![]() The act of breaking the connection makes 'R' go from very low to very high. ![]() When you disconnect the wire, even for fractions of sections, the inductance tries to hold 'I' constant. Now, remember the inductance of the system is going to try and maintain that current. The current 'I' is equal to V/R, which is the battery voltage (9V) divided by the the resistance of the wire and battery. What this means is that once a wire connection is made across the battery terminals current starts flowing through the wire. To answer this question, you'll need to know Ohm's Law: V=IR, as well as inductance which "stores" current or rather, resists changes in current. I'm not sure if inductive kickback is strong enough with a 9 V battery to cause a spark by itself, but it would help current to flow after the plasma path has formed. The voltage can increase enough to cause dielectric breakdown of air and allow current to flow through it.Īttempting to open an inductive circuit often forms an arc, since the inductance provides a high-voltage pulse whenever the current is interrupted Metal vapour (from the contacts) form plasma, which temporarilyĪlso, when a flowing current is interrupted, it will cause inductive kickback, where the collapsing magnetic field causes an increase in voltage, to try to maintain the existing current. Noticeably by this mechanism in a darkened room. The current becomes constricted to these small hot spots,Ĭausing them to become incandescent, so that they emit electrons As the contactsĪre separated, a few small points of contact become the last to ![]() With a gap often produces a low-voltage spark or arc. While lower voltages do not, in general, jump a gap that is presentīefore the voltage is applied, interrupting an existing current flow As the contact is being broken, a connection is made through very small pieces of metal (microscopic features), which have enough current through them to vaporize, the ions of which then support a current through the air briefly.
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