We have the following circuit:

enter image description here

A neon lamp and a inductor are connected in parallel to a battery of 1.5 $V$. The inductor has a 1000 loops, a length of $5.0 cm$, an area of $12cm^2$ and a resistance of $3.2 \Omega$. The lamp shines when the voltage is $\geq 80V$.

  • When the switch is closed, $B$ in the inductor is $1.2\times 10^{-2} T$.

  • The flux then is $1.4 \times 10^{-5} Wb$

(calculated myself, both approximations).

You open the switch. During $1.0 \times 10^{-4} s$ there is induction. Calculate how big the current through the lamp is.

My textbook provides me with the following answer:

$U_{ind} = 1000 . 1.4 \times 10^{-5} / 1.0 \times 10^{-4} = 1.4 \times 10^{2} V$.

$ I = U/R_{tot} = 1.4 \times 10^{2} / (3.2+1.2) = 32A$

My concerns:

  • How do we know that $1.4 \times 10^{-5}$ is $|\Delta \phi|$? This is the flux in the inductor while the switch is closed, but when you open it doesn't induction increase/decrease the flux? Or will the flux just become 0 and hence give us $1.4 \times 10^{-5}$ ?

  • Why do we have to take the $R_{tot}$? What does the resistance of the inductor have to do with the lamp?

p.s. - This question can't be asked on electronics SE, since their site doesn't allow for such a question.

  • $\begingroup$ @PatEugene Haha I'm quite certain we are not supposed to use a time dependent differential equation, at least not for a couple of years! $\endgroup$ Commented Jan 18, 2013 at 23:08
  • $\begingroup$ Yeah ok so this problem, is like I said a little silly. It seems like you have to assume the current drops to zero in the given time and therefore so does the flux. This gives you the first part. $\endgroup$
    – PatEugene
    Commented Jan 18, 2013 at 23:40

3 Answers 3


When you close the switch the inductor "charges", gaining magnetic energy and hence an associated flux. When you open the switch, there is a potential energy associated with the inductor, and hence it will "discharge", generating a current in the circuit. So under the assumption that all the flux discharges, then $\Delta \phi$ will be $1.4 \times 10^{-5}$ $Wb$.

Now that there is a current flowing in the circuit, the current will see all the resistances in the circuit, not just the ones in front of it (since the circuit is closed and the sums of the sources and potential drops around the whole circuit must be zero.)

One can consider only the resistance of the lamp if the resistance of the inductor was zero. But since it has a finite resistance (you could think of it like the internal resistance of a battery) you will have to consider the internal resistance of the inductor in series with the resistance of the lamp.

  • $\begingroup$ But that means that the resistance through the loop is equal to the resistance through the lamp? If so, does this always hold for parallel components? In other words, do parallel components always have the same resistance? $\endgroup$ Commented Jan 22, 2013 at 22:11
  • $\begingroup$ Wait, I might get it now: if the switch is open the current flows through the loop and lamp? $\endgroup$ Commented Jan 22, 2013 at 22:26
  • $\begingroup$ I think I get the logic! $\endgroup$ Commented Jan 22, 2013 at 22:26
  • $\begingroup$ Wait, does your final comment mean the question is wrongly stated? $\endgroup$ Commented Jan 22, 2013 at 22:30
  • $\begingroup$ @YlykCoitus - You're confusing yourself. The resistance of the inductor and of the lamp are in series, not parallel. Therefore they should be added up, like any two series resistances. I just thought it would be easier to think of it as a perfect inductor (zero resistance) in series with a resistance, since your inductor does have a resistance. When the switch is open, current doesn't flow through the bottom part of the circuit (with the battery), so the only source of voltage is the energy that was stored in the inductor. contd in next comment. $\endgroup$
    – Kitchi
    Commented Jan 23, 2013 at 5:31

Yeah ok so this problem, is like I said a little silly. It seems like you have to assume the current drops to zero in the given time and therefore so does the flux. This gives you the first part, the induced voltage across the inductor. For the second part, it seems we simply have to apply Kirchoff's first loop rule and Ohm's law to find the current in the loop. This all seems very odd to me, because we are assuming the current is changing in order to induce a voltage, but also a single value for the current. Really, the current should be time dependent and the induction occurs for all time, not simply a finite amount. For the sake of completing this homework problem, we are done, but in reality we have to solve a differential equation and end up with exponential behavior.

  • $\begingroup$ So the reason we have to take $R_{tot}$ is because 'Kirchoff's first loop rule'? I still don't quite understand why you have to take the total resistance if you want to calculate the current through the lamp $\endgroup$ Commented Jan 19, 2013 at 9:01

I think the other answers cover your question about the change in flux. As to why $U/R_{tot}$ is used instead of $U/R_{lamp}$... it's not quite clear whether you're expected to find the current at a particular point on the circuit, or what point that might be. If I had to pick one current to characterize the circuit, it would be the current through a bit of wire that didn't have any parallel counterparts (or equivalently the sum of the currents through a set of parallel wires). The voltage drop has to be equal across both the lamp and the inductor, so the current is just $I_{tot}=V/R_{tot}$. If after this you're interested in the current through the lamp, you can find it easily by calculating the current through the inductor individually ($I=V/R_{ind}$), then $I_{lamp}=I_{tot}-I_{ind}$.

In your original question you don't say anything about needing to find the current through the lamp, but in one of your comments you imply that that's the quantity you're after - is that actually what the original problem statement asks for?

  • $\begingroup$ Yes, the current through the lamp! I am so sorry I didn't notice it, wow! $\endgroup$ Commented Jan 22, 2013 at 19:41
  • $\begingroup$ Right - well that just adds one extra step, as I outlined above. The 32A your textbook provides as an answer is, I believe, the total current (e.g. through the switch). $\endgroup$
    – Kyle Oman
    Commented Jan 22, 2013 at 20:25
  • $\begingroup$ Well, that's their answer to the question (the current through the lamp) $\endgroup$ Commented Jan 22, 2013 at 20:31

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