This depends on how the circuit is wired.
If there is only a single pair of wires on the circuit, then the blowing of either bulb will break the circuit, and both bulbs will go out. OTOH, if a separate pair of wires is run to each bulb from/to the power source and ground, then one bulb can blow, and the other one will stay lit because it still has power. In that circumstance, however, there are really multiple circuits; one for each bulb.
AnswerIf the lamps are connected in series then, if one lamp burns out, the other lamp will stop working. If the lamps are connected in parallel, then a burnt-out lamp will not affect the operation of the other lamp.
The remaining bulb will be brighter than it was when both bulbs were working - due to the increased voltage.
It depends what is the outcome of burning. If that light has created open circuit, circuit will break and there will not be any current. But if there is short circuit means overall resistance is decreased. Thus current will increase. Besides overall brightness will also increase. Thus, if in a series lesser the number of bulbs more will be brightness.
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.
then the other light bulbs in the circuit go out. However, in the case of modern Christmas tree lights in series, the bulbs are designed to short out when they blow, so that the other bulbs do not go out. They do get brighter, however, and this can lead to cascade failure, but that is why there is often a fuse in the plug.
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.
Its because it is in a parallel circuit and the dead bulb stops the power from continuing on to the other bulbs on the circuit.
In a parallel circuit nothing would happen. All the other light bulbs would remain on since there is an alternative path for current to flow. In a series circuit the entire circuit would be de-energized and all the bulbs would go out.
A metal of a certain ionization potential.A light source that provides light of that exact energy potential.a circuit that has all the other components needed.
parallel circuit: Providing that the breakage does not result in a short circuit the other bulbs will still light. series circuit: If the breakage results in a short circuit through the bulb the other bulbs will light more brightly. If the breakage results in a breakage of the connection through the bulb then the other bulbs will not light.
It depends on the circuit. If it is a constant-current circuit, any light bulbs connected in parallel with it will become brighter. If it is a constant-voltage circuit like a typical household circuit, nothing will happen. Any connected in series with it will go out.
A switch simply opens a circuit, stopping the flow of electricity. For example: a simple circuit would be two wires from a battery, one of which goes to a switch, and the other goes to a light bulb. A third wire goes from the other side of the switch to the other conductor of the light bulb. With the switch closed the circuit is complete and the light goes on. Open the switch and the circuit is broken and the light goes off.
A switch simply opens a circuit, stopping the flow of electricity. For example: a simple circuit would be two wires from a battery, one of which goes to a switch, and the other goes to a light bulb. A third wire goes from the other side of the switch to the other conductor of the light bulb. With the switch closed the circuit is complete and the light goes on. Open the switch and the circuit is broken and the light goes off.