Question

The choke of a tube light work on the principal of

Answer 1

ANSWER-:


When the switch is ON, full voltage will come across the tube light through ballast and fluorescent lamp starter. No discharge happens initially i.e. no lumen output from the lamp.

At that full voltage first the glow discharge is established in the starter. This is because the electrodes gap in the neon bulb of starter is much lesser than that of inside the fluorescent lamp.

Then gas inside the starter gets ionized due to this full voltage and heats the bimetallic strip that is caused to be bent to connect to the fixed contact. Current starts flowing through the starter. Although the ionization potential of the neon is little bit more than that of the argon still due to small electrode gap high voltage gradient appears in the neon bulb and hence glow discharge is started first in starter.

As voltage gets reduced due to the current causing a voltage drop across the inductor, the strip cools and breaks away from the fixed contact. At that moment a large L di/dt voltage surge comes across the inductor at the time of breaking.

This high valued surge comes across the tube light electrodes and strike penning mixture (mixture argon gas and mercury vapor).

Gas discharge process continues and current gets path to flow through the tube light gas only due to low resistance as compared to resistance of starter.

The discharge of mercury atoms produces ultra violet radiation which in turn excites the phosphor powder coating to radiate visible light.

Starter gets inactive during operation of tube light.

HOPE IT WILL HELPS YOU!