
#1
Feb1113, 06:14 PM

P: 67

1. The problem statement, all variables and given/known data
http://imgur.com/9fk2QXV 2. Relevant equations V=IR 3. The attempt at a solution I am stuck on part d, i have found the answers for part c to be 12,6 and 8 for I1,I2, and I3 respectively. For part d they seem to say that I3 is just equal to I1+I2, but i am confused as to why. If the switch is opened after a long time the inductors should act like wires so it would just be a fairly simple circuit should it not? How can you tell it should be I1+I2=I3 and not I3+I2=I1? thanks for any help 



#2
Feb1113, 06:34 PM

P: 2,861

If the switch is closed for a long time energy is stored in the inductors. Remember the rule... Capacitors try to maintain constant voltage Inductors try to maintain constant current... The equation for an inductor is V = Ldi/dt. Switching the current to an inductor off suddenly means you are trying to make di/dt = ∞ so what happens to the voltage? If the inductor can't get current from the battery (because the switch has been opened) where will it come from? 



#3
Feb1113, 06:49 PM

P: 67





#4
Feb1113, 07:21 PM

HW Helper
P: 4,716

DC circuit questionThat negative sign is crucial to an understanding of what is going on. 



#5
Feb1113, 07:24 PM

P: 67




#6
Feb1113, 07:27 PM

HW Helper
P: 4,716





#7
Feb1113, 07:30 PM

HW Helper
P: 4,716





#8
Feb1113, 08:10 PM

P: 67

Would this then be the case if there are 4 parallel paths, 2 with inductors and 2 with only resistors, would the two inductorless paths share the total current coming from both of the 2 paths WITH inductors? 



#9
Feb1113, 08:36 PM

HW Helper
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#10
Feb1213, 06:28 AM

P: 2,861

When the switch is opened the inductors behave like current sources (until the magnetic energy stored in their core dissipates).
So L1 tries to maintain 12A. L2 tries to maintain 6A. Apply KCL and you get 18A flowing through R3 (or 18A using your definition for the direction). Consider what happens if there is no R3 in the circuit. They still try to maintain 12A and 6A but there is nowhere for the current to go if R3 is removed. What happens to the voltage if you open circuit a current source? This effect is actually used in DCDC step up voltage converters. 


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