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THE SOURCE-FREE RL CIRCUIT
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In the RL circuit of Fig. 7-5, assume that at t 0 the current is I0 . For t > 0, i should satisfy Ri L di=dt 0, the solution of which is i Aest . By substitution we nd A and s: A R Ls est 0; The initial condition i 0 A I0 . Then i t I0 e Rt=L The time constant of the circuit is L=R.
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EXAMPLE 7.4 The 12-V battery in Fig. 7-6(a) is disconnected at t 0. for all times. Find the inductor current and voltage v
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s R=L
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for t > 0
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CHAP. 7]
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FIRST-ORDER CIRCUITS
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Fig. 7-5
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Fig. 7-6 Assume the switch S has been closed for a long time. The inductor current is then constant and its voltage is zero. The circuit at t 0 is shown in Fig. 7-6(b) with i 0 12=4 3 A. After the battery is disconnected, at t > 0, the circuit will be as shown in Fig. 7-6(c). For t > 0, the current decreases exponentially from 3 A to zero. The time constant of the circuit is L=R 1=100 s. Using the results of Example 7.3, for t > 0, the inductor current and voltage are, respectively, i t 3e 100t v t L di=dt 30e 100t i t and v t are plotted in Figs. 7-6(d) and (e), respectively. V
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FIRST-ORDER CIRCUITS
[CHAP. 7
ESTABLISHING A DC CURRENT IN AN INDUCTOR
If a dc source is suddenly applied to a series RL circuit initially at rest, as in Fig. 7-7(a), the current grows exponentially from zero to a constant value with a time constant of L=R. The preceding result is the solution of the rst-order di erential equation (8) which is obtained by applying KVL around the loop. The solution follows.
Fig. 7-7
Ri L
di V0 dt
for t > 0;
i 0 0
Since i ih t ip t , where ih t Ae Rt=L and ip t V0 =R, we have i Ae Rt=L V0 =R The coe cient A is found from i 0 A V0 =R 0 or A V0 =R. The current in the inductor and the voltage across it are given by (9) and (10) and plotted in Fig. 7-7(b) and (c), respectively. i t V0 =R 1 e Rt=L di v t L V0 e Rt=L dt for t > 0 for t > 0 9 10
THE EXPONENTIAL FUNCTION REVISITED
The exponential decay function may be written in the form e t= , where  is the time constant (in s). For the RC circuit of Section 7.2,  RC; while for the RL circuit of Section 7.4,  L=R. The general decay function f t Ae t= t > 0 It is seen that
is plotted in Fig. 7-8, with time measured in multiples of .
f  Ae 1 0:368A
CHAP. 7]
FIRST-ORDER CIRCUITS
Fig. 7-8
that is, at t  the function is 36.8 percent of the initial value. It may also be said that the function has undergone 63.2 percent of the change from f 0 to f 1 . At t 5, the function has the value 0.0067A, which is less than 1 percent of the initial value. From a practical standpoint, the transient is often regarded as over after t 5. The tangent to the exponential curve at t 0 can be used to estimate the time constant. In fact, since slope f 0 0 A 
the tangent line must cut the horizontal axis at t  (see Fig. 7-9). More generally, the tangent at t t0 has horizontal intercept t0 . Thus, if the two values f t0 and f 0 t0 are known, the entire curve can be constructed.
Fig. 7-9
At times a transient is only partially displayed (on chart paper or on the face of an oscilloscope), and the simultaneous values of function and slope needed in the preceding method are not available. In that case, any pair of data points, perhaps read from instruments, may be used to nd the equation of the transient. Thus, referring to Fig. 7-10, f1 Ae t1 = which may be solved simultaneously to give  and then A in terms of  and either f1 or f2 . t2 t1 ln f1 ln f2 f2 Ae t2 =
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