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A 50 mH inductor is in series with a 10 Ohm resistor and a battery with an emf of 25 V. At t = 0 the switch is closed. Find Q: How long it takes the current to rise to 90% of its final value? |
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Answer» Solution :We need the time taken for I to reach`0.91_(0) = 0.9 e/R.` `0.91_(0)= I__(0)(1-e^(-t/t)` From this we FIND that exp`(-t/zeta)( 0.1)`, which may be expressed as` (-t/zeta) = ln(0.1)` .THUS, `t=-zeta ln(0.1)= 11.5 xx 10^(-3)s` |
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