Arithmetic of Electricity: A Practical Treatise on Electrical Calculations of All Kinds Reduced to a Series of Rules, All of the Simplest Forms, and Involving Only Ordinary Arithmetic ...N.W. Henley Publishing Company, 1909 - 162 стор. |
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Результати 1-5 із 8
Сторінка 17
... fractions of the circuit . Ohm's Law applies to these cases also . EXAMPLES . An electric generator of unknown resistance main- tains a difference of potential of 10 volts between its terminals connected as described . The terminals are ...
... fractions of the circuit . Ohm's Law applies to these cases also . EXAMPLES . An electric generator of unknown resistance main- tains a difference of potential of 10 volts between its terminals connected as described . The terminals are ...
Сторінка 20
... fractions will express the proportional cur- rents as fractions of one . If the total amperage is given , it is to be multiplied by the fractions to give the amperes passed by each branch . The solution can also be done in decimals ...
... fractions will express the proportional cur- rents as fractions of one . If the total amperage is given , it is to be multiplied by the fractions to give the amperes passed by each branch . The solution can also be done in decimals ...
Сторінка 21
... fractions . A current of .71 amperes passes through two branches of a circuit . One is a lamp with its con- nections of 115 ohms resistance ; another is a resist ance coil of 275 ohms resistance . What current passes through each branch ...
... fractions . A current of .71 amperes passes through two branches of a circuit . One is a lamp with its con- nections of 115 ohms resistance ; another is a resist ance coil of 275 ohms resistance . What current passes through each branch ...
Сторінка 69
... fraction ( 1.4 ) etc. ) of the current required apply Rule 53 . 4 ' 5 ' 8 ' Rule 51. Case A. Divide the required difference of po- tential of the outer circuit by the voltage of a single cell diminished by the product of the required ...
... fraction ( 1.4 ) etc. ) of the current required apply Rule 53 . 4 ' 5 ' 8 ' Rule 51. Case A. Divide the required difference of po- tential of the outer circuit by the voltage of a single cell diminished by the product of the required ...
Сторінка 72
... fraction ( 1 ) Case C. applies . Solution : Cell energy = 1.75 = .375 watts . Ex- ternal energy = 50 × 1 × 20 : .375 ) X 410,666 cells . Solution by Rule 53 : = = 1000 watts . ( 1000 + Voltage required taking = lamps in series 20 x 50 ...
... fraction ( 1 ) Case C. applies . Solution : Cell energy = 1.75 = .375 watts . Ex- ternal energy = 50 × 1 × 20 : .375 ) X 410,666 cells . Solution by Rule 53 : = = 1000 watts . ( 1000 + Voltage required taking = lamps in series 20 x 50 ...
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Загальні терміни та фрази
50 cents 50 ohms 50 volts air gaps alternating current ampere turns ance appliances armature core battery C. G. S. units calculated calories capacity cells in parallel cells in series circular mils coefficient of leakage coil conductance conductor copper coulombs cross-sectional area cycles per second difference of potential distance divided drop dynamo efficiency electromotive force energy engines EXAMPLE expressed external circuit factors farad field resistance fractions given gives gram heat developed horse power horse-power inductance internal resistance joules kilogram-meter lamps in parallel leads length lines of force lines per square magnet core magnetic circuit meter microfarad microhms mils in diameter motor multiplied number of cells number of turns Ohm's law ohms resistance permeability permeance proportion ratio of conversion reciprocal reduced reluctance Rule 55 Rules 30 sectional area shunt Solution specific resistance square centimeter square inch tion turns of wire voltage weight wrought iron
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Сторінка 50 - This loss as previously explained, is equal to the square of the current multiplied by the resistance. The...
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