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.I thought that this one was easy. first time around I used i(Io)(e^-Rt/L) --- 7.1421e^(-90t/(55x10^-3)) 2.53926105e-4this is wrong. Then based on someone's advise I got on here, I used the equation i(Io)e^(t/-RL) and I got an answer of 5.34.also wrong! Please help! What am I doing wrong? What's the answer??
The power in an inductor is proportional to the rectangular of the present. At part the present it holds a million/four the volume of power. It held forty J on the starting, it now holds 10 J so the difference is 30 J
you are right i Io e^(-Rt/L) this when solved for t gives 6.5 * 10^-4 s
7.14 21 e^ - (R/L)t e^-(90/0.055) t 0.34 e^- (1636.36 t) 0.34 ln both sides : - 1636.36 t - 1.0788 hence t 6.59 x 10^-4 s
