frequently termed, the inductance of a coil, is measured by the number of volts of counter E. M. F. when the current changes at the rate of one ampere per second. (Atkinson.)

The unit of inductance is the henry.

Rule 84. The inductance of a coil is found by multiplying together 12.5664, the total number of turns in the coil, the number of turns per centimeter length, the sectional area of the core of the coil in square centimeters, and the permeability of the magnetic circuit; divide the resulting product by 1,000,000,000.

EXAMPLE.

Find the inductance of the coil specified in the preceding example.

Solution: The factors are the same as before, omitting the current and time. Product of 12.5664 X 300 X 5 X 25 X 200 is 94,248,000. Dividing this by 1,000,000,000 gives the inductance .094248 henrys.

The E. M. F. of self-induction may be computed when the inductance, the current and the time taken for the current to reach its maximum are known.

Rule 85. To find the E. M. F. of self-induction divide the product of the inductance and current by the time of current rise.

EXAMPLE.

Using again the data of the foregoing examples, find the counter E. M. F., the inductance being

.094248 henrys, the current 5 amperes, and the time .02 second.

Solution: 5 X .094248.47124; dividing this by .02 gives 23.562 volts, as before.

self-induction

Rule 86. The resistance due to equals 6.2832 times the product of the frequency and the inductance.

EXAMPLE.

Find the inductive resistance of a circuit whose frequency is 60 cycles per second and the inductance is .05 henry.

Solution: 6.2832 X 60 X .05 = 18.8496 ohms.

Ans.

The time constant of an inductive circuit is a measure of the growth or increase of the current. It is the time required by the current to rise from zero to its average value. The average value of an alternating current is .634 times its maximum value. It must not be confused with its effective value, which is .707 times the maximum.

The average value may be obtained by multiplying the effective value, as shown by instruments, by .897.

Rule 87. To find the time constant of a coil or circuit, divide its inductance by its resistance.

EXAMPLE.

What is the time constant of a coil whose induct ance is 3.62 henrys and resistance is 20 ohms. Solution: 3.62 ÷ 20 = .181 second. Ans.

CHAPTER XII.

CONDENSERS.

A condenser, though it will allow no current to pass through it, yet it will accumulate or store up a quantity of electricity depending on various factors which the following rules will show:

Rule 88. The quantity stored equals the product of the E. M. F. applied and the capacity of the condenser.

Q = EC.

Rule 89. The capacity of a condenser equals the quantity stored divided by the applied E. M. F.

E

Rule 90. The E. M. F. applied to a condenser equals the quantity stored divided by its capacity.

The quantity stored in a condenser is measured in coulombs (i. e., ampere-seconds); the E. M. F. in volts, and the capacity in farads. Condensers in practical use have, however, so small a capacity that

it is usually stated in microfarads and the quantity in microcoulombs.

EXAMPLES.

A battery of 30 volts E. M. F. is connected to a condenser whose capacity is one half microfarad. What quantity of electricity will be stored?

Solution: 30 volts X .0000005 farads .000015 coulombs. This solution could also be given directly in micro-quantities, thus: 30 volts X 1⁄2 microfarad 15 microcoulombs.

A condenser is charged with 7.5 microcoulombs under an E. M. F. of 15 volts. What is its capacity? Solution: 7.5 microcoulombs 15 volts - .5 microfarad. Ans.

What E. M. F. is required to charge a condenser whose capacity is .1 microfarad with 21 microcoulombs of electricity?

Solution: 21 microcoulombs.1 microfarad = 210 volts. Ans.

By connecting condensers in parallel the resulting capacity is the sum of their individual capacities. When they are connected in series the resulting capacity equals 1 divided by the sum of the reciprocals of their individual capacities. It will be noticed that these laws of condenser connections are the inverse of those for the parallel and series connection of resistances.

When applying a direct current to a condenser, as in the above examples, it flows until the increasing charge opposes an E. M. F. equal to that of the charging current.

With an alternating current a charge would be surging in and out of the condenser, so that a real current will be flowing on the charging wires in spite of the fact that the actual resistance of a condenser, in ohms, is practically infinite.

Rule 91. The alternating current in a circuit having capacity equals the product of 6.2832, the frequency, the capacity, and the applied voltage.

EXAMPLE.

Find the current produced by an E. M. F. of 50 volts and a frequency of 60 cycles per second in a circuit whose capacity is 125 microfarads.

Solution: The capacity 125 microfarads equals .000125 farads.

6.2832 X 60 X .000125 X 50 2.3562 amperes. Ans.

Rule 92. The alternating E. M. F. required to be impressed upon a circuit of a given capacity in order to produce a certain current is equal to the current divided by 6.2832 times the product of the capacity and the frequency.

EXAMPLE.

Find the E. M. F. necessary to produce an alternating current of 50 amperes at 50 cycles per second in a circuit of 80 microfarads capacity.

Solution: 6.2832 X .000080 X 50 = .0251328