# Circuit settles to steady state and switch is opened

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1. Mar 17, 2016

### j3dwards

1. The problem statement, all variables and given/known data
The circuit consists of three identical light bulbs and two identical coils, connected to a DC current source.

The ohmic resistance of the coils is negligible. The system is left to settle into a steady state and then the switch S is opened. Describe and explain in as much detail as you can the brightness of the bulbs shortly after the switch is opened.

2. Relevant equations
V=IR?

3. The attempt at a solution
I think this is more a worded question rather than involving equations.

So I'm thinking that after the switch is opened, no more current can flow. But the current that has just left the current source, which is passing through the coils, still moves until the charge has completed its circuit and moved to the negative terminal (or in actual fact electrons move from negative to positive - but i'm considering the conventional case). So does this mean that the light bulbs still all remain at equal brightness until the current has passed?

2. Mar 17, 2016

### Staff: Mentor

Can you provide a diagram of the circuit? There are many different ways to connect seven components...

3. Mar 17, 2016

### j3dwards

Yes, I have attached the diagram drawing

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• ###### curcuit.png
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4. Mar 17, 2016

### Staff: Mentor

Okay,
Note that the supply is a battery so it's technically a voltage source, not a current source. Yes you get current from a voltage source, too, but the primary characteristic of a voltage source is that it provides a constant voltage and any amount of current required to maintain that voltage, while a current source does the opposite: it will provide a fixed current no matter how much voltage it needs to produce to do so. That's just an aside though.

When the switch opens no current can flow in the battery branch of the circuit, so no charge will be flowing back to the battery. No current can flow where there isn't a complete circuit. However, at the moment that the switch opens there will still be current flowing through the inductors and energy stored in their magnetic fields. That's what inductors do... they store energy in magnetic fields when there is current flowing through them.

What you need to do is compare the currents of interest in the circuit prior to the switch opening to those that will exist the instant after the switch opens.

5. Mar 17, 2016

### j3dwards

I'm confused - how is there a current flowing through the inductors at the moment that the switch opens if you just said no current can flow? Is this because, like you said, the coils have stored energy and this energy is producing in emf which in turn creates a current?

Also, i'm guessing that the current in the instant that the switch has opened will be much less than the original current? Hence the lights will be dimmer momentarily and then they will black out?

6. Mar 17, 2016

### Staff: Mentor

I said that there will be no current in the battery branch. There are still three other branches that have complete circuits.

Yes, the stored energy in the coils will continue to drive current. They will produce an emf that drives it.
No, it is a basic property of inductors that they refuse to change their current flow instantaneously. Inductance works for current much the way mass works for velocity: they provide inertia. Capacitors have a similar property in that they resist instantaneous changes in their potential difference.

I'd suggest that you begin by making a sketch of the circuit where you assume that the battery has some emf V and the light bulbs all have the same resistance R, and write in expressions for the currents in terms of V and R throughout the circuit when it's in steady state. Then do the same again for the instant after the switch opens.

7. May 1, 2016

### Vibhor

Hi gneill ,

Consider a complete circuit consisting of a resistor , inductor and a battery and a switch . A steady current flows through the circuit . Now the switch is opened . If the " current doesn't change instantaneously " as you have stated then EMF induced across the inductor would be zero , as E = -Ldi/dt . di = 0 but dt ≠ 0 (although very small ) . But this isn't the case . Rather a large EMF would be induced .

Sorry , if I have misunderstood you . Please help me clear this doubt .

8. May 1, 2016

### Staff: Mentor

The di and dt are infinitessimals. It's not correct to assign di=0. You can say that di approaches zero.

What happens (for ideal components) is that while the current is still essentially the original value, the ratio di/dt in that instant is infinite: The current is trying to shut off in time dt → 0.

With real-world components the result is an inductor emf sufficient to keep the current flowing, typically by forcing a path between the opening switch contacts (air dielectric breakdown and resulting arcing). This is one way that real switches can degrade over time as the contacts become pitted and charred by the arcing, causing them to exhibit higher resistance. This leads to ongoing heating issues in the contact connections, perhaps melting/fusing the contacts or causing a fire.

9. May 1, 2016

### Vibhor

Doesn't this mean current changes instantaneously and suddenly decreases to zero (when the switch is opened) ?

10. May 1, 2016

### Staff: Mentor

In practical terms it seems instantaneous. But it's really a case of a mathematical limit. This is why we take care to specify times $t = 0^-$ and $t = 0^+$ and never just $t = 0$ when analyzing the details.

11. May 1, 2016

### Vibhor

With reference to post#7 , what would be the current i in the inductor after time Δt , where Δt is the time required to open the switch ?

I believe it will be 0 . Do you agree ?

12. May 1, 2016

### Staff: Mentor

I agree, if the definition of "switch opening time" is based on the time when the current flow stops

The phrase "Δt is the time required to open the switch" is pretty vague if you think about it. What does it mean for contacts that are separating? Is the severing of contact instantaneous or gradual as the area of physical contact decreases over a short time? The latter is more likely in a real switch. What else might come into play once the contacts are physically separated?

In a real switch, during the contact separation interval the contact resistance will rise as contact area and pressure decreases and eventually physical contact is lost. But the gap between the two contacts will start out infinitessimal, so it will take a very small (read:infinitessimal) potential difference to exceed the breakdown potential of the gap; Current continues to flow. As the gap widens the potential difference rises to compensate and the arc grows longer. All this time stored energy is being lost as heat in the resistances and to a lesser extent as electromagnetic radiation from the circuit. The current will be changing over time.

At some point the arcing may stop but current will continue to flow! You see, the switch contacts, being metal plates separated by a gap, form a capacitor. So the circuit will then look like an RLC circuit that's been "kick started" with an initial current. The C will be tiny but nonzero. The result is that the circuit behaves as a damped RLC circuit, and will be either overdamped, critically damped, or underdamped, and the current will behave accordingly. There may be interrupted periods of arcing if the ringing is large, and currents can flow in either direction!

This continues until the energy available in the circuit is exhausted, lost to heat in the resistances (again, ignoring electromagnetic radiation).

So determining the current in the inductor at some particular time after the switch is opened is problematical if Δt is commensurate with what is called the "switch opening time". By defining the switch opening time to be the settling time of the circuit, then you can definitely state that after time Δt the current will be zero.