An inductor that has a resistance of 80 k-Ohms is connected to an ideal battery of 84V. 0.8 milliseconds after the switch is thrown the current in the circuit is .5355mA. Calculate the inductance. Answer in Units of H.I've been having a serious problem with this question. I figured out that there's a voltage drop (the voltage drops to 42.84V). But i'm not entirely sure where to go from there (or if that's even right!). Any help is greatly appreciated as this is an important question that'll be on my final exam.Thanks.
after the switch is thrown ???? what is this language? is it switched on or off R-L circuitswitched on equation of emf isRI + L dI/dt E LdI/dt [E - RI] LdI/dt R [(E/R) - I] dI /[(E/R) - I] [R/L] dt its solution is [at t0, I 0 (switched on) I (E/R)[1 - exp(- R*t/L)] ------------------- (1) this is how currentgrows in LR circuit R 80000 ohm, E 84 V, t 0.8*10^-3 sec, I 0.5355*10^-3 A 0.5355*10^-3 [84/80000][1 - exp(- 80000*0.8*10^-3/L)] 0.51 [1 - exp(- 64/L)] exp(- 64/L) 1 - 0.51 0.49 flipexp(64/L) 1/0.49 2.041 taking natural log 64/L ln[2.041] 0.7134 L 64/0.7134 L 89.7 H
