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Consider the following circuit: Circuit

I have a bulb connected to a battery and then the wire as shown is earthed from two points. Now, since across the bulb potential difference will be zero, no current will flow through it.
But my doubt is if suppose I current flows from positive terminal of the battery, then it will pass through earth, but why will the current return to wire?

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  • $\begingroup$ I don’t know who downvoted you but I like this question. It’s important to remember that whenever you ground something, that point can act as a source/sink in terms of current. Also remember to think about what you mean by “ground”. For example, what would you consider as the potential difference between ground and the negative part of the battery? $\endgroup$
    – Max
    Jan 1 at 18:43
  • $\begingroup$ What it would be? $\endgroup$
    – Shekhar Dangi
    Jan 2 at 2:15

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You have a loop composed entirely of wire from the positive to negative battery terminal. A "short circuit". A large current will flow. Neither the bulb nor the Earth connection are involved.

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  • $\begingroup$ What if the earth connection is a different ground? Then couldn't it also act as a source/sink for both ends? At least if all the connections are made instantaneously. Just as a response to "nor the Earth connection are involved". $\endgroup$
    – Max
    Jan 2 at 22:32
  • $\begingroup$ @Max The Earth is just another conductor, like a wire. $\endgroup$
    – John Doty
    Jan 2 at 22:56
  • $\begingroup$ If you were to take an ideal voltage meter and measure the voltage between the Earth and the negative terminal, what would it read? What bothers me is: A lightning strike causes charge to flow into the Earth right? How does this differ from connecting say a pole of the battery to the Earth? If Earth is treated as a conducting wire, the first intuition is that no charge will flow. But on the other hand, do surface charges not redistribute in the wire to cancel the initial field in the transient state? For an "infinite" conductor, wouldn't this drain the battery? $\endgroup$
    – Max
    Jan 2 at 23:26
  • $\begingroup$ @Max What a voltmeter connected between two points in your circuit would read is poorly determined. If the battery is weak, it'll read close to zero for any two points. If the battery is strong enough to make this untrue, the wire will probably melt before you can make the measurement. $\endgroup$
    – John Doty
    Jan 3 at 1:53
  • $\begingroup$ @Max You should get a voltmeter, make some circuits, and try to understand them. For safety, use a weak battery (alkaline is good) or a current-limited power supply. No lithium. Analysis on paper without experience with reality is confusing. $\endgroup$
    – John Doty
    Jan 3 at 1:58
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The earth connection is meaningless. You are putting a direct short across the battery so the potential will be essentially zero.

Now if you had connections to earth at two locations, as in the diagram below, then there would be a potential difference across the bulb, the amount depending on the relative resistance of the path through earth at the location, resistance of the bulb, and internal battery resistance.

Hope this helps.

enter image description here

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  • $\begingroup$ Will the entire current return to wire in the second case? $\endgroup$
    – Shekhar Dangi
    Jan 2 at 2:17
  • $\begingroup$ I don't understand what you are asking. What do you mean by "entire current"? And what is the "second case"? $\endgroup$
    – Bob D
    Jan 2 at 2:23
  • $\begingroup$ I mean the case when two points are separately earthed. And I am trying to ask when it is earthed why would the current come back to wire? Can't it just flow into earth? $\endgroup$
    – Shekhar Dangi
    Jan 2 at 3:20
  • $\begingroup$ Because the electric field produced by the battery in the earth acting on the electrons is directed between the two grounding electrodes. Assuming the soil conductivity is locally uniform, the path of least resistance is directly between the electrodes. $\endgroup$
    – Bob D
    Jan 2 at 13:20

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