For the circuit in Figure 13-24, what is the current at 0.1 ms and 0.6 ms?

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Accepted Answer

The RL time constant for the circuit is

[latex] au =\frac{L}{R}=\frac{1.0 \ mH}{10 \ K\Omega } = 0.1 \ ms[/latex]

If the square-wave generator period is long enough for the current to reach its maximum value in [latex]5 au [/latex], the current will increase exponentially and during each time constant interval will have a value equal to the percentage of the final current given in Table 13-1:

NUMBER OF TIME CONSTANTS	APPROXIMATE % OF FINAL CURRENT
1	63
2	86
3	95
4	98
5	99 (considered 100%)

. The final current is

[latex]I_{F}= \frac{V_{S}}{R}= \frac{2.5 \ V}{10 \ k\Omega }= 0.25 \ mA[/latex]

The current at 0.1 ms is

i = 0.63(0.25 mA) = 0.158 mA

At 0.6 ms, the square-wave input has been at the 0 V level for 0.1 ms, or [latex]1 au [/latex]; and the current decreases from the maximum value toward its final value of 0 mA by 63%. Therefore,

i = 0.25 mA - 0.63(0.25 mA) = 0.092 mA

Question 13.8: For the circuit in Figure 13-24, what is the current at 0.1 ...

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