In Figure 13-28, what is the current at each microsecond interval for one complete cycle of the input square wave, Vs, After calculating the current at each time, sketch the current waveform.

Question

In Figure 13-28, what is the current at each microsecond interval for one complete cycle of the input square wave, [latex]V_{s}[/latex], After calculating the current at each time, sketch the current waveform.

Accepted Answer

[latex] au =\frac{L}{R}=\frac{560 \ \mu H}{680 \ \Omega } = 0.824 \ \mu s [/latex]

When the pulse goes from 0 V to 10 V at t = 0, the final current is

[latex]I_{F}= \frac{V_{s}}{R}= \frac{10 \ V}{680 \ \Omega }= 14.7 \ mA[/latex]

For the increasing current,

[latex]i= I_{F}(1-e^{-Rt/L})= I_{F}(1-e^{-t/ au })[/latex]

[latex]At \ 1\mu s : i= 14.7mA(1-e^{-1\mu s/0.824\mu s})= 10.3 \ mA[/latex]

[latex]At \ 2\mu s : i= 14.7mA(1-e^{-2\mu s/0.824\mu s})= 13.4 \ mA[/latex]

[latex]At \ 3\mu s : i= 14.7mA(1-e^{-3\mu s/0.824\mu s})= 14.3 \ mA[/latex]

[latex]At \ 4\mu s : i= 14.7mA(1-e^{-4\mu s/0.824\mu s})= 14.6 \ mA[/latex]

[latex]At \ 5\mu s : i= 14.7mA(1-e^{-5\mu s/0.824\mu s})= 14.7 \ mA[/latex]

When the pulse goes from 10 V to 0 V at t = 5 μs , the current decreases exponentially. For the decreasing current,

[latex]i= I_{i}(e^{-Rt/L})= I_{i}(e^{-t/ au })[/latex]

The initial current is the value at 5 μs, which is 14.7 mA.

[latex]At \ 6\mu s : i= 14.7mA(e^{-1\mu s/0.824\mu s})= 4.37 \ mA[/latex]

[latex]At \ 7\mu s : i= 14.7mA(e^{-2\mu s/0.824\mu s})= 1.30 \ mA[/latex]

[latex]At \ 8\mu s : i= 14.7mA(e^{-3\mu s/0.824\mu s})= 0.38 \ mA[/latex]

[latex]At \ 9\mu s : i= 14.7mA(e^{-4\mu s/0.824\mu s})= 0.11 \ mA[/latex]

[latex]At \ 10 \mu s : i= 14.7mA(e^{-5\mu s/0.824\mu s})= 0.03 \ mA[/latex]

Figure 13-29 is a graph of these results.

Question 13.11: In Figure 13-28, what is the current at each microsecond int...

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