# Homework Help: Inductor circuit

1. Apr 15, 2017

### superslow991

1. The problem statement, all variables and given/known data
For the circuit shown in the figure, the inductors have no appreciable resistance and the switch has been open for a very long time.

(Image of circuit)

(a) The instant after closing the switch, what is the current through the 60.0-Ω resistor?
(b) The instant after closing the switch, what is the potential difference across the 15.0-mH inductor?
(c) After the switch has been closed and left closed for a very long time, what is the potential drop across the 60.0-Ω resistor?

2. Relevant equations
V=IR
I=V/R

3. The attempt at a solution
Part A- i assume since we are finding the current through the 60 ohm resister the 10 and 60 ohm resistors will be in series so for the current i just use I=V/R? I ignore both inductors and the 30 ohm resistor right? if so why? I also believe the circuit will act like an open circuit but im at a loss for that.

Part B- Apparently the potential difference between the 15 mH inductor is equivalent to the potential difference between the 60 ohm resistor but why is that so?

Part C- I realize after some time has passed when the switch is closed the current will not fluctuate as much so there would be no change in current and the inductor behaved like a short circuit where no voltage is induced so the voltage would be 0? still not sure on this.

Last edited: Apr 15, 2017
2. Apr 15, 2017

### kuruman

Sorry, there is no figure to look at.

3. Apr 15, 2017

### superslow991

Really? you try copy and pasting the gyazo link?

4. Apr 15, 2017

### Staff: Mentor

5. Apr 15, 2017

### superslow991

Fixed thanks

6. Apr 15, 2017

### kuruman

Thank you for fixing the figure.
(a) What is the expression for the current through a charging inductor? What happens at t = 0?
(b) What is the voltage across the 30 Ω resistor? Hint: How much current is flowing through it?
(c) If dI/dt is zero, the voltage across an inductor is zero. That means you can replace it with a short.

7. Apr 15, 2017

### superslow991

(a)- I figure I = V/R and if t=0 doesnt that mean the circuit is open?
(b)- I think no current goes through the 30 ohm resistor so there potential difference is 0?

8. Apr 15, 2017

### kuruman

Correct on both counts. A charging inductor acts like an open circuit at t = 0 and like a short a long time later.

On edit: To clarify, in (b) the potential difference across the 30 Ω resistor is zero. If that's the case, the potential difference across the 15.0 mH inductor is the same as ... ?

Last edited: Apr 15, 2017
9. Apr 15, 2017

### superslow991

the same as the 60 ohm resistor?

Also im not sure why no current goes through the 30 ohm resistor.

Last edited: Apr 15, 2017
10. Apr 16, 2017

### cnh1995

Yes.
Only at t=0, because

11. Apr 16, 2017

### superslow991

ok Thanks but what about for part b why would the potential difference across the 15 mH be the same for the 60 ohm resistor?

12. Apr 16, 2017

### cnh1995

Because the voltage across the 30 ohm resistor is zero. It can be treated as a wire with zero resistance.

13. Apr 16, 2017

### superslow991

I get that but like what's the correlation with the 70 ohm resistor and the 15 mH?

14. Apr 16, 2017

### kuruman

You mean the 60 ohm resistor. Look at the circuit imagining that the 30 ohm resistor is replaced with a straight wire. How is the voltage across the inductor related to the voltage across the 60 ohm resistor?

15. Apr 16, 2017

### cnh1995

Replace the inductors with open switches and apply KVL to the rightmost loop containing the 60 ohm and 30 ohm resistances and the open switch (15mH inductance).

16. Apr 16, 2017

### superslow991

Parallel?

17. Apr 16, 2017

### kuruman

Yes, so how does it compare with the voltage across the 60 ohm?

18. Apr 16, 2017

### superslow991

So if it's parallel they should have the same resistance so it would just tact on the 60 ohm resistor?

19. Apr 16, 2017

### cnh1995

20. Apr 16, 2017

### superslow991

Oh ok thanks a lot for the help