An 8.0 mH inductor and a 2.0 ohm resistor are wired in series an ideal battary. A switch in the circuit is closed at a time 0 sec. at which time current is zero the current reaches half its final value at a time of ofThe answer is 2.8 ms, How?????
Let Vo be the voltage across the battery and VL be the voltage across the inductor then for your circuit: VL Vo e-(tR/L) If you divide both sides by R you get the current relationship instead (Ohms law) VL/R Vo/R e-(tR/L) this equals:- I Io e-(tR/L) or you can simply start with this line instead. for the current to reach half its value I/Io 1/2 I/Io e-(tR/L) 1/2 e-(tR/L) Now, to get rid of the exponential part we take NATURAL logs (ln on your calculator) of both sides giving:- ln(1/2) -(tR/L) this gives t as:- t-L/R ln(1/2) t -[(8x10^-3)/2 x (-0.6931) t 0.00277 t 2.8 ms
