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I have the following DC circuit. A cell and a resistor (lightbulb) is serially connected and both ends are grounded. I see the lightbulb is on and hence the current flows.

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It is because the electric potential is zero at both grounds. My question is how the earth maintains it. Is it because the earth is conducting?

It cannot be a high resistive conductor, then it forms a closed loop with large resistence and there will be a little current. Then is the earth a almost perfect conductor? Or is there any way to think about how it maintains the zero potential?

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  • $\begingroup$ Remember for resistors made out of a cylinder of uniform material, $R=\rho\frac{d}{A}$. The earth isn't really a cylinder, but even if its $\rho$ is not very low, its $A$ is quite large compared to other conductors you might be familiar with. $\endgroup$ – The Photon Sep 23 at 18:38
  • $\begingroup$ Thanks @ThePhoton. So roughly the earth has large cross sectional area making the resistivity low. $\endgroup$ – user112002 Sep 24 at 8:23
  • $\begingroup$ high cross section area makes the resistance low, not the resistivity. $\endgroup$ – The Photon Sep 24 at 14:22
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It is not clear how you made the earth connections.

If it was via soil then soil is a conductor but soil resistivity is extremely variable and depends on many factors eg moisture, temperature and chemical content.

Assuming that $3\,\rm V$ is the working voltage of your light bulb then the fact that the light bulb emits light shows that the resistance of the soil between the two earth positions is much smaller than the resistance of the light bulb.

If the connection to earth was made via the earth connections on power supplies then it is not a surprise that the resistance across the earths was low as those two earthing points would have been connected with copper wire.

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  • $\begingroup$ Thank you. The situation is as you described. So the earth is as good conductor as copper wire, and really there is current flow through the earth. It is amazing. Maybe the earth just tries to maintain charge equilibrium by redistributing the charges. $\endgroup$ – user112002 Sep 24 at 8:27

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