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As current(DC) flows from positive to negative terminal which is actually flow of electrons which in turn flow from negative to positive terminal. So the question is which side of battery starts conducting current first? According to me it should be negative side as current is flow of electrons which originate from negative terminal. Application of above concept is in following circuit where question is which bulb fuses first? According to me bulb B should fuse first

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  • $\begingroup$ The wire is 'full' of mobile electrons (even if the cell is disconnected). Electrons from the wire enter the positive plate as soon as the cell is connected so there is no need to 'wait on electrons from the negative plate to get there'. Is this what your question is about? $\endgroup$ – Alfred Centauri Jun 18 '17 at 14:42
  • $\begingroup$ So should both bulbs fuse simultaneously? And how does the fuse saves the appliances if it's so? $\endgroup$ – tejasvi88 Jun 18 '17 at 14:44
  • $\begingroup$ The point of your question is unclear; are the 'bulbs' fuses and if so, are you asking which one will open first? I suggest editing your question to provide more context. $\endgroup$ – Alfred Centauri Jun 18 '17 at 14:47
  • $\begingroup$ In my comment first fuse refers to blowing of bulb and second fuse refers to fuse as an appliance to protect from short circuit. $\endgroup$ – tejasvi88 Jun 18 '17 at 14:49
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Propogational effects of the electrical signal are only meaningful when the wire has finite capacitance and inductance. If it does then the electric signal travels according to the telegrapher's equations with a speed given by $$u=\frac{1}{\sqrt{LC}}$$ neglecting resistance. Now coming to your question, it is not just the circuit that matters but also how the electric disturbances are introduced i.e. how the terminals are connected to the circuit. If both the terminals are connected simultaenously the disturbances will travel from both ends towards the bulbs (the voltage of each disturbance will be different of course, and differ by Voltage of the battery) and since the bulbs are symmetrically placed both will light up at the same time. In general which bulb will light up first depends on the position of the bulbs and the timing of the cpnnection of each terminal. Also I must emphasise that in realistic situations this will happen negligible timescales and even wothin those timescales it will be complicated and varying waveform that exists in the wire, not a simple travelling step.

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  • $\begingroup$ Fine answer, but one question. Suppose both bulbs will blow up at voltage >10V, so in above case which bulb will blow first will depend on the procedure of initialization of circuit? $\endgroup$ – tejasvi88 Jun 18 '17 at 16:48
  • $\begingroup$ If both terminals are connected simultaneously both will blow up at exactly the same time. $\endgroup$ – alex Jun 18 '17 at 16:53
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Current flows simultaneously at all points in a circuit. It flows at both terminals at the same time.

If the bulbs were identical they would blow at the same time. However, one is likely to have slightly more resistance than the other. This bulb will heat up quicker. Heating increases the resistance which further increases the rate at which it heats up compared with the other bulb. This bulb will burn out first, breaking the circuit and stopping the flow of current. The other bulb might be intact.

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  • $\begingroup$ Good answer, precisely explains what is asked. Thanks $\endgroup$ – tejasvi88 Jun 20 '17 at 16:55

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