# Significance of sliding switch in a LR circuit

The following image and statement are from my textbook Concepts of Physics on the chapter "Electromagnetic Induction", topic 38.6 "Growth and decay of current in an LR circuit", sub-topic "Decay of current":

Consider the arrangement shown in figure (38.11a). The sliding switch $$S$$ can be slid up and down. [...] The special arrangement of the switch ensures that the circuit through the wire $$Aa$$ is completed before the battery is disconnected.

I don't understand the reason behind the usage of a sliding switch. What happens if we use the following switch instead of a sliding switch?:

It's stated that the sliding switch ensures that electrical contact is made with the end $$a$$ before it looses contact with $$b$$. What if we conduct the experiment the other way round? I think both methods should make no difference, however, it would be helpful if you could explain why we need to short circuit the battery first and then remove it out of the circuit?

This arrangement using a make before break switch can be used as a basis of an arrangement to illustrate the decay of current in an inductor $$L$$ and resistor $$R$$ in series circuit.

The time constant of such a circuit is $$\dfrac LR$$.
With a conventional switch, break before make, there will be a time where the total circuit resistance $$R$$ is very large which would make the time constant $$\frac LR$$ very small.
This being the case one would find that the current through the inductor would drop very rapidly during the time the switch completed the circuit with the induced emf $$\mathcal E = L \frac {dI}{dt}$$ being so large that an arc is produced across the contacts.

In your circuit there is a continuity in terms of the resistance in the circuit as the switch contacts are changed the only difference being the internal resistance of the cell.

An circuit which shows the decay of current/voltage in an $$LR$$ circuit and uses a "conventional" switch is shown below with the circuit resistance changing very little as a result of switching.

• "With a conventional switch, break before make, there will be a time where the total circuit resistance R is very large which would make the time constant LR very small." Why sir? Commented Sep 7, 2023 at 8:30
• The time constant is $\frac LR$, inductance divided by resistance, so as $R$ gets bigger that fraction gets smaller. Commented Sep 7, 2023 at 8:44
• No no I know that, I am talking about why R gets bigger? Commented Sep 7, 2023 at 9:06
• There will be an air gap between the switch contacts and air is not a very good conductor. Commented Sep 7, 2023 at 9:26
• Thank you very much sir Commented Sep 7, 2023 at 10:59

For an inductor, the rate of change of the current is proportional to the voltage. That is, if you try to stop the current in the inductor suddenly, as you do if you simply open circuit it, it will produce an indefinitely large (practical details depend on the materials and geometry) spike in the voltage in response.

The principle of the circuit above is that by sliding the switch so that the current can continue to flow. It will change, but as the field starts to collapse, the voltage drives the current through the resistor - leading to a continuous change in the current, and no voltage spike.

Open circuiting an inductor is like short circuiting a capacitor.