3
$\begingroup$

Let's say that the earth has 0V and everything is ideal, does current still flow and why? My illustration of the circuit

$\endgroup$
  • $\begingroup$ Are you connecting one or both terminals to Earth? Is it an open or closed circuit? Can you provide a sketch? $\endgroup$ – my2cts Dec 28 '18 at 13:12
  • $\begingroup$ @my2cts Hello I added a sketch to better describe my question. $\endgroup$ – boboboy131 Dec 28 '18 at 14:07
  • $\begingroup$ That's just a short circuit so yes a current will flow. The current will be the battery EMF divided by the wire resistance plus the internal resistance of the battery. The fact you have connected the wire to earth makes no difference, $\endgroup$ – John Rennie Dec 28 '18 at 14:12
  • $\begingroup$ I am not sure if your sketch matches your question. If you remove the connection to earth the current will still flow, since the rest of the system is a closed circuit. The connection to the earth doesn't make a difference in deciding if there is a flow of a current or not. $\endgroup$ – Javatasse Dec 28 '18 at 14:32
  • $\begingroup$ This is a wrong circuit... You have created a short circuit regardless of that pointless connection to the earth. It just kills your battery :)). You should connect the electrodes to the earth separated by an interval... read my answer for more details. $\endgroup$ – Ramtin Dec 29 '18 at 1:47
0
$\begingroup$

An ideal voltage source is a two-terminal device that maintains a fixed non-zero(in case of finite circuit resistance) voltage drop across its terminals.

In your diagram, name the positive terminal $A$, the negative terminal $B$ and the earthing point $C$.

Funny thing is, if everything in the diagram is 'ideal', then is gives a contradiction. Since both $A$ and $B$ are at the same potential as $C$(which is chosen to be zero for convenience), the emf of the voltage source is just $V_A - V_B = 0$ which makes no sense(as if there is short-circuit in place of the emf source).

However, a realistic interpretation would be to assume the voltage source to be a finite source of energy, whereby, initially(before connecting the circuit), $V_A \ne 0$ and $V_B = V_C = 0$, and after connection, only for a finite time, will current flow, through the circuit $AC$ and not $BC$. Current will flow, only as long as the chemical reactions in the battery maintain a non-zero potential $V_A$.

P.S: One may argue, w.r.t some model that the potential $V_A$ can only be asymptotically 0, and never actually reach 0 in finite time. We neglect such trivialities in interpreting the overall working of the circuit. Feel free to point out any mistakes.

$\endgroup$
-3
$\begingroup$

The resistivity of the earth is about $\rho_{earth} \sim 1 k\Omega m$ (the accurate value depends on the soil).

So when connecting the battery to two electrodes separated by a distance $d$ the ground, a current flows through your circuit according to the Ohm's law: $$ I= d\ V_{battery}/\rho_{earth}$$

$\endgroup$
  • 3
    $\begingroup$ How can you give a single resistance to the Earth? What assumptions go into this? For example, how much Earth are you talking about? $\endgroup$ – Aaron Stevens Dec 28 '18 at 12:55
  • $\begingroup$ @AaronStevens Stevens I added more explanation so that you can understand it better... $\endgroup$ – Ramtin Dec 29 '18 at 1:48
  • $\begingroup$ I cannot understand why people are giving down marks... this should be the correct answer !!! $\endgroup$ – Ramtin Dec 29 '18 at 1:54
  • 1
    $\begingroup$ Because you haven't provided any explaination for why you're equations or constants are valid. You're also treating the earth as having uniform linear resistivity which doesn't make much sense. $\endgroup$ – Alex S Dec 29 '18 at 14:45

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.