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I am a beginner in electronics and i am learning on my own. I am currently reading this book "Electronics for dummies" and after reading some pages of this book i have a LOT of questions. I know about conventional and electronic current but what i don't understand is why do we still use conventional current to model circuits. For example, take this circuit that was shown in the book

Current Flow Demonstration

As we can see conventional current is used in this figure. The book stated that it doesn't make any difference whether you use conventional or electronic current, the result will be the same. Here are some of my questions in response to this:

  1. Assume that a circuit was built exactly as modeled in this figure. If we open the switch and make this circuit an open circuit the bulb should still light up since electrons will flow from the negative terminal to the bulb and there is no switch to stop the electrons from reaching the bulb.This is one of my objection to using conventional current in modeling circuits.
  2. Now assume that there's a resistor in place of that switch and an LED in place of that bulb (i am sure i saw such a circuit somewhere). I can suppose that the resistor is there to prevent high voltage from reaching the LED so that it would not fry but again assuming that we make a circuit exactly like explained above the LED would still fry up because the current would reach the LED first and then the resistor. So why place a resistor up near the positive terminal, it should have been placed near the negative terminal so that the resistor dim the current before it reaches the LED.

I am sorry if i asked something stupid or childish but these are genuine questions that i have. Also, it would be helpful if you provide me with some additional resources to learn electronics. Thank you :)

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  • $\begingroup$ I'm not quite sure that I understood your question, but if it helps, almost all our formulas for electromagnetism, along with other less common concepts, use sign conventions from conventional current, not electronic current(if the term you used for the flow of electrons is true; I've never heard it before). $\endgroup$ – user191954 Jun 10 '18 at 9:55
  • $\begingroup$ But why do they do so? This was my primary question $\endgroup$ – Muneeb Jun 10 '18 at 9:58
  • $\begingroup$ Related: physics.stackexchange.com/q/17109/2451 $\endgroup$ – Qmechanic Jun 10 '18 at 10:02
  • $\begingroup$ @Muneeb It's historical: when Benjamin Franklin was studying static electricity, he declared that positive charges were flowing in a certain direction, but later we found that it was actually negative charges going in the other direction. It was just that he had no way of knowing and hence coined the terminology and decided the conventions in a manner which we now find counterintuitive. There's no physical significance. $\endgroup$ – user191954 Jun 10 '18 at 10:11
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Good fundamental questions here describing the usual doubts people reach in electronics.

why do we still use conventional current to model circuits

Keep this in mind: Negative charges moving one way correspond to positive charges moving the other way.

The current arrow represents positive charge flow. Always.

You are in the beginning of electronics studies and only encounter rather simple circuits at the moment. Therefore you only see electronic current (i.e. electrons being charge-carriers). But there are many other possible charge-carriers:

  • negative electrons in metallic conductors (metal wires, usual circuitry),
  • positive "holes" in semiconductors (PV solar panels, thermocouples, transistors),
  • positive and negative ions in a mix in conductive fluids or electrolytes (batteries, fuel cells, the human body),
  • etc.

All such charge-carriers can appear in an electric circuit. A mix of positive charge moving one way and negative the other way in different parts. Luckily, a negative charge moving one way always corresponds to a positive moving to other way - an electron leaving a spot leaves that spot more positive than it was before, corresponding to it gaining a net positive charge. So, due to this equivalency, someone has back in time decided to simplify all talk about current and chose that whenever we talk about current, we mean the direction a positive charge would move.

This same consensus was made for several other topics as well, such as electric field direction, magnetic field direction etc.

  1. [...] If we open the switch and make this circuit an open circuit the bulb should still light up since electrons will flow from the negative terminal to the bulb and there is no switch to stop the electrons from reaching the bulb.

This is true - for a very short while. When the battery (voltage source) is turned on, electrons will move from negative terminal and as far away as they can. Through the bulb, yes.

But soon the first electron reaches the end and can move no further. More and more arrive and accumulate there. As you know, like charges repel each other, so the more that accumulate there, the more do they prevent further electrons from arriving. This slows down the current until it stops.

This is the reason no steady current can flow in an open circuit. Current can flow momentarily, but not at any constant steady rate. And the momentary flow in the beginning might not cause enough power on the light bulb filament to heat it up to glow.

  1. Now assume that there's a resistor in place of that switch and an LED in place of that bulb [..]. I can suppose that the resistor is there to prevent high voltage from reaching the LED so that it would not fry but again assuming that we make a circuit exactly like explained above the LED would still fry up because the current would reach the LED first and then the resistor.

Again, a LED and a light bulb will both "fry" (the electrode/filament will melt) at a too high temperature. Although large current happens momentarily, and although large power may happen momentarily as a result of that, there is not enough energy transferred to heat up the filaments to the limiting temperature. Or at least not for long enough time for the filament to actually melt.

When light bulbs do burn out, though, it usually does happen when turning it on of this exact reason with momentarily large currents. You may have experienced turning on a lamp just to see it light up in a big flash and blowing right away.

So why place a resistor up near the positive terminal, it should have been placed near the negative terminal so that the resistor dim the current before it reaches the LED.

There is no difference between the positive and negative "side" of the electric circuit. While electrons will be moving away from the negative terminal and follow the description I made above, they will in a similar way move towards the positive terminal at the other side. Remember that electrons are already present in the metallic wire before hand; they are not "sent out" from the battery only. So the exact same explanation and effect can be used there. There is no "safe area" in such a circuit.

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I'm having trouble understanding the fine details of your question, but there are a couple of points I would like to stress.

  1. A circuit must be continuous for any current to flow. Current only flows if there is a continuous path from one pole of the battery to the other. If you open the switch, no current will flow anywhere in the circuit. It will not flow from one from one pole of the battery to the lamp.

  2. For simple circuits like yours, the current is identical at every point in the circuit. Hence, inserting a resistor anywhere will reduce the current everywhere in the circuit.

  3. Whether you are looking at conventional current or real current (negative electrons) my first two points are always valid. Hence it does not matter what type of current you use.

Note that with circuits that are more complicated and have multiple branches, the currents in the branches will not necessarily be the same.

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  • $\begingroup$ If you think mine is the correct answer to your question, then please accept it by clicking the check-mark next to it. $\endgroup$ – hdhondt Jun 10 '18 at 10:30
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The circuit must be closed for the electrons to flow along the wires: if you leave the switch open, there is no current at all passing through the bulb, so it doesn't light up.

The electrons flow toward the point where the electrostatic potential is higher: that point is the positive terminal of the battery or of the generator (whatever that is). If the two terminals of the battery are not connected, the electrons do not "see" the point with higher potential, so they don't travel in that direction; in fact in an ideal case* they don't flow at all, and the bulb stays off.

You could apply the same reasoning to the second case: if the current would flow opposite to the direction of the LED (since it allows the current to flow only in one of the two directions, depending on how it is oriented) then the charges cannot pass through, so in this case you can replace the LED by an open switch.

Therefore you can see that no matter which convention of the current you choose, the electric charges cannot flow along the circuit either way.


*The ideal case I refer to is the one where the positive terminal is disconnected and there's no way the electrons can, say, "jump" through the air between the loose end of the wire and the battery to close the circuit.

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  • $\begingroup$ Sorry but i didn't understand one of your points. You said electrons flow towards the point where potential is higher but isn't the negative terminal of battery at higher potential? $\endgroup$ – Muneeb Jun 10 '18 at 10:04
  • $\begingroup$ If we're talking about the electrostatic potential (not the potential energy) then no, the negative terminal has a lower potential, as the name suggests. But if we're talking about the potential energy, then we have to take the potential multiplied by the charge of the electron, that is a negative quantity: therefore we find that at the negative terminal the electrons have a higher potential energy. The electrons flow toward the point where the potential is higher or the potential energy is lower, but that's the same thing. $\endgroup$ – yellowquark Jun 10 '18 at 10:08

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