# Currents in an series electric circuit

1. Jan 26, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data
Is the current at all points the same? If not then what is the order from highest to the lowest?

http://img141.imageshack.us/img141/2418/currenthn9.th.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
2. Jan 26, 2009

### AEM

You didn't really identify those nice round circles for us, but let me ask you this question: If the current weren't the same in all parts of the circuit, where would it go? Can it just disappear without a trace?

3. Jan 26, 2009

### -EquinoX-

well the circles are light bulbs, sorry for the drawings.. hmm.. the currents can be not the same if it's in parallel right? in the end points of the battery of course it will be the same

4. Jan 26, 2009

### LowlyPion

The current could not be the same if electrons were being radiated, but I think it's only photons that emit when electrons change state. On the presumption that your bulbs retain all the electrons that might escape the filament inside its glass bound vacuum, then I would think there is no current loss from the bulbs.

Being in series suggests that the current must be the same along the path of the current flow doesn't it?

5. Jan 26, 2009

### AEM

Yes, if the light bulbs have different resistance, and they are in different branches of a parallel circuit the current will be different through each bulb. (We're neglecting any losses of electrons that might "boil off" as Lowlypion mentioned). As you point out, the total current returning to the battery will be the same as the current leaving the battery.

In a series circuit we take the current to be the same through each circuit element, while in a parallel circuit we take the voltage drop across each circuit element to be the same.

6. Jan 26, 2009

### -EquinoX-

I am just asking this question in general, the question basically is whether if in a series circuit will all current have the same values, say that i pick two random points in the circuit

Last edited: Jan 26, 2009
7. Jan 26, 2009

### AEM

8. Jan 26, 2009

### -EquinoX-

one more follow up question, the two bulbs shown above is set in series, what if it's set in paralell? will the bulb have the same brightness as the one set in series?

9. Jan 27, 2009

### AEM

With just two bulbs, you can only have them in series or in parallel. Think of the circuit as being made up of loops hooked together. Each loop is a branch of the circuit. With parallel circuits you have more than one loop (or branch). If you have two identical bulbs connected in parallel they will have the same brightness. If this is not clear, make a drawing to go with your last question and try it again.

10. Jan 27, 2009

### -EquinoX-

yes I know the one in parallel will have the same brightness, however is it the same brightness as the one with one bulb in series

11. Jan 27, 2009

### AEM

I think you are asking this question: If I have two circuits --one with one light bulb connected to a battery (this is a series hook up)-- and one circuit with two bulbs in parallel (you have to have two elements to have a parallel circuit), then Will all bulbs be of equal brightness?

If that is your question, then the answer is yes. The brightness is related to the Power (energy/time) delivered to the bulbs and is given by $$P = i^2 R$$ where i is the current and R is the resistance. In this case the same current will flow through all three bulbs.

But maybe you're asking this question: I have two circuits -- one with two bulbs in series and one with two bulbs in parallel, will all bulbs be of equal brightness? Here the answer is no. There will be less current flowing through the bulbs in the series circuit and so the energy per unit time that is converted into heat and thus into light by the bulbs will be less. The current through the bulbs in the series circuit is given by the voltage (of the battery) divided by the total resistance. The total resistance of the two bulbs will equal to 2R. On the other hand in the parallel circuit the current through each bulb will be I = V/R and so will be twice as large so these bulbs will be brighter.

( I would have put this into better equations but was having trouble with tex)