How does a simple circuit work?

[tellmesomething]

Homework Statement

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Relevant Equations

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I am going to reason this out for a simple circuit with a battery (with no internal resistance) and connecting wires which can be considered perfect conductors. Please let me know if I am thinking correctly, I am quite confused by the overall working.

An isolated battery has two electrodes which are dipped in an electrolyte, the ions of the electrolyte have a force acting on them which is non electrostatic, this force helps charge the two electrodes and this ability to do work obviously comes from the chemical energy stored in the batter. Aas the electrodes get charged an electric field is developed and after some time the force on the ions due to the electrostatic nature of the electrodes and due to the non electrostatic force becomes equal.

Now you connect this to connecting wires. The positive terminal is at a higher potential and the negative terminal is at a lower potential for a proton, however since free electrons are what move the opposite is true and we get a movement of electrons from the negative terminal to the positive terminal through the wires, which means this is work done by the system i.e fall in potential energy. However we know that perfect conductors do not have field lines inside them so I am confused as to how they sense this potential and move..

Do I make any sense?

 

[Drakkith]

However we know that perfect conductors do not have field lines inside them so I am confused as to how they sense this potential and move..

Don't get confused over various simplifications you may encounter. They are simplifications and should not be expected to make sense when you look into the details of things. Obviously real conductors are not perfect and have no trouble developing an electric potential between their ends.

 

[kuruman]

However we know that perfect conductors do not have field lines inside them . . .

We know that to be the case when conductors are in static situations such as having fixed charges outside them or in cavities inside them. In these cases, the entire conductor is an equipotential and the gradient of the potential, a.k.a. the electric field, is zero at all points inside the conductor.

If the terminals of a battery are connected at two different points on the conductor, the conductor is no longer an equipotential and the electric field can no longer be zero. The potential difference drives an electric current between the points of contact. For as long as the battery is connected and does its job to maintain that potential difference the current will be there. High potential energy charge carriers are injected into the conductor at one end and low potential energy charge carriers are collected at the other end. The difference in potential energy is dissipated as heat inside the conductor. When the battery is disconnected, the electric field inside the conductor reverts back to zero.

I should mention the generation of eddy currents inside a conductor without the need for a battery by placing it inside a time-varying external magnetic field.

 

[tellmesomething]

We know that to be the case when conductors are in static situations such as having fixed charges outside them or in cavities inside them. In these cases, the entire conductor is an equipotential and the gradient of the potential, a.k.a. the electric field, is zero at all points inside the conductor.

If the terminals of a battery are connected at two different points on the conductor, the conductor is no longer an equipotential and the electric field can no longer be zero. The potential difference drives an electric current between the points of contact. For as long as the battery is connected and does its job to maintain that potential difference the current will be there. High potential energy charge carriers are injected into the conductor at one end and low potential energy charge carriers are collected at the other end. The difference in potential energy is dissipated as heat inside the conductor. When the battery is disconnected, the electric field inside the conductor reverts back to zero.

I should mention the generation of eddy currents inside a conductor without the need for a battery by placing it inside a time-varying external magnetic field.

Okay so only in electrostatic equilibrium, there are no field lines inside a conductor?

Also you say that the potential energy converts to heat when the current goes through however if there was a resistor connected in between it , circuit looking like below.

We were taught that before reaching the resistor all points have same potential as that at the positive electrode. Then while going through the resistor it drops by iR( I being the current, R being the resistance of the resistor. Then the potential becomes $V_{+ve} - iR=V_{-ve}$
But how can the current flow if the points before the resistor are "equipotential " which howsoever you established in your answer is not..

 

[kuruman]

We were taught that before reaching the resistor all points have same potential as that at the positive electrode.

That is an idealization like ignoring air resistance in projectile motion. You know it's there but it's a good approximation (and simplification) to ignore it. All conducting wires in a circuit (unless they are superconducting) have some non-zero resistance which is normally ignored. Lead wire resistance may be included in cases where it makes a difference for example when doing a four-point measurement.

 

[tellmesomething]

I

That is an idealization like ignoring air resistance in projectile motion. You know it's there but it's a good approximation (and simplification) to ignore it. All conducting wires in a circuit (unless they are superconducting) have some non-zero resistance which is normally ignored. Lead wire resistance may be included in cases where it makes a difference for example when doing a four-point measurement.

But if theres no potential difference between two points before the resistor how would the charge "move"? What force would make it move if the work done in moving from near the battery to near the resistor is zero I.e potential diffeence is zero

 

[kuruman]

There is an electric field therefore there is a potential difference.

 

[tellmesomething]

There is an electric field therefore there is a potential difference.

So there is a potential difference but we ignore it? Because its very small?

 

[tellmesomething]

Also if we extrapolate this case to a battery with an internal resistance r, would that change anything? I know that the potential difference reduces and becomes less than the emf of the cell, this would mean that the work done by the system to move it from the positive to the negative electrode is less than the work done on the system to move it from the negative electrode to the positive electrode if there were no internal resistance..? Why is that..is it because the internal resistance takes some potential energy and converts it into kinetic energy which gets converted to heat? @kuruman

 

[tellmesomething]

Also case 3 ) when we have a battery with emf 6V and we charge it by providing it a current of 2ampere and 7.2V , I dont understand how this charging works, the source creates a potential difference of 7.2 V and our charge travels through some wire and enters the battery which apparently requires 6V to move through the battery and 1.2 volts is dropped due to the resistance inside the battery? I dont get it, how is this charging of the battery
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There is an electric field therefore there is a potential difference.

 

[jbriggs444]

So there is a potential difference but we ignore it? Because its very small?

Yes. That is a good way to think about it.

Real life wires have a tiny resistance so that a tiny potential difference can result in a huge current.

We idealize this to perfect wires with zero resistance so that a zero potential difference is consistent with an arbitrarily large current.

 

[tellmesomething]

Yes. That is a good way to think about it.

Real life wires have a tiny resistance so that a tiny potential difference can result in a huge current.

We idealize this to perfect wires with zero resistance so that a zero potential difference is consistent with an arbitrarily large current.

Ohkay

 

[Drakkith]

Also case 3 ) when we have a battery with emf 6V and we charge it by providing it a current of 2ampere and 7.2V , I dont understand how this charging works, the source creates a potential difference of 7.2 V and our charge travels through some wire and enters the battery which apparently requires 6V to move through the battery and 1.2 volts is dropped due to the resistance inside the battery? I dont get it, how is this charging of the battery

There is a minimum voltage needed to charge a battery. That minimum voltage is typically more than the cell outputs on discharge. So a 12 volt lead-acid battery will need something like 12.9 volts to charge at all. Higher voltages result in a larger current and a reduced charge time, but too high of a voltage will charge the battery too fast and damage it.

Ignore the 6-volts in your post here. That's the discharge voltage. The charging process requires some voltage to function at all and the 'excess' voltage simply makes the charging reactions happen faster. In other words, a higher voltage generates a larger current and that larger current is used in the chemical reactions to charge the battery.

In terms of voltage drop, a charging battery doesn't behave any different from a resistor. If you put your battery in series with a voltage source and nothing else, then the battery MUST drop all of the voltage, as there's nothing else in the circuit to do so.