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Brightness of lamp (power) depends on two factors:

  1. Current which flows through the lamp.
  2. voltage (p.d) across the lamp.

In the figure: When the variable resistance increases, what happens to the brightness of both lamps P and Q?

The model answer states that the brightness of both lamps don't change.

I know that the voltage across the lamps and variable resistor is the same since all components connected in parallel. But what about the current, it is impossible to stay the same!!!!

Can anyone please explain why they must still have the same brightness before and after change of the variable resistance value?

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  • $\begingroup$ How is it impossible for the current to stay the same in this situation? I believe you may be assuming that there is a fixed current in this system, when that was never stated. Or, to put it another way, why would you expect the current to change if the potential difference across each bulb is the same, and the bulbs didn't change, so they should have the same internal resistance. $\endgroup$
    – JMac
    Commented Jan 29, 2019 at 21:23
  • $\begingroup$ Since the variable resistor increases, so the total resistance increases and therefore current decreases. So of course there is a different current will flow through lamps which mean different brightness $\endgroup$
    – Rami ki
    Commented Jan 29, 2019 at 22:06
  • $\begingroup$ @Ramiki An increase in the variable resistor only decreases the current that V delivers to the variable resistor. So the total current delivered by V does decrease. But only to the variable resistor and not to the lamps P and Q. It is only resistance in SERIES with the lamps P and Q that can decrease their current. The variable resistance is in PARALLEL with P and Q. Hope that helps. $\endgroup$
    – Bob D
    Commented Jan 29, 2019 at 22:19
  • $\begingroup$ @Ramiki Your voltage source isn't current limited. It can deliver whatever current it needs to achieve that voltage difference. Having a higher total resistance is irrelevant since it doesn't change the voltage across the fixed resistors. $\endgroup$
    – JMac
    Commented Jan 29, 2019 at 22:22
  • $\begingroup$ I'm really sorry but please explain this. I increase the variable resistance value to extreme values so the total current which flow through the circuit must be very low, of course the current which flow through the lamp will not be the same before I increase the variable resistance value....I can't imagine how the current will be the same $\endgroup$
    – Rami ki
    Commented Jan 29, 2019 at 22:49

3 Answers 3

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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.

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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.

$$I_P=\frac{V}{R_P}; \qquad I_Q=\frac{V}{R_Q};$$

The variable resistance allows changing the current through that resistance. $I_{var}=V/R_{var}$

But the total current is $I_P+I_Q+I_{var}$. 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 $I_P+I_Q+I_{var}=I_{Total}$. Only $I_{var}$ and $I_{Total}$ 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.
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  • $\begingroup$ Ok "the total current and current which flows through variable resistor would be variable but the current which flow through two lamps the same", I totally agree with you, but my problem is that I compare the current which flows through two lamps before and after changing variable resistor value, in my opinion there is of course changing in total current and the current flows in the two lamps will be divided in different values comparing to the current before changing the variable resistance value $\endgroup$
    – Rami ki
    Commented Jan 29, 2019 at 22:27
  • $\begingroup$ @Ramiki Please, read again the last sentence before the "some more comments". It's saying that the variation of the resistor affects the total current, but not the bulbs. $total=fixed part + manual variation$, so the total part varies as much as the one of the resistor. $\endgroup$
    – FGSUZ
    Commented Jan 29, 2019 at 22:59
  • $\begingroup$ It is the total current what depends on the circuit, not the other way around. $\endgroup$
    – FGSUZ
    Commented Jan 29, 2019 at 23:03
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Brightness of lamp (power) depends on two factors: 1)Current which flows through the lamp. 2) voltage (p.d) across the lamp.

First of all, obviously this is not true for all types of lamps because it says the brightness (lumens) of a lamp depends only on wattage. We know than an LED light of equal brightness to a tungsten filament light is much lower in wattage. I'm sure you mean this only in the context of changes in the given circuit.

I know that the voltage across the lamps and variable resistor is the same since all components connected in parallel. But what about the current, it is impossible to stay the same!!!!

Well, the current in P and Q will depend on the impedance of P and Q and the voltage across them which, in this case, is the same (V). But why would you think changing the variable resistor would change the current through P and Q? The variable resistor has no effect on the voltage across P and Q (assuming, and this is important, that the voltage source V is an ideal source and not effected by load current).

Hope this helps.

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  • $\begingroup$ Since the variable resistor increases, so the total resistance increases and therefore current decreases. So of course there is a different current will flow through lamps which mean different brightness!!! $\endgroup$
    – Rami ki
    Commented Jan 29, 2019 at 22:14
  • $\begingroup$ @Ramiki Total current delivered by V decreases only because an increase in the variable resistor decreases current in the variable resistor. It does not decrease the current in P and Q. Only an increase in resistance IN SERIES with P and Q will decrease their current. Hope that clarifies. $\endgroup$
    – Bob D
    Commented Jan 29, 2019 at 22:23

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