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help please
A lamp is marked 1.8w in normal brightness it carries a current of 600mA

1) what is the related voltage?​

Sagot :

Answer:

Explanation:

If you are not told otherwise, you have to assume that the voltage source is ideal, meaning that it can maintain its voltage regardless of what resistances (or other loads) are in the circuit, and that it has the means to provide whatever current is necessary to maintain that voltage.

0

Okay, it's an interesting question. Your problem is right: the brightness stays the same.

First of all, I can't resist saying that I would expect 0 brightness haha. Because the circuit, as it is drawn, has no power sources. I guess the upper element is meant to be a voltage source. However, if you write "V" in the circle, that's the symbol of a volt-meter, not a power source. Consider changing the symbol for a battery or a voltage source.

After that comment, let's suppose the circuit consists in two lamps, a resistor and a voltage source, all of them in parallel.

The key is that

A voltage source provides a fixed voltage, but the current it provides is unknown, as it depends on the circuit to which it is attached.

That's the first idea. The second one is:

A lightbulb can be regarded as a resistance, at first level approximation.

That means that we have 3 resistances in paralell.

Two of those resistances are fixed (light bulbs). Constant voltage, and constant resistance, imply constant current.

[tex]IP=VRP;IQ=VRQ;[/tex]

The variable resistance allows changing the current through that resistance. Ivar=V/Rvar

But the total current is IP+IQ+Ivar. This is the current powered by the source. But the source is a voltage source, not a current one. A voltage source can have any current allowed by its components.

What is happening is that the current supplied by the source varies, in order to keep the voltage constant, which is the source's mission.

In other words,

Keeping the voltage constant requires a sacrifice: the current supplied by the source might vary.

But, as the voltage is the same for the bulbs, and intensities of the bulbs depend on the bulbs themselves (I=V/R), and they do not depend on the source, then of course the power dissipated in the bulbs is the same, and thus they have the same brightness.

This means that IP+IQ+Ivar=ITotal. Only Ivar and ITotal can vary. The bulbs' ones are fixed by Ohm's law.

Some more comments:

Of course, this is the ideal case. There is actually a range of currents that the source and the components can handle. Ideally, a voltage source can provide any current needed. In real life, this is not true. Real sources have a maximum current value.

Also, bulbs have a resistance that depends on the current, because their R depends on temperature.

If we didn't consider light bulbs as resistance, it would be a little more complicated, and different.

Hope this helps if not im sorry