Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wanted to ask under what conditions will charges not flow in a closed circuit. Or when is current through the circuit zero even when an EMF is applied? Like in the case of potentiometer, we say that we are measuring the emf of the battery because current through the secondary circuit is 0? So what condition are we fulfilling here so that current is 0?

share|cite|improve this question

Basically, it is infinite resistance; consider Ohm's law

$$ I = V/R $$

if you let $R$ get arbitrarily large, then the current goes to zero.

share|cite|improve this answer
Other than that? Like the case of potentiometer?Thank you – Harsh Feb 13 '13 at 14:13
In the case of the potentiometer, we assume its resistance $R_p$ is 'infinitely' large. This is not true of course, but if $R_p$ is much greater than the resistances of any other apparatus in the circuit, it's a decent approximation. You might want to check out the wikipage. – Wouter Feb 13 '13 at 15:09

Current is flow of charges - (1)

Charges flow whenever there is an external $\vec E$ i.e. an $\vec E$ besides the charge's own field .

So talking about electrostatic field $'\vec E$ $'$ which is present in DC circuits,considering ideal DC circuits, $\vec E$ is present whenever there is a non zero potential difference between two points.

Now you can use this fact to see whenever current will not flow in a circuit

Elaborating some applications

$1.$So when there is $0$ resistance in circuit, as $V=IR$, so as $R$ is $0\implies $ $V=0$. So, no current .

But you applied an external source of finite $V$ in the circuit, then what happened? The ideal circuit theory doesn't restrict that nothing can move faster than light. In ideal circuit theory, instantaneous events can occur, so as soon as you connected the terminal, in $0$ time, current flowed and all the charge on the positive plate went to the negative plate and potential difference between the two terminals of the battery vanished .

Of course, this isn't what happens in the real world, that is why we had to change the theory because the earlier theory predicted this would happen .

$2.$ A capacitor and battery in steady state, positive plate of the capacitor has the potential as the positive terminal of the battery which we don't know and same for the negative plate and negative plate of the capacitor.

Now, as there is no potential difference between plates and terminal (positive and negative both) connected respectively through conductors . So, no current will flow in the circuit.

Now a new case in which current won't flow even if $V\neq0$, is the case when $R \rightarrow \infty$ . In this case $I=0$ by Ohm's law.

Physically imagining this, the resistance of the conductor is so high that inspite of having an external $\vec E$ source which applies a force on the charged particle, the resistance is so high, that the charge particle bumps so much, it isn't able to move .

Try imagining pushing a block through a wall. Usually the block moves when there's no wall or the wall's not that strong but when the wall is infinitely rigid, then no matter how much force you apply you simply can't break through.

Of course you can say I'll apply infinite $V$ as well then .

Well in that case current can be finite as is evident from Ohm's Law that $\infty/\infty$ can be finite .

Now summing these two points.

Current won't flow

Firstly,If $V=0$ between two points or you can say when there's no $\vec E$ to accelerate the charged particles.

Second when $R\rightarrow\infty$ that means there is no way present for the charged particle to move to.

You can work out cases of potentiometer, meter bridge/wheatstone bridge . Using these two principles .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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