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I had a doubt. This may be really basic, but yeah.

Say there's a bulb. The bulb is connected to a battery by two wires. One wire connects the negative terminal of the battery to the bulb (let's call it wire A), while the other connects the bulb to the positive terminal of the battery (wire B). The switch connected to wire B is left open. Current was initially flowing through the circuit. Now for a tiny interval current will flow up to the bulb but then won't have a path for completion.

I know that the circuit is incomplete and that the bulb will not glow.

But electrons do flow through the bulb initially, right? What effect will that have on it?

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  • $\begingroup$ Isn't breaking "wire B" essentially what a light switch does? $\endgroup$
    – Kyle Kanos
    Commented Feb 1 at 18:39
  • $\begingroup$ I think this would be better worded as each terminal has its own switch and both are initially open. What happens if only switch A is closed? $\endgroup$
    – KDP
    Commented Feb 15 at 6:51

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Basically, the entire circuit up to the point of the break goes to the potential of the positive battery terminal. So yes, at the initial connection there is a transient response where the electrons in the conductor rearrange themselves in a way so as to form an equipotential surface of the conducting material, i.e. lead and bulb. Since both leads are attached to the battery the effect of the break will be to form a miniscule air gap capacitor in series with the bulb.

So the transient behavior will come from an analysis of the equivalent circuit which is a constant voltage source connected to a series circuit comprising a switch to stand for connection, a resistor accounting for the resistance of the leads, the bulb's resistance and the tiny air gap capacitance of the break. To be more accurate one might include any stray capacitance and inductances of the bulb itself, which will be, in general, specific to the design and manufacture of bulb. So, ignoring stray reactance of the bulb, one applies Kirchoff's voltage law to get the following equation for the equivalent series circuit: $$V(t)=RI+{1\over C}\int_{t_0}^t I(t)\; dt,$$ differentiating with respect to time yields, $${dV(t)\over dt}=R{dI(t)\over dt}+{I(t)\over C}.$$ Since the voltage $V(t)=V_B$, is a constant, one has the following first order ordinary linear differential equation with constant coefficients that may be solve for the current $I(t)$ as a function of time: $${dI(t)\over dt}+(RC)^{-1}I(t)={V_B\over R}.$$ No doubt the solution will have a damped exponential form so that there is a brief (transient) current that rapidly dies to the $I(t)=0$ steady state solution.

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  • $\begingroup$ Thanks so much for your response. $\endgroup$
    – entropy
    Commented Feb 2 at 3:46

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