 1 in. by 1/2 in.
=360. Directions.= (A) Repeat (B) and (C) of Exp. 142, recording the reading of G V in each case.
(B) Compare the results with those of Exp. 142, remembering that the _internal_ resistance is larger than before. Is the current as strong with small plates as with large plates when the external resistance is small? When the external resistance is small (the 5-turn coil, for example), should the cell have a high or low internal resistance to produce the greatest effect upon the needle?
=EXPERIMENT 144. To find whether the changes in current strength, due to changes in internal resistance, are as great when the external resistance is large, as they are when the external resistance is small.=
_Apparatus._ Same as for Exp. 142, 143, also the rheostat containing the two meters of G-s wire (Exp. 121).
=361. Directions.= (A) Arrange as in Fig. 109, the external resistance being 2 meters of No. 30 G-s wire in series with G V. The 2-F C in the Fig. is replaced, however, by the simple cell as in Exp. 143.
(B) Find the effect upon the strength of the current of moving the plates about when but 1 in. of acid is in the tumbler.
(C) Nearly fill the tumbler with acid and repeat (B), taking readings with plates near each other and as far apart as possible. Lift them nearly out of the acid and take the reading.
(D) Still increase the external resistance of the circuit by adding coils of wire or the meter of No. 28 G-s wire and repeat. Is the strength of the current greatly affected by _slight_ changes in the internal resistance when the external resistance is large?
_=362. Discussion.=_ We shall study, by means of figures, how changes in internal resistance affect the strength of the current.
Let R stand for the total external resistance of a circuit, and r for the total internal resistance of the cell or cells; ohm's law, then, will be expressed by
C = E / (R + r)
=EXAMPLE.= Let us take a circuit (A) when the external resistance, R, is small, and (B) when R is large compared with r, E being taken as 1 volt in both cases.
(A) Let R = 1, and r = 2; substituting these values in the formula above, we have:
C = 1 / (1 + 2) = 1 / 3 = .33+ ampere.
Now let the internal resistance, r, be slightly increased from 2 to 3 ohms; the value of C then becomes 1/4 ampere, as R + r = 4. The change in C, then, is the difference between 1/3 and 1/4; and this expressed in decimals becomes .33 - .25 = .08 ampere.
(B) Let R = 200 ohms, and r = 2 ohms as in (A). Substituting these values we have,
C = 1 / (200 + 2) = 1 / 202 = .00495 ampere.
Increasing r from 2 to 3, as before, etc., we find that C = 1 divided by 203 = .00492 ampere.
The above shows clearly (A) that the value of C is changed considerably by changes in r when R is _small_, and (B) that changes in r produce very slight changes in C when R is _large_. Review your results of Exps. 142-144. (See Elementary Electrical Examples.)
_=363. Arrangement of Cells and Current Strength.=_ We have seen that internal resistance affects current strength. In joining cells, then, attention must be given to the internal resistance as well as to the E. M. F. of the combination.
_=364. Cells in Series.=_ It has been shown by careful experiments that the E. M. F. of two cells joined in series (Fig. 110) is equal to the sum of the E. M. F. of each. Ten cells, joined in series, have ten times the E. M. F. of one cell, provided they have the same E. M. F. As the Zn of one is joined to the Cu of the other, the current is obliged to pass through one solution after the other; that is, the internal resistance of the two in series is equal to the sum of their internal resistances. Ten cells, joined in series, have ten times the internal resistance of one cell, provided they have equal internal resistances.
_=365. Cells Abreast.=_ When the positive plates are joined together and the negative plates are also joined together (Fig. 111), the cells are said to be _abreast_, _in parallel_, or in _multiple arc_. It has been shown that two cells of equal strength, joined abreast, have the same E. M. F. as one cell. The two Cu plates, being joined, must have the same potential; all the Zn plates have the same potential, so the difference of potential at the terminals of the combination is the same as that at the terminals of a single cell.
In two cells abreast (Fig. 111) the current has two liquid paths, side by side, to get from Cu to Zn; this makes the internal resistance one-half that of one cell, provided their internal resistances are equal. Ten cells, of equal internal resistance, when joined abreast, have one-tenth the internal resistance of one cell.
=EXPERIMENT 145. To find the best way to join two similar cells when the external resistance is small.=
_Apparatus._ Two simple cells using dilute sulphuric acid, with copper and zinc elements, as in Exp. 112; galvanoscope, G V; connecting wires, etc. Have the zincs well amalgamated. Remove them from the acid as soon as readings are taken.
=366. Directions.= (A) Partly fill the tumblers with the acid. Join the cells in series (Fig. 110), then connect wire 1 (Fig. 110) with the left-hand binding-post of G V, and wire 2 with the middle one, thus putting the 5-turn coil into the circuit. Take the reading of G V.
(B) Join the cells in multiple arc (Fig. 111), connecting them as in (A) with G V. Write down the reading, and compare it with that found in (A).
(C) Take the reading with but 1 cell joined to G V.
=EXPERIMENT 146. To find the best way to join two similar cells when the external resistance is large.=
_Apparatus._ Same as for Exp. 145, also the rheostat containing 2 metres of No. 28 or 30 G-s wire. Arrange the G-s wire in series with the 15-turn coil of G V, as shown in Fig. 109, two simple cells being used, however, instead of 2-F C as shown.
=367. Directions.= (A) Take the reading of G V when the two cells are in series (Exp. 145), the external resistance being the 15-turn coil and G-s wire.
(B) Join the cells in parallel and take the reading, using the same external resistance as in (A).
(C) Increase the external resistance by adding coils of wire or 2 metres of No. 28 G-s wire and repeat (A) and (B). What does the experiment show?
(D) Take the reading with 1 cell and large external resistance.
_=368. Best Arrangement of Cells.=_ It will be seen by experiments that with a given number of cells the strongest current is produced when they are arranged so that the internal resistance of the battery nearly equals the external resistance of the circuit.
When the external resistance is small, the internal resistance may be kept down by joining the cells in parallel; and, although the E. M. F. is also kept small, the value of C will be larger than it would be with a larger internal resistance and a larger E. M. F.
When the external resistance is large, the internal resistance can be made large by joining the cells in series. The advantage comes, however, from having a large value of E. A large resistance can not hold back a current of large E. M. F. By joining the cells in series the value of E is made large, and the value of C becomes large even though there is an increased internal resistance. (See Elementary Electrical Examples.)