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A 20 mH coil is connected in series with a `2 k Omega` resistor and a 12 V battery. Calcualte (i) time constant of the circuit, (ii) time during which current decrys to 10% of its maximum value. |
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Answer» Here, `L = 20 mH = 20 xx 10^(-3) H`, `R = 2 k Omega = 2 xx 10^(3) ohm, V = 12 V` (i) Time constant `= (L)/(R ) = (20 xx 10^(-3))/(2 xx 10^(3)) = 10^(-5) s` (ii) For decay of current, `I = I_(0) e^(-(R )/(L) t)` `(I_(0))/(10) = I_(0) e^(-10^(5) t)` `:. e^(10^(5) t) = 10` `10^(5) t = 2.303 log_(10) 10` `t = (2.303)/(10^(5)) = 2.303 xx 10^(-5) s` |
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