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Circuit Containing Inductance OnlyAuthor: E.E. Kimberly When a circuit containing only inductance L, such as that of Fig. 8-10 (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. 8-10 (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 half-cycle, ϕ makes a total change of 2ϕmax from ϕ = 0 to ϕ = +ϕmax to ϕ = 0.
If, however, the circuit is closed at T0, as in Fig. 8-10 (c), -ϕmax does not exist at that instant and a sinusoidal change of 2ϕmax must occur in the first half-cycle beginning with ϕ = 0. Therefore, in the first half-cycle, the flux must rise to 2ϕmax. Inasmuch as the flux is proportional to the current i which produces it, i must also rise to 2Imax in the first half-cycle. Under these ideal conditions i would vary sinusoidally between zero and 2Imax 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 steady-state 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,
(8-15) The term cos μ is the transient component of the current, and the term is the steady-state component.In Fig. 8-10 (c) the circuit was closed when t=0 and v=0 with the slope positive. Therefore, ϕ=0 and
(8-16)
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