The current of one bulb (two bulbs shorted) would be about1 three times the current of three bulbs.
1 I say "about" because resistance is a function of temperature, and running three times the current through one bulb will make that one bulb much hotter, increasing its resistance. It might also burn out the bulb.
By adding more light bulbs
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.
Brighter in parallel. In series the voltage is divided between the two bulbs, thus the current will only be half so that the power of each bulb will only be one quarter (of 5 watts) in the series set-up.
You break the circuit and they both extinguish (go out).Answer: The circuit will open up causing the current to quit flowing to both bulbs therefore there will be no lighted bulbsCommentYou also have the full supply voltage appearing across the empty lampholder!
A parallel circuit is one in which there is more than one current path. An example might be several light bulbs in a house, all of which provide a path for electric current. The advantage is that any of the lamps can be turned on or off without affecting the current flow through any other lamp. A series circuit is one in which the current has only a single path through multiple components. An example might be two resistors, one after the other. Both will have the same current through them but they may have varying voltages depending on their resistance. Another example of a series circuit is a houehold lamp and a light switch. When the light switch is open, no current flows through it and therefore, no current can flow through the lamp either. When the switch is closed, current will now flow through the switch and the lamp.
Since the SAME electrons have to go through both light bulbs, the current in both light bulbs will be the same (Kirchhoff's current law).
As the number of bulbs in a series circuit increases, the current decreases. As the number of bulbs in a parallel circuit increases, the current increases.
Bulbs in a parallel circuit draw the same amount of current, so each will display the same brightness. Bulbs in a series circuit share the current so all bulbs will appear dimmer.
the same current flows through both light bulbs
It is unclear what type of circuit you are referring to, so I'll give both answers.parallel, current increases until too many bulbs have been added, then circuit breaker pops and current drops to zero.series, current decreases and all bulbs dim.
All the bulbs will go out. In a series circuit, the current at all points is the same. This is Kirchoff's Current Law. If you loosen or remove a bulb in a series circuit, the current at that bulb becomes zero, and by Kirchoff's Current Law, the current in every part of the circuit also becomes zero.
They don't unless you speaking about a parallel circuit in which total currect would be the sum of all the currents in each light bulb (The more light bulbs, the more current draw) If you're talking about a series circuit, nothing at all happens to the current, as in a seires circuit current is constant throughout the entire circuit (voltage changes). In a case such as this the more light bulbs in the circuit, the less the voltage becomes across those bulbs (furthest from the source), thus they will become dimmer due to lower power (P=IE).
The total current in the circuit will decrease.
By adding more light bulbs
If all three bulbs are in parallel, then there are three current pathways.
The meaning of "differently lighted up" is very unclear. As long as all the bulbs in a series circuit are connected to the circuit, and all of the bulbs have good filaments with no holes in them, current will flow in the circuit. Depending on the ratings of each bulb, the current may not be enough to cause all of them to glow visibly, but there will certainly be a current. That may or may not suit your definition of "work".
As more light bulbs are added in a series circuit, the effective resistance of the circuit increases. That causes the current leaving the source to decrease.