[Physics] the current through the lamp

electric-circuitselectromagnetismhomework-and-exercisesinduction

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.

Best Answer

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.

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