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An inductor `(L = 20 mH)`, a resistor `(R = 100 Omega)` and a battery`(varepsilon = 10 V) ` are connected in series. Find (a) the time constant, (b) the maximum current and (c ) the time elapsed before the current reaches 99% of the maximum value. |
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Answer» Here, `L = 20 mH = 20 xx 10^(-3) H, R = 100 ohm`. `E = 10 V` Time constant, `tau = (L)/(R ) = (20 xx 10^(-3))/(100) = 2 xx 10^(-4) s` Maximum current, `I_(0) = (E)/(R ) = (10)/(100) = 10^(-1) A` Using `I = I_(0) (1 - e^(- t//tau))` `0.99 I_(0) = I_(0) (1 - e^(-t//tau))` `:. e^(-t//tau) = 0.01`. Solve to get `t = 0.92 xx 10^(-3) s` |
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