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A coil of inductance 8.4 mH and resistance 6Omegaare connected to a 12 V battery. At what time the current in the coil will be 1.0 A ? |
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Answer» Solution :`i_0 = E/R = 2 ,` given current , `i=(i_0)/(2)` `therefore i= i_0 (1-e^(t/TAU)) rArr 1/2 = 1-e^(t/tau)` `rArr e^(t/tau) = 2 therefore t= tau ln 2 = L/R ln 2` `i.e., t = (8.4xx 10^(-3) XX 0.6931)/(6) ~~ 1 xx 10^(-3)s` |
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