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I have a circuit in where there are 3 different bulbs (circular, oval, and rectangular, the circular is the dimmest, the rectangular is more brighter, and the oval is the brightest. Now , I am wondering is the current the same through the whole circuit and is it different at each bulb? Or how does the current in each bulb compares to the current through the battery?

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Current is an amount of charge flowing every second. No new charge suddenly comes into existence during the circuit, so charge flowing from the battery must be shared.

The current can be equal in all of the branches (through all light bulbs). But what if one light bulb resists charge passing through a bit more than the others? Then the others have more charge passing through. You can set up Ohm's law for each light bulb:

$$U=R_1I_1\\U=R_2I_2\\U=R_3I_3$$

They all have the same voltage (potential difference) across them, because they are all directly attached to the battery ends. So if one has a higher resistance than another, $R_1>R_2$, then the current has to decrease to still makes Ohm's law fit, $I_1<I_2$.

Larger resistance reduces the amount of charge that passes through, if the "pressure" or "push" on the charge (the voltage) is unchanged.

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  • $\begingroup$ okay so you are saying that the current will decrease through each light bulb since there is a different resistance for each one? & what do you mean by " directly attached to the battery ends"? Also can we say that the current flowing trough the battery is the same ? $\endgroup$ – Dee Feb 24 '17 at 17:37
  • $\begingroup$ @Dee 1) Yes. 2) "directly attached to the battery ends" means that each light bulb has a wire going straight to the battery ends (without any other components in the way that would reduce the voltage). 3) No, not necessarily. Clip the wires, and no current flows; remove all the resistances and an enormous current flows (maybe high enough to melt the wires). $\endgroup$ – Steeven Feb 24 '17 at 17:41
  • $\begingroup$ Alright that makes sense , now what would be the difference between the current passing is each bulb versus the current through the battery? If you don't mind explaining ? $\endgroup$ – Dee Feb 24 '17 at 17:47
  • $\begingroup$ @Dee Good question. All the resistors correspond to one resistance, which the battery feels. And Ohm's law still holds for the battery with this equivalent resistance,$$U=R_{equivalent}I$$If you can find $R_{equivalent}$, you can find the current $I$ "through" the battery. And $R_{equivalent}$ can be calculated in steps: Are the resistors in series (one after another on the same wire), you add them up directly,$$R_{equivalent}=R_1+R_2+R_3+\cdots$$ Are they in parallel (your case), you add them up reversely,$$\frac1R_{equivalent}=\frac1R_1+\frac1R_2+\frac1R_3+\cdots$$ $\endgroup$ – Steeven Feb 24 '17 at 18:01
  • $\begingroup$ Alright so then what can we infer about the current in the battery? Would it be that the current going through the battery also decreases since it feels the resistance that is add up from all the 3 bulbs? $\endgroup$ – Dee Feb 24 '17 at 18:06

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