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Circuit Containing Inductance OnlyAuthor: E.E. Kimberly When a circuit containing only inductance L, such as that of Fig. 810 (a), is Closed, the current must produce a voltage drop in the circuit that is equal at every instant (including 2 = 0) to the applied voltage v. The counter emf must be sinusoidal to match the applied voltage. To produce a sinusoidal emf, the rate of change of flux linkages in the inductance must be a sinusoidal function of time from the instant at which the circuit is closed. If the circuit is closed at time ti when the voltage is at its maximum value, as indicated in Fig. 810 (5), the conditions of i=0 and ϕ=0 are the same as they would be if the circuit had been closed many cycles earlier with no transient persisting. The current and flux then start from zero and vary sinusoidally without a transient component. In the first halfcycle, ϕ makes a total change of 2ϕ_{max} from ϕ = 0 to ϕ = +ϕ_{max} to ϕ = 0.
If, however, the circuit is closed at T_{0}, as in Fig. 810 (c), ϕ_{max} does not exist at that instant and a sinusoidal change of 2ϕ_{max} must occur in the first halfcycle beginning with ϕ = 0. Therefore, in the first halfcycle, the flux must rise to 2ϕ_{max}. Inasmuch as the flux is proportional to the current i which produces it, i must also rise to 2I_{max} in the first halfcycle. Under these ideal conditions i would vary sinusoidally between zero and 2I_{max} indefinitely, and the current could not be called a transient. In practice, however, there is always some resistance in the circuit which causes the current to diminish and there is established a steadystate variation which is symmetrical about the time axis. When the circuit is closed at any instant,
The angle μ is that corresponding to the instant at which the circuit is closed after v = 0. From this equation
By integration, When t=0, i=0 and the constant of integration is Therefore,
(815) The term cos μ is the transient component of the current, and the term is the steadystate component.In Fig. 810 (c) the circuit was closed when t=0 and v=0 with the slope positive. Therefore, ϕ=0 and
(816)


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