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Current Transient when RL Circuit is Closed

Author: E.E. Kimberly

If the resistance and inductance of Figs. 8-1 and 8-2 be connected in series, as in Fig. 8-3 (a), the rate of rise of current i will be affected by both circuit impedance elements simultaneously; and the result will be similar to that shown in Fig. 8-3 (b) when the switch is closed. At any instant after the switch is closed the voltage V must accomplish two things. It must provide one compromiss, which is the iR voltage drop across R1 and a second component,which is equal to the induced voltage of the inductance L.

Therefore, '

ee_001-231.png

ee_001-232.png (8-1)

To derive the equation for the current at any instant after the circuit is closed, it is convenient first to rewrite the voltage equation as follows:

ee_001-233.png

Multiplying by

ee_001-234.png

and dividing by

ee_001-235.png

we obtain

ee_001-236.png (8-2)

This reduces to

ee_001-237.png

The current then at any time after the switch is closed is the steady-state value less the inductive transient component

ee_001-238.png

as shown in Fig. 8-3 (&).

ee_001-239.png

The transient component of current approaches zero exponentially with time. Theoretically, the circuit current does not reach its final value of V/R until time is infinite. In most instances, however, it will attain practically its steady-state value in a second or less.

Example 8-1. - A constant emf of 10 volts is applied to a coil having a resistance of 5 ohms and an inductance of 0.01 henry. How much time will elapse after the switch is closed until one-half of the steady-state value of current is reached?

Solution. - Steady-state current iM = 10/5 = 2 amp. By equation (8-2),

Hence, and

ee_001-240.png




Last Update: 2010-10-06