Current flowing in an open circuit

[laser]

Homework Statement

conceptual

Relevant Equations

V=IR

1. Consider this diagram. Does the bulb light up? The equation I=V/R tells me no, but if there is a potential difference between the two points, surely current will flow and the bulb will light? As there is nowhere for the electrons to go, I assume it will only light for a very short period of time?

2. Consider the same diagram but without a potential difference between the two points. Does the battery send current for a very very short period of time? I ask this because how does the electrons know there is no potential difference / infinite resistance. I read somewhere that the electric field travels at the speed of light to check paths and if they are open then no current will flow. But by that point some electrons would have flowed.

 

[Orodruin]

You cannot imagine the point you say to imagine has a potential to have a potential. The potential difference to the ground is zero since no current is flowing in the circuit.

 

[BvU]

'Conceptual' is a nice term...

Your circuit diagram is conceptually equivalent to a light bulb circuit with an opened switch. Any short-term effects around the moment of opening or closing the switch cannot be derived from studying this minimal circuit.

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[laser]

You cannot imagine the point you say to imagine has a potential to have a potential. The potential difference to the ground is zero since no current is flowing in the circuit.

But what if there is a potential difference between one end of the battery and the open circuit? eg by charging it

 

[BvU]

'Conceptual' is a nice term...

Your circuit diagram is conceptually equivalent to a light bulb circuit with an opened switch. Any short-term effects around the moment of opening or closing the switch cannot be derived from studying this minimal circuit.

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[laser]

Any short-term effects around the moment of opening or closing the switch cannot be derived from studying this minimal circuit.

Could you expand on that?

What would allow us to derive these short-term effects?

 

[Delta2]

The two open ends of the circuits (the one end where you say imagine that ... and the other end opposite to it) can be modeled as a capacitor with very low capacitance C.

A current will initially flow but because the capacitance is very small this capacitor will be charged immediately (I guess you know the basic of an RC circuit) so this current will have very small duration and it will start at value E/R where E the emf of battery and R the resistance of the bulb but it will decay very fast to zero (we know this happens in about t=10RC but because C is really really small this time is very small too).

BUT perhaps you had in mind another circuit where we have the positive pole of a battery connected to one end of the bulb and the other end of bulb connected to ground. There seems to be a potential difference between the two ends of bulb so why the bulb doesnt light? Is this what you really wanted to ask?

About your 2. , it is another thing the electric field and another thing the current. Electric field can exist without conducting path, but current needs a conducting path to flow.

 

[BvU]

Extensive knowledge of electromagnetism and all details of all parts of the circuit. 'All' to be taken litterally.
Conceptually we make do with simplifications (first-order approximations in the form of capacitances, (self-)inductances, dimensions, etc etc).



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[laser]

BUT perhaps you had in mind another circuit where we have the positive pole of a battery connected to one end of the bulb and the other end of bulb connected to ground. There seems to be a potential difference between the two ends of bulb so why the bulb doesnt light? Is this what you really wanted to ask?

Thanks, your reply makes sense. Regarding this, no - I just drew ground in my diagram to reference where I was taking zero potential to be. However, it is an interesting question I considered during the day. I would say that because the Earth is so big, it would ground the wire. E.g. If you ground the negative terminal, electrons would move from the negative terminal to the ground, so I guess the bulb could flicker? But not for too long!

 

[hutchphd]

Whern you draw a schematic diagram of a circuit, certain approximations are tacitly assumed by all who view it.

1. Circuit elements are isolated and ideal
2. The connecting wires are ideal (they have only resistance and it is zero)
3. We can ignore relativistic causality

Prof Walter Lewin knows that this is not trivially understood by all iin the business.