if you have 2 bulbs and one burns out, the other wont shine, if you remove 1 bulb, you are breaking the circuit, and the current cant flow to the second bulb, so no it wouldn't shine.
When a bulb fuses the circuit is broken if it is on a serial circuit. If the bulb is on a parallel circuit, only the fused bulb will go out, any other bulb would remain lit. On a serial circuit, until the bulb is replaced by a new one, the circuit is not able to be used.
The question is ambiguous, however one possibility is a parallel circuit, which would permit one light bulb to remain lit while the other light bulb was switched off. By contrast, if the light bulbs were connected in a series circuit, switching one light bulb off would cause both lights to go off.
A series circuit is like the old string of Christmas Tree lights. If any bulb burns out none of the others light because the circuit is now open. The same would happen if you used series wiring in a house. Another draw back is the number of devices in series governs how much voltage would be across each device.
It doesn't matter where the bulb is in respect to the battery, as long as the circuit is complete, the bulb will light up.
Nothing. That's why it's a parallel circuit. If it was a series circuit, then the first bulb would go out.
Unscrewing any bulb in a series circuit turns them all off. This is the same as opening the switch that controls them.
When you unscrew a light bulb in a series circuit, the circuit will break and all the other light bulbs in the circuit will turn off. This is because in a series circuit, the current flows through each component in succession, so removing one component interrupts the flow of current to the rest of the circuit.
A parallel circuit will not effect the other bulb. If the bulbs are in a series circuit the other bulb will not turn on.
Without seeing the circuit involved no real answer can be given. I would guess you have a diagram of the circuit, tracing it out should answer your question.
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.
The first lamp will go out. ANSWER: any lamps in series the first or the last if unscrewed will cause all lamps to extinguish no matter whether the source potential is in series or parallel
In a series circuit, if a light bulb is missing or broken, the circuit becomes incomplete, and the electrical current cannot flow. As a result, the other bulbs in the series will not light up. All components in a series circuit must be functioning for the entire circuit to operate.
All other light bulbs will turn off due to the series circuit being broken. The electrons cannot travel all the way through, hence the current will also not flow in the circuit, switching off all the other bulbs too.CommentIt's worth pointing out that the full supply voltage will then appear across the lamp holder -so take care!
If the bulb's filament is broken, the whole circuit becomes open. If the bulb is in a series circuit or is the only bulb, the electricity would no longer flow. If the bulb is in a parallel circuit, along with other bulbs, then only the blown bulb would go out, and the other bulbs would still work.
If you unscrew a light bulb in a series circuit, it will cause an open circuit, which will break the flow of current in the circuit. As a result, all other bulbs in the series circuit will also turn off because there is no longer a complete path for the electricity to flow.
It would not be dangerous to leave a light bulb partially unscrewed, however, there is some possibility that it will fall out of the socket. I would advise that if you don't want the bulb to be on, and you don't have a light switch, unscrew the bulb completely and put it away until you need it.