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A 50 mH inductor is in series with a 10 ...

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?

Text Solution

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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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