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Postby TGTS0907129» Current growth and decay in inductive circuit

Describe the growth and decay of current in an inductive circuit


Consider the circuit which includes a source of terminal potential different V and negligible internal resistance a, resistor R of negligible inductance, and inductor of inductance L and negligible resistance. When the switch S is throwing to position 1, the current is zero at the instant the switch is closed. At that instant the current starts to rise so that there is an induced emf eL = -L(?i/?t). the induced emf opposes the rise in current; i.e., when ?i/?t is positive eL is negative, a counter emf. We shall represent the counter emf vL as

As the current rises, the rate of changes of current is not constant, but at each instant the sum of the iR drop in potential and the instantaneous counter emf vL must equal the potential difference V of the source.
V = iR + vL

The solution for the current i is a calculus problem. This solution is an exponential equation,

At the instant t = 0, i = 0. As the current V/R. Whether the growth is fast or slow depends upon the relative values of L and R. The exponent of the logarithmic base e must be a pure number. Thus L/R has the dimensions of time and is called the time constant of the circuit. Let t = L/R. Then, i = (V/R)  (1 - e-1) = (V/R)(1 - 1/e). The time constant is the time required for the current to rise to the final steady value if it continued to rise at the final steady value if it continued to rise at the initial rate. If the inductance of the circuit is low, the current rises very rapidly and very soon approaches close to the steady value.

By TGTT24071251 on 9/23/2015 11:04:46 PM
TGTS0907129 on 9/23/2015 10:36:29 PM

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