Circuit Question Involving Resistors and Inductors

1. Aug 5, 2014

Jordan1361

1. The problem statement, all variables and given/known data

For the circuit below, if L=2.0 H, emf=10 V, R1=10 ohms, R2= 20 ohms, what is the current through each component

a) at the instant the switch is closed
b)a long time after the switch is closed
c) If after a long time the switch is reopened, what is the time constant for the current?

2. Relevant equations

V=IR

3. The attempt at a solution

For a) I think the resistors are in series so equivalent resistance would be 30 ohms. Then I used V=IR to find a current of 0.33 A.

Could someone please check if a) is right and show me the steps to do the other parts. Thanks!

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Last edited by a moderator: Aug 5, 2014
2. Aug 5, 2014

Staff: Mentor

The current you calculated for right after the switch is closed is correct. What about the current a long time after the switch is closed? What is an inductor's impedance at DC?

3. Aug 5, 2014

Jordan1361

After a long time, the inductor will act like a piece of wire with no resistance so current would still be flowing through R1 but zero current would be in R2. The current in R1 would be I=10/10=1 A. Am I correct?

For part c, I know that the time constant is tau=L/R so do I just use L=2.0 H and R=0 ohms?

4. Aug 5, 2014

Staff: Mentor

Yes, very good.

When the switch is opened, what is in parallel with the inductor? The inductor current cannot change instantly, so it will decay with what time constant?

5. Aug 5, 2014

Jordan1361

R2 is in parallel with the inductor. I know that current decrease is represented by
I(final)=I(initial)*e^(-t/τ).

6. Aug 5, 2014

Looks good!

7. Aug 5, 2014

Jordan1361

So the time constant is τ=-t/ln(Ifinal/Iinitial)
There is no value for part c?