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I think the current will go through earth wire but my physics teacher told us the current will continue to flow through the circuit.


closed as off-topic by John Rennie, user197851, user191954, Qmechanic Mar 5 at 8:52

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    $\begingroup$ Hi Qiao and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. $\endgroup$ – John Rennie Mar 5 at 6:52
  • $\begingroup$ Does D says the potential becomes negative? $\endgroup$ – TechDroid Mar 5 at 7:03
  • $\begingroup$ No it doesn't. How much do you know about electrostatics. Did your physics teacher explain in details what an electric potential means and how it works? $\endgroup$ – TechDroid Mar 5 at 7:06
  • $\begingroup$ the potential at earth wire is 0 $\endgroup$ – Qiao Yunxi Mar 5 at 7:08
  • $\begingroup$ For me to answer this question, you have to rephrase the question conceptually so it doesn't seem like you're asking us to solve your homework for you. $\endgroup$ – TechDroid Mar 5 at 7:16

Current in a circuit is driven by voltage differences. Imagine you would disconnect the point Q from ground. Then it is clear that the battery drives a current $I = 0.5$ A from "+" to "-" through the resistors, and there is a voltage drop $\Delta U = R\times I$ across each resistor in the circuit. However, in this setting, it is undefined, what the potential difference between point Q and ground is. It could be any arbitrary value.

When you connect Q to ground, then you define the potential difference between this point and ground to be zero. This will not affect the voltage difference between "+" and "-" of the battery, hence the current in the circuit stays the same. But it defines the potential between "+" and ground, and between "-" and ground. For example, without the ground connection the voltage between "+" (point P) and ground could be 110 V, and that between "-" (point S) and ground could be 100 V. With the ground connection, the voltage between "+" (point P) and ground is +2 V, and the voltage between "-" (point S) and ground it is -8 V, such that the voltage between point Q and ground is zero.

  • $\begingroup$ and then, because the voltage between point Q and the ground is zero, no current flows along that wire. $\endgroup$ – RogerJBarlow Mar 5 at 10:58
  • $\begingroup$ @RogerJBarlow: well, I did not want to say that, because it may be misleading. If you take a point slightly left of Q and one slightly right of Q, there is also no voltage, but there is still a finite current. $\endgroup$ – flaudemus Mar 5 at 11:55
  • $\begingroup$ Yes, you're right, I was trying to be helpful but didn't appreciate the subtlety $\endgroup$ – RogerJBarlow Mar 5 at 20:13

It appears that you are asking why it is that the answer is not option $C$ on the assumption that if the current starts at the positive terminal of the battery (node $P$) it then flows through the $4\,\Omega$ resistor and out through the earth contact at node $Q$ so no current flows through the other two resistors?

This logic is flawed.

If this is a conventional circuit then the current is actually a flow of electrons coming out of the negative terminal of the battery and flow from right to left through the resistors.
Using your logic there would be no current through the $4\,\Omega$ resistor!

However it is the conservation of charge (Kirchhoff's first law) which gives the simple answer.

The current out of the positive terminal of the battery must equal the current in to the negative terminal of the battery as there can be no accumulation or loss of charge within the battery.

If some current leaks out of the circuit through the earth connection at node $Q$ how does that current get back into the circuit to make sure that the currents in and out of the battery are the same?

In this problem the purpose of the earth label at node $Q$ is to give you a value for the potential of node $Q$ so that then the potentials of the other nodes can be assigned.


I think this is an exercise in high school.

Your question is why does not the current leave the circuit and enter the ground. And the answer is that the current will continue to flow through the circuit.

If we ignore how the battery works and treated it as a device that keeps an invariant potential difference between these two ends. Then there is a static electric field near the battery, no matter if there is the circuit. Then you can imagine that put the circuit to connect two ends of the battery, together with the earth connected with the Q point. We know that there are "free" electrons in the circuit (metal wire) and the earth. They will be driven by the electric field of battery, so the current has been constructed. The system will arrive at the "equilibrium" (in fact static) state, if there is current that is no zero in the circuit, it should be continuous, that is to say in any time interval, the number of electrons who leave the battery is equal to the number of electrons who enter it. So there are only two possible states, the current is zero in the whole circuit or circuit does not leave the circuit and enter the earth. We consider these two states now:

  1. No current in the circuit means any two points on the wire are at the same potential because it is a conductor. But we know that it is impossible for there is a fixed 10V difference between the two points of the battery.

  2. This is the only possible state the circuit should be at. And is the answer to this exercise.

However, you may ask, why the current should be continuous when the system get static? That is because if it does not, the battery will accumulate the charge by time or lose charge. If it accumulates the charge and gets a higher potential than other parts, the electron (with the negative charge) will be attracted to the battery (Coulomb interaction) and then it cannot gather more charge, vice versa. This looks like the current is continuous in macroscopic.

You may also ask, why the system must get the static state. And the answer is similar to the one to the question above, there is a negative feedback in the system because of the electric interaction. The point is that in metal (in fact the conductor), the electric field will drive the electrons move. In the classical view, electrons meet the ions and the interaction between them is similar to the damping of the water flow. The electric field driving and the damping force lead the system to static at last. It is similar to the behavior of 2nd order equation: $x''=F-x'$ , or the motion of an object under wind resistance.

Finally, why there should be earth connects to the point Q? In high school exam or homework, it performs the reference level of the potential: 0V. Sometimes it is also an infinity charge pool. You may meet such exercise in the future study and it should not bother you.

  • $\begingroup$ I'm sorry to bother but your answer is a bit of a jumble and I don't feel good down voting. Where does the static come from? There are some of your statements that are right, but overall, it's not. $\endgroup$ – TechDroid Mar 5 at 8:10
  • $\begingroup$ Sorry I did not explain how the system gets static detailly enough but skip it with the word "negative feedback". I edit it to show it better in the answer! $\endgroup$ – Zhenduo Wang Mar 5 at 8:21
  • $\begingroup$ By static you mean the battery is emptied and current doesn't flow anymore? $\endgroup$ – TechDroid Mar 5 at 8:38
  • $\begingroup$ No I mean the current is not dependent on time. Maybe I should use the word "steady". $\endgroup$ – Zhenduo Wang Mar 5 at 8:44
  • $\begingroup$ Does that relate in any way to the question? The question, from my comprehension, is based on potential of Q relative to the negative terminal of the battery, how does that have to do with time, steadiness, or "static". $\endgroup$ – TechDroid Mar 5 at 9:11

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