resistance of the outer circuit is 150 ohms and it maintains a current of 16 amperes. What is its duty?

Solution: The total electrical H. P. is found by Rules 30 and 43 to be 162 X 160.7 ÷ 746 = 55.1

H. P. Duty

= 55.1 58.0 = 95%.

The result must always be less than unity; if it exceeded unity it would prove that there had been an error in some of the determinations.

Rule 45. The commercial efficiency of a generator is the quotient obtained by dividing the electric energy in the outer circuit by the mechanical energy expended in turning the armature. e. H. P. (outer circuit) m. H. P.

C. Eff. =

EXAMPLES.

What is the commercial efficiency of the dynamo just cited?

Solution: The electrical H. P. of the outer circuit is found by the same rules to be 162 × 150 ÷746 = 51.5 commercial efficiency 51.5 58.0 = 88.8%.

=

Rule 46. The resistance of the outer circuit is to the total resistance, as the commercial efficiency is to the duty.

EXAMPLES.

Take the case of the generator last given and from its duty calculate the commercial efficiency. Solution: 150: 160.7::x:95.0.. x = 88.8 or

CHAPTER VIII.

BATTERIES.

GENERAL CALCULATIONS OF CURRENT.

A BATTERY is rated by the resistance and electromotive force of a single cell, which factors are termed the cell constants. In the case of storage batteries, whose susceptibility to polarization is very slight, the resistance is often assumed to be neglible. It is not so, and in practice is always knowingly or otherwise allowed for.

From the cell constants its energy-constant may be calculated by Rule 31, as equal to the square of its electro-motive force divided by its resistance. This expresses its energy in watts through a circuit. of no resistance.

There are two resistances ordinarily to be considered, the resistance of the battery which is designated by R or by n R if the number of cells is to be implied and the resistance of the external circuit which is designated by r.

Rule 47. The current given by a battery is equal to its electro-motive force divided by the sum of the external and internal resistances.

Six cells in parallel.

Six cells in series.

Six cells-two in parallel, three in series.

Six cells-three in parallel, two in series.

ARRANGEMENT OF BATTERY CELLS.

EXAMPLE.

A battery of 50 cells arranged to give 75 volts E. M. F. with an internal resistance of 100 ohms sends a current through a conductor of 122 ohms resistance. What is the strength of the current? Solution: Current 75 ÷ (100+122)= .338 ampere. This rule has already been alluded to under Ohm's law (page 14).

=

ARRANGEMENT OF CELLS IN BATTERY.

In practice the cells of a battery are arranged in one of three ways. a: All may be in series; b: all may be in parallel; c: some may be in series and some in parallel, so as to represent a rectangle, s cells in series by p cells in parallel, the total number of cells. being equal to the product of s and p.

Other arrangements are possible. Thus the cells may represent a triangle, beginning with one cell, followed by two in parallel and these by three in parallel and so on. This and similar types of arrangement are very unusual and little or nothing is to be gained by them.

Rule 48. The electromotive force of a battery is equal to the E. M. F. of a single cell multiplied by the number of cells in series.

Rule 49. The resistance of a battery is equal to the number of its cells in series, multiplied by the resist, ance of a single cell and divided by the number of its cells in parallel.

R.

battery 8 R

=

EXAMPLES.

A battery of 50 gravity cells 1 volt, 3 ohms each is arranged 10 in parallel and 5 in series. What is its resistance and electromotive force?

The same battery is arranged all in parallel; what is its resistance and E. M. F.?

Solution: This gives one cell in series.

Resistance = 1 X 3 ÷ 50:

= .06 ohms.

E. M. F. =1X1=1 volt.

The same battery is arranged all in series; what is its resistance?

Solution: This gives one cell in parallel.

Resistance 50 × 3 = 150 ohms.

1

E. M. F. = 50 X 1 = 50 volts.

The current given by a battery is obtained from these rules and from Ohm's law.

EXAMPLE.

150 cells of a battery (cell constants 1.9 volts, ohm) are arranged 10 in series and 15 in parallel. They are connected to a circuit of 1.7 ohms resistWhat is the current?

ance.

Solution: The resistance of the battery

.333 ohms. The E. M. F.

=

10 X 1.9 = 19 volts.

Current 19+ (.333 +1.7)= 9.34 amperes.