Calculate the RL time constant for Figure 13-21. Then determine the current and the time at each time-constant interval,measured from the instant the switch is closed.

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Accepted Answer

The RL time constant is

[latex] au =\frac{L}{R}=\frac{10 \ mH}{1.2 \ K\Omega } = 8.33 \ \mu s [/latex]

The current at each time-constant interval is a certain percentage of the final current. The final current is

[latex]I_{F}= \frac{V_{S}}{R}= \frac{12 \ V}{1.2 \ k\Omega }= 10 \ mA[/latex]

Using the time-constant percentage values from Figure 13-1,

[latex]At \ 1 au : i= 0.63(10 \ mA)= 6.3 \ mA;t= 8.33 \ \mu s[/latex]

[latex]At \ 2 au : i= 0.86(10 \ mA)= 8.6 \ mA;t= 16.7 \ \mu s[/latex]

[latex]At \ 3 au : i= 0.95(10 \ mA)= 9.5 \ mA;t= 25.0 \ \mu s[/latex]

[latex]At \ 4 au : i= 0.98(10 \ mA)= 9.8 \ mA;t= 33.3 \ \mu s[/latex]

[latex]At \ 5 au : i= 0.99(10 \ mA)= 9.9 \ mA \cong 10 \ mA ;t= 41.7 \ \mu s[/latex]

Question 13.7: Calculate the RL time constant for Figure 13-21. Then determ...

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