The failed bulb breaks the circuit so no current can flow - so the other bulb goes out (but is OK).
series
Lights at home are connected in parallel. (Switches are connected in series with the lights in order to turn them on and off.) This way, each light receives the same voltage. Placing the lights in series would divide the voltage between the lights, reducing the power of each, and it would have the effect that one light burning out would cause all the others on the same circuit to stop working.
its a series connection
The answer to this depends on where the ammeter is in the circuit. Assuming the ammeter is in series with the bulb and no other objects are attached, then the current is 0.4 Amps. Otherwise, your question is unanswerable without more information.
Actually, I have seen this in a respected electronics magazine (Elektor) quite some time ago: A diode (selected to be capable to handle the amps, the voltage and the heat from the bulb - if placed inside the lighting fixture) is used to "cut off" one half of the AC sine wave and thus sending pulses of 120V to an incandescent light bulb, simulating a simple dimming device. The proposal was meant to run 240V bulbs at 50% of their brightness but I am pretty certain this concept can be used to run 120V bulbs from 240V power. One other way of connecting 120V bulbs to 240V would be to connect two 120V bulbs OF THE SAME WATTAGE in SERIES, so the two bulbs form their own voltage divider. Not sure what happens if one bulb burns out - I would expect the other one to blow too, as a dying light bulb represents zero Ohms (plasma/arcing) for a few milliseconds before it goes dark forever.
In a series circuit, all bulbs are necessary to complete the circuit. If one bulb goes out, the circuit is broken, so none of the bulbs would light up.
It goes back to that source
It is called a series circuit. In a series circuit, if one light bulb goes out, it breaks the circuit and causes all the lights to go out.
When a particle passes through a light bulb, it interacts with the atoms in the filament of the bulb. This interaction can cause the particle to lose energy, which results in the emission of light. The emitted light is what we perceive as the light produced by the light bulb.
If one of the light bulb goes out, then the rest go out too.
It goes dark It is time for bed well that a stupid answer! here's an answer a surge of electricity goes on the wire and then boom light!
It goes dark It is time for bed well that a stupid answer! here's an answer a surge of electricity goes on the wire and then boom light!
Electrical goes in. Electromagnetic (heat and light) come out.
bulb goes crazy
The energy that goes in is electrical energy; the energy that leaves it is visible light, as well as heat.
The other light bulb still continues to shine if one light bulb is taken out of a parallel circuit. In a series circuit if one light bulb is removed the other light bulb goes black. This happens because the circuit is incomplete.
This has to do with the potential difference(voltage) reaching each bulb.If the bulbs are placed in series, the voltage will be divided among them(for example, 9V battery powering two identical bulbs will result in each bulb getting 4.5V). Thus, they would each only be One fourth as bright than if there was a single bulb as the power is Voltage squared. If Voltage is halved then power goes one fourth.If the bulbs are placed in parallel, the current(amperage) is divided instead, resulting in each bulb getting the same voltage. This will result in both bulbs having brightness equal to a single bulb.