If the wiring is in series, the entire circuit is cut and the current is zero. If the wiring is in parallel, the current decreases. I'm too occupied to write out formulas and examples on hypothetical scenerios, but the current decreases with lessened loads. C'mon! You didn't know that??...
Answer: it will burn out **Explain:**The same current flows through each part of a series circuit. If the circuit is broken at any point there won't be any current that will flow. In this case, if one of the bulbs blew out, the other bulb would not be able to light up because the flow of electric current would have been interrupted. #Carryonlearning
You know if current is flowing in a bulb circuit because, if there is enough power (voltage times current), the bulb will illuminate. If there is current, but not enough power to illuminate the bulb, you will need to measure the current with an ammeter to see if there is any current.
A series circuit is where there is only one path for the current. As a result, and as a direct consequence of Kirchoff's current law, the current at every point in a series circuit is the same. The two bulbs have the same current flowing through them.
When Susan removed the motor from the circuit, the load on the circuit decreased, leading to an increase in current. This increase in current could cause the bulb to shine brighter, as it receives more electrical energy. However, if the current exceeds the bulb's rated capacity, it could also risk damaging the bulb or shortening its lifespan. Therefore, the effect on the bulb depends on the extent of the current increase.
The whole circuit fails - because the action of the bulb blowing cuts the circuit.
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.
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.
The resistance is increased, the voltage across each bulb is decreased and the current through the circuit is reduced.
The failed bulb breaks the circuit so no current can flow - so the other bulb goes out (but is OK).
current in series always stays the same
If a wire is connected in parallel with a bulb in a circuit, the intensity of the bulb may decrease because the current can take the path of least resistance through the wire, reducing the current flowing through the bulb. Conversely, if the wire is connected in series with the bulb, the total resistance in the circuit increases, leading to a decrease in overall current and a reduction in bulb intensity. The specific impact on intensity depends on the configuration of the circuit.
If one bulb burns out in a series circuit, the other bulb will also turn off since there is no longer a complete path for the current to flow. In a parallel circuit, the other bulb will continue to function normally as each bulb has its own separate path for the current to flow.
There are a few possible different results. One thing that happens in EVERY possible situation is that the bulb you unscrew is dark after you unscrew it. -- If the two bulbs are configured in either a series or a parallel arrangement and the power is off, then both bulbs are dark before you unscrew one, and nothing changes after. -- If the power is on and the two bulbs are configured in parallel, then the one you don't unscrew continues to glow after the other one is gone. -- If the power is on and the two bulbs are configured in series, then BOTH bulbs go out when you unscrew only one of them.
Possibly. If it still works when you turn it on, the of course it is using power. If it does not work when it otherwise would, then it is definitely not using power, just as if you didn't have the bulb there at all. <<>> With the voltage on the circuit and the light bulb in the circuit, the bulb will glow. If you unscrew the bulb until it goes out the potential voltage will still be in the circuit but no current will flow so no power will be used. In this scenario the unscrewing of the bulb will be the same as using a switch in the circuit to interrupt the current flow.
If one bulb in a series circuit goes out, then current can't flow anywherein the circuit. A circuit in which current can't flow is an open circuit.
Answer: it will burn out **Explain:**The same current flows through each part of a series circuit. If the circuit is broken at any point there won't be any current that will flow. In this case, if one of the bulbs blew out, the other bulb would not be able to light up because the flow of electric current would have been interrupted. #Carryonlearning
You know if current is flowing in a bulb circuit because, if there is enough power (voltage times current), the bulb will illuminate. If there is current, but not enough power to illuminate the bulb, you will need to measure the current with an ammeter to see if there is any current.