When breaking a circuit transmitting a given power, how do I minimize sparks/arcing?

Question

If I have a pair of wires delivering a given power, $P$, and I have the option of breaking the circuit in two places breaking the circuit in which place will minimize air sparking/arcing: the high voltage/low current place, or the low voltage/high current place? My own instinct is breaking the circuit on the low current side produces the least arcing, and thus damage/carbon scarring to electrical contacts because trying to break a circuit imposes a high $\frac{\operatorname{d}I}{\operatorname{d} t}$, causing arcing. Breaking a high voltage circuit, though, means it takes longer before the electric field between the conductors ($\vec{E} \sim \frac{\Delta V}{\Delta x}$) becomes small enough to be below the breakdown level.

What is the correct answer? Is there some optimal balance, or is one of the above factors always sub-dominant? Or does it depend on the circuit in question (e.g. the inductance of the circuit being broken)?

Answer 1

Since you posted in a physics forums, I assume you want a physics answer. In that case I think you're asking about the following circuit:

In the above circuit, I dont think it matters which side you close, you get the same back-voltage.

What would happen if we opened either switch $S_1$ or $S_2$? Well since $N_1$ and $N_2$ are different, the inductance $L_1$ and $L_2$ will be different (someone want to confirm that the inductance looks different on different sides of a transformer?). The back-emf will be: $$ \mathcal{E}_1 = -L_1 \frac{dI_1}{dt} $$ $$ \mathcal{E}_2 = -L_2 \frac{dI_2}{dt} $$

It is this back-emf that causes arcing when opening mechanical switches. If we assume you close both switches in the same time, then we can approximate $dI/dt = -I/\Delta t$. Dividing the back voltages:

$$ \frac{\mathcal{E_1}}{\mathcal{E_2}} = \frac{L_1 I_1}{L_2 I_2}$$

The ratio $L_1 / L_2$ will probably be just the ratio of the turns $N_1 / N_2$ which equals the ratio of the voltages $V_1 / V_2$ so it all cancels out:

$$ \frac{\mathcal{E_1}}{\mathcal{E_2}} = \frac{V_1 I_1}{V_2 I_2} = \frac{P_1}{P_2} = 1 $$

Therefore, same back-voltage in either case.

Answer 2

It is normally the current that may wear and destroy mechanical contacts. With inductivities, the bigger problem might be switching off instead on.

The emf of a transformer is not equal on both sides - as can be seen with the old car ignition transformers, where some 10 Thousands of volt can be generated, while the primary side does see only some hundreds of volts at the time of switching off.

A switch on the side with lower current should be better, even if the emf is higher, since the initial arcing electrons are fewer, and the air gap takes much of the kinetic energy via ionization, radiation and other energy transfer on a longer distance. If there are too many electrons crossing the gap, the contact points could become so hot that new electrons are produced out of the metal via thermal effects that will make it difficult to really stop the current, the contacts could be even welded together. For tiny currents, there would be no arcing, but a glow discharge effect only. But since most of these effects are non-linear, there could be cases where switching the higher current could be better- it depends on current, voltage, metal, pressure, reflectivity of UV in the contact zone etc.

As for a circuit diagram, i.e. for pure circuit theory with ideal elements resistor, switch, transformer/inductivity and (series) connections, there may be a contradiction involved. An ideal switch has an infinite dI/dt by definition, while an inductivity has a finite dI/dt, i.e. a current through an inductivity can not be a falling part of a step function. As mentioned in other statements above, in reality a path for the exponentially decreasing current must be offered when a current of an inductivity is switched off, e.g. via a capacitor across the switch's contacts.