An inductor of negligible resistance and an inductance of 0.2 H is connected in series with a 330 Ω resistor to a 12V d.c. supply. Determine: (a) the time constant of the circuit (b) the voltage drop across the inductor after two time constants (c) the voltage drop across the resistor after three time constants (d) the resistance of a 0.2 H coil used to replace the inductor if the circuit’s time constant falls to 0.55 ms. I have answer a) using formula t = L/R L is 0.2 and R being 330, t giving 0.0006 seconds ( or 0.6 ms ) Im unsure of b) and c) but will spend time on those shortly. Part d) looked simular to a) I thought maybe use the same formula (t=L/R) but make R the subject, that would be R=t/L. I plugged in 0.00055 for t and 0.2 for L, giving an answer of 363.63? time was given as 0.55 ms so converted to seconds giving 0.00055. Am I on the right track or have I gone wrong?