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Help understanding current through series circuit 
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#1
Sep2410, 05:29 PM

P: 106

This isn't a specific homework question, it's just something I'm having trouble understanding or visualising. Take this L.E.D. circuit for example, which works according to our textbook:
The current flows from the  to the +, so clockwise in this circuit. To me, it looks like the current would leave the battery and 12V would flow straight to the LED light and melt it before it gets a chance to reach the resistor. So based on that, I'd think that this circuit wouldn't work... but I also know that current is equal at all points in a series circuit, so the resistor must have affected the current before it reaches the LED? If I worked it out right then the resistor has 10.5V and 52.5Ω. I don't understand how it can take 10.5V and lower the current if the LED is before it in the circuit, can someone please explain this? I'm sure I'm just thinking about it the wrong way! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Sep2510, 02:37 AM

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First: the current flows out from the positive terminal of the battery and in at the negative one, so it is anticlockwise in the circuit shown.
Do not visualize the current as a car, reaching at different points of the circuit at different times. It is more similar to a long train all round the loop. When the engine starts, it pulls the first carriage, but as soon it starts to move, it pulls the next one and so on... practically, the whole train starts to move at the same time. When you switch on a battery, the electrons near the negative pole of the battery are pushed away and get closer to the electrons farther away and push them, those will push the electrons in front of them and so on, round the loop to the positive pole where the excess electrons are absorbed by the battery. Meanwhile new electrons enter at the negative pole, so the electron density does not change in time , only "the push" travels along the circuit: The potential difference between the terminals sets up an electric field in the loop almost without time delay: it is the electromagnetic field that travels, not the electrons, they are set into motion by the electric field.The current is the same at every point of a series circuit. In the LED, there is mechanism at the pn junction which allows a certain current and voltage to set up. Do not worry about it, just use the given data. ehild 


#3
Sep2510, 08:06 AM

P: 106

So to sum up, connecting the + and  terminals of a battery together creates an electromagnetic field around the whole circuit, and it's this field that causes the electrons to begin moving. The field is created almost instantly around the circuit, causing all electrons to be "pushed" at the same time when the battery is switched on. (which explains why the current is equal at all points). Is this correct? Also, you mentioned the flow is anticlockwise. You later said the electrons are pushed from the  terminal towards the + terminal, so wouldn't that mean the current is flowing clockwise? I was also taught that it flows from  to +. 


#4
Sep2510, 08:57 AM

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Help understanding current through series circuit
The flow of electrons is clockwise, but by definition, the electric current is the amount of charge flowing through the crosssection of the conductor in unit time. Charge is not the same as charged particles.
Imagine you have two boxes connected by a pipe, and the boxes contain positively and negatively charged particles. Removing positive particles with charge q from the left box and making them travel through the pipe to the right box, the right box gains q positive charge and the left one loses q charge, that is, it becomes more negative as before. q charge was transferred from the left box to the right one, by flowing through the pipe. If all this happened in t s, we say that the I=q/t current flew through the pipe. Assume that negative particles move from the right box to the left one. The particles move in the opposite direction as before, but the result is the same: the left box becomes more negative and the right box more positive. We can say that q charge was transferred from the left box to the right one, the current is I=q/t and flows in the direction of the charge transfer, from left to right. ehild 


#5
Sep2510, 09:51 AM

P: 106

If I understood that correctly, then the charge is said to transfer in the direction of which ever side becomes more positive? The current is just a measure of the amount of charge flowing through a section over a given time (or I=Q/t), and not really something that's flowing in any direction?
So the current isn't really flowing clockwise like I thought, it's the electrons that are being pushed clockwise by the electromagnetic field, and the current is just a measurement of how quickly the charge is being transfered. 


#6
Sep2510, 12:26 PM

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Yes, you got it!
ehild 


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