The choke works in conjunction with a starter . When you turn on the light , voltage is applied to the choke, and when the switch opens after a few seconds, it creates a high voltage across the contacts at each end of the tube. This starts the ionization of the gas , which glows. Once ionized, it continues producing light even when the voltage applied is only 120 VAC, the power line voltage.
choke provides the sudden change in the voltage ionisation process of the lamp
once thegas in the lamp gets ionisation then the gas starts conducting
once the conduction gets started then the starter in the tube light will get
gets out of the circuit
then after if starter is took out of the tube then it doesnt affect the lighting
The idea of the choke is to avoid a sudden surge of energy, since the tube has a very small resistance. Without a choke, it would burn out.
to decrease the current
AnswerTo limit the current flowing through the lamp, once the gas in the tube has ionised.
It momentarily produces an inductive kick in the form of high voltage, around 1000 volts, so that the electrical discharge will pass along the column of the tube
No it will not. If you need increase the supply voltage and remove the choke.
If one light bulb in a series circuit fails, all the other light bulbs will go out, until the failed bulb is replaced and the series circuit is completed again.If one light bulb in a parallel circuit fails, all the other light bulbs will still work.
All of the light bulbs in the series circuit would go out.
It is very beneficial to have a parallel circuit... for example: 1. If one light in a parallel circuit goes out, the other light bulbs will remain lit Whereas if a light in a series circuit goes out, all bulbs will go out 2. If not all light bulbs are needed on, you can turn them off with the remaining light bulbs staying on
To make any electrical circuit work it has to be complete. Source of power to the load, the load itself and a return path from the load back to the source. The source in this case being the distribution panel. Any breaks in this complete path will cause the current to stop flowing and the device to not work. A light fixture and light switch work on this principle, open the circuit and the light goes out.
The choke works as an inductor(filter) induced 850V to 1100V in starting,by strarter make and break of the circuit. After glowing the tube light choke work as inductance and provide 110v to circuit, here the function of starter comes. It helps the chocke to brek from the circuit and to 110v from 850V.thankining U "Nikki"
No it will not. If you need increase the supply voltage and remove the choke.
Same choke tube as a 1200 or 1400 pump
A T8 LED tube has to be used with a ballast to bring the correct voltage to the tube for it to operate correctly. Removing the ballast from the circuit will prevent the tube from operating.
signal light
Tail light doesn't. Work with new bulb in
In that case, the entire circuit won't work.
Either close the circuit, or complete the circuit, should work here.
Once started the fluorescent tube no longer needs the starter. In fact starters are designed to electrically "remove" themselves from the circuit when the fluorescent tube is conducting.
UV bulbs are usually fluorescent tubes with the normal white phoshor powders removed. Heaters at each end of the tube excite electrons to 'boil' off the wires. A high voltage is applied via an inductor or choke across the tube ends which sets up a flow of electrons through the tube called a plasma. This is the 'electric blue' of a miniature lightning bolt and emits UV light.
If one light bulb in a series circuit fails, all the other light bulbs will go out, until the failed bulb is replaced and the series circuit is completed again.If one light bulb in a parallel circuit fails, all the other light bulbs will still work.
Flashlight by light, battery, circuit, a hood and a reflective cup, circuit storehouse, switch, etc, the basic principle is the current from the battery, and then to the circuit, after adjusting circuit voltage and current, and then output to the light source for lighting.