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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Сторінка 16
... ohms resistance ? Solution : By Ohm's law , Rule 1 , the current .25 ÷ ( 71⁄2 +5 ) = .02 amperes . = A battery of 51 volts E. M. F. and 20 ohms resist- ance has opposed to it in the same circuit a 16 ARITHMETIC OF ELECTRICITY .
... ohms resistance ? Solution : By Ohm's law , Rule 1 , the current .25 ÷ ( 71⁄2 +5 ) = .02 amperes . = A battery of 51 volts E. M. F. and 20 ohms resist- ance has opposed to it in the same circuit a 16 ARITHMETIC OF ELECTRICITY .
Сторінка 17
... ance has opposed to it in the same circuit a battery of 26 volts E. M. F. and 25 ohms resistance . A current of % ampere is maintained in the circuit . What is the resistance of the wire leads and con- nections ? = - Solution : The ...
... ance has opposed to it in the same circuit a battery of 26 volts E. M. F. and 25 ohms resistance . A current of % ampere is maintained in the circuit . What is the resistance of the wire leads and con- nections ? = - Solution : The ...
Сторінка 19
... ance . Rule 8. In divided circuits , each branch passes a portion of a current inversely proportional to its re- sistance . EXAMPLES . = A portion of a circuit consists of two conductors , A and B , in parallel of A = 50 , and B 75 ohms ...
... ance . Rule 8. In divided circuits , each branch passes a portion of a current inversely proportional to its re- sistance . EXAMPLES . = A portion of a circuit consists of two conductors , A and B , in parallel of A = 50 , and B 75 ohms ...
Сторінка 20
... ance of each branch for a denominator of a fraction having 1 for its numerator . In other words , for each branch write down the reciprocal of its resistance . Then reduce the fractions to a common denominator , and add together the ...
... ance of each branch for a denominator of a fraction having 1 for its numerator . In other words , for each branch write down the reciprocal of its resistance . Then reduce the fractions to a common denominator , and add together the ...
Сторінка 21
... ance coil of 275 ohms resistance . What current passes through each branch ? Solution : The proportions of the current are as ts : or reduced to a common denominator and to their lowest terms : 5. Proceeding as before , and taking the ...
... ance coil of 275 ohms resistance . What current passes through each branch ? Solution : The proportions of the current are as ts : or reduced to a common denominator and to their lowest terms : 5. Proceeding as before , and taking the ...
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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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