A certain circuit consists of an inductor of 55 mH in series with a resistor of 90 Ω. At a moment when the current in the circuit is 21 A, a switch in the circuit is opened. How long will it take for the current to fall to 7.14 A? Answer in units of s.Okay, easy smeasy I thought. i made an expression 7.1421(1-e^(-90t/55*10^-3))Simply algebra, so I got t2.53926x10^-4. This is wrong! please show me the errors of my ways. Do the entire problem. Show work. I need to get this down to I can tackle the tougher problems.And I based my expression off of the equation i(Io)(1-e^(rt/L)). Thanks! will choose best answer!
This is easy. You used the wrong formula. The formula is for current INCREASING from zero and asymptoting to Io. But it isn't accurate even for that. The actual formula is i Io (e ^- t/(rL)) Note the features. The larger the value of r the less that is lost (or gained ) per second. The larger the value of L the less that is lost (or gained) per second so these must be in the denominator. The larger the time the more that is lost so it must be in the numerator. The negative sign in e ^ - x means that as x becomes larger ( more time taken) the answer becomes smaller. If we had 1 - e^ -x then it would asymptote to 1 not to zero. This would be the case for increasing the current to a maximum.
