In Figure 13-27,determine the inductor current 30μs after the switch is closed.
Question 13.10: In Figure 13-27,determine the inductor current 30μs after t...
The RL time constant is
\tau =\frac{L}{R}=\frac{100 \ mH}{2.2 \ K\Omega } = 45.5 \ \mu sThe final current is
I_{F}= \frac{V_{S}}{R}= \frac{12 \ V}{2.2 \ k\Omega }= 5.45 \ mAThe initial current is zero. Notice that 30 μs is less than one time constant, so the current will reach less than 63% of its final value in that time.
i_{L}= I_{F}(1-e^{-Rt/L})= 5.45 \ mA(1-e^{-0.66})= 5.45 \ mA (1-0.517)=2.63 \ mARelated Answered Questions
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