# Inductance in DC Networks

1. Apr 30, 2014

### debreets

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?

Last edited by a moderator: Apr 30, 2014
2. Apr 30, 2014

### Rellek

They are asking for the resistance of a new inductor if it has the same inductance and it is put in series with the 330 Ohm resistor. So, you have a number that is on the way to getting the right answer.

3. Apr 30, 2014

### debreets

I'm confused as I beleive iv worked out the resistance of the inductor at 363.63 ohms, so would the resistance of the coil not be the same?

4. Apr 30, 2014

### Rellek

No, because this new circuit is still factoring in the 330 Ohm resistor that is connected when you are calculating the total resistance.

5. May 2, 2014

### debreets

Ok, so, 363 (reistance of ciruit) - 330 (resistance of resistor in circuit) = 33 ohm (resistance of coil)
Think iv got it now,