. Hawkins electrical guide. Questions, answers & illustrations; a progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications; a practical treatise. ven in (3),(4) and (5) RE• A 51 STAN> OHM CE cc INDUCTIVE COIL NDEN&EMICR0FW R V V 2 V $ HENRY 125 ADt 60 R H n OMI ilC DROP IN DUCTAJ1 CE DROP CAPACITY DROP 8AMPERE5 n IMPRES&EDPRESSl/RE Fig. —Diagram of circuit containing 25 ohms resistance. .15 henry inductance, and 125microfarads capacity, with current of 8 amperes at 60 frequency. Eim=
. Hawkins electrical guide. Questions, answers & illustrations; a progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications; a practical treatise. ven in (3),(4) and (5) RE• A 51 STAN> OHM CE cc INDUCTIVE COIL NDEN&EMICR0FW R V V 2 V $ HENRY 125 ADt 60 R H n OMI ilC DROP IN DUCTAJ1 CE DROP CAPACITY DROP 8AMPERE5 n IMPRES&EDPRESSl/RE Fig. —Diagram of circuit containing 25 ohms resistance. .15 henry inductance, and 125microfarads capacity, with current of 8 amperes at 60 frequency. Eim= ^R,It+(2r/LI-^-^)J= 1 ^R + ^/L-^-)* (7) Ques. What does the quantity under the square rootsign in equation (7) represent? Ans. It is the impedance of a circuit possessing resistance,inductance, and capacity. ALTERNATING CURRENT DIAGRAMS 1,099 Ques. Why? Ans. Because it is that quantity which multiplied by thecurrent gives the pressure, which is in accordance with Ohmslaw. EXAMPLE.—An alternator is connected to a circuit having, as infig. 1,341, 25 ohms resistance, an inductance of .15 henry, and a capacityof 125 microfarads. What pressure must be impressed on the circuit toallow 8 amperes to flow at a frequency of 60?. V INDUCTANCE DROPf 453 VOLTS RESULTANT DROP OR DROP DUE TO EXCESS CAPACITY. 282 VOLTS OHMIC DROP Fig. 1,342.—Diagram for finding the pressure necessary to be impressed on the circuit shownin fig. 1,341. to produce a current of S amperes. The ohmic drop is E0 = RI = 25 X 8 = 200 volts. The inductance drop is E,-=27r/LI= volts The capacity drop is I 8 E,= 2x/G = 170 volts. IT.\]VKIXS ELECTRICITY Substituting the values thus found, impressed pressure = VE20 + (Ei —Ee)* = V200!+(452-170jl = V2002+282» = Vl19524= volts. THE POWER FACTOR 1,101 CHAPTER XLVIIITHE POWER FACTOR The determination of the power in a direct current circuit isa simple matter since it is only necessary to multiply together thevolt
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