DC circuit question


by hahaha158
Tags: circuit
hahaha158
hahaha158 is offline
#1
Feb11-13, 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
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CWatters
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#2
Feb11-13, 06:34 PM
P: 2,861
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?
Yes and no.

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?
hahaha158
hahaha158 is offline
#3
Feb11-13, 06:49 PM
P: 67
Quote Quote by CWatters View Post
Yes and no.

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?
Do they get it from I3? In that case shouldn't I1 and I2 share the current in I3 (8A) rather than I3 get the current from I2+I1 (18A)?

NascentOxygen
NascentOxygen is offline
#4
Feb11-13, 07:21 PM
HW Helper
P: 4,715

DC circuit question


Quote Quote by hahaha158 View Post
For part d they seem to say that I3 is just equal to I1+I2, but i am confused as to why.
Noting the directions of the currents you have marked on the schematic, I hope they are saying i3 = (I1 + I2)

That negative sign is crucial to an understanding of what is going on.
hahaha158
hahaha158 is offline
#5
Feb11-13, 07:24 PM
P: 67
Quote Quote by NascentOxygen View Post
Noting the directions of the currents you have marked on the schematic, I hope they are saying i3 = (I1 + I2)

That negative sign is crucial to an understanding of what is going on.
http://imgur.com/1bWhh8A

here is the solution
NascentOxygen
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#6
Feb11-13, 07:27 PM
HW Helper
P: 4,715
Quote Quote by hahaha158 View Post
In that case shouldn't I1 and I2 share the current in I3 (8A) rather than I3 get the current from I2+I1 (18A)?
It boils down to which element is most willing to allow an instantaneous change in its current: an inductor or a resistor?
NascentOxygen
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#7
Feb11-13, 07:30 PM
HW Helper
P: 4,715
Quote Quote by hahaha158 View Post
http://imgur.com/1bWhh8A

here is the solution
As you can see, their I3 is in the opposite direction to your I3! So they won't need the negative sign, but for the direction you marked I3 you will need the negative. (Direction is just as important as magnitude when describing voltages and currents.)
hahaha158
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#8
Feb11-13, 08:10 PM
P: 67
Quote Quote by NascentOxygen View Post
It boils down to which element is most willing to allow an instantaneous change in its current: an inductor or a resistor?
So basically you are saying that because path 3 has the least overall resistance, then that will be the path of the total current?

Would this then be the case if there are 4 parallel paths, 2 with inductors and 2 with only resistors, would the two inductor-less paths share the total current coming from both of the 2 paths WITH inductors?
NascentOxygen
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#9
Feb11-13, 08:36 PM
HW Helper
P: 4,715
Quote Quote by hahaha158 View Post
So basically you are saying that because path 3 has the least overall resistance, then that will be the path of the total current?
The word "resistance" has a specific meaning in electronics, so is not appropriate for the way you have used it here. A characteristic of an inductor is that its current doesn't like to be changed rapidly. A resistor has no problems with its current changing in magnitude (or in direction) in an instant.
Would this then be the case if there are 4 parallel paths, 2 with inductors and 2 with only resistors, would the two inductor-less paths share the total current coming from both of the 2 paths WITH inductors?
Yes. And because the inductor currents don't like to change direction, their currents direction dictate the direction of current in the rest of the circuit.
CWatters
CWatters is offline
#10
Feb12-13, 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 DC-DC step up voltage converters.


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