In the circuit of Figure 10.57 (a), the excitation $v_{s}(t)$ is a pulse as shown in Figure 10.57 (b).

a. Compute $i_{L}$ for 0 < t < 0.3 ms

b. Compute and sketch $i_{L}$ for all t > 0

Question

In the circuit of Figure 10.57 (a), the excitation [latex] v_{s}(t) [/latex] is a pulse as shown in Figure 10.57 (b).

a. Compute [latex]i_{L} [/latex] for 0 < t < 0.3 ms

b. Compute and sketch [latex] i_{L} [/latex] for all t > 0

Accepted Answer

a. For this circuit [latex] v_{s}(t) = 24[u_{0}(t) - u_{0}(t - 0.3)] [/latex] and since we are not told otherwise, we will assume that [latex] i_{L}(0^{-}) [/latex] = 0 . For 0 < t < 0.3 ms the circuit and its Thevenin equivalent are as shown below

Then,

[latex] i_{L}(t) = i_{Lf} + i_{Ln} = 16 /8 + A_{1}e^{-(R/L)t} = 2 +Ae^{-8000t} [/latex]

and at t = 0

[latex] i_{L}(0) = i_{L}(0^{-}) = 0 = 2 + A_{1}e^{0} [/latex]

or [latex]A_{1} [/latex] = -2 and thus for 0 < t < 0.3 ms

[latex] i_{L}(t) = 2 - 2e^{-8000t} [/latex] (1)

b. For t > 0.3 the circuit is as shown below. For this circuit

[latex] i_{L}(t) = A_{2}e^{-(R/L)(t - 0.3)} = A_{2}e^{-8000(t - 0.3} (2) [/latex]

and [latex] A_{2} [/latex] is found from the initial condition at t = 0.3 ms, that is, with (1) above we get

[latex] i_{L}|_{_{t = 0.3 ms}} = 2 - 2e^{-8 × 10³ × 0.3 × 10^{-3} } = 2 - 2e^{2.4} = 1.82 A [/latex]

and by substitution into (2) above

[latex] i_{L}|_{_{t = 0.3 ms}} = 1.82 = A_{2}e^{-8000(0.3 - 0.3)} [/latex]

or [latex] A_{2} = 1.82 . [/latex] Therefore for t > 0.3 ms

[latex] i_{L}(t) = 1.82e^{-(t - 0.3 ms)} [/latex]

The waveform for the inductor current [latex]i_{L}(t) [/latex] for all t > 0 is shown below.

Question 10.p.3: In the circuit of Figure 10.57 (a), the excitation vs(t) is ...

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