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If I have two lamps, both of them have different power values, and voltage value is constant , I've been told that the lamp that has less power would be brighter and make more light than the lamp that has a higher power rating in the series circuits.

I searched for the reason but I didn't find an answer.

Floris
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loody
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1 Answers1

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Suppose you have 2 bulbs, one rated 240V 60W and the other rated 240V 100W. These ratings mean that when you apply a voltage of 240V to each, they emit 60W and 100W of light power respectively.

The 100W bulb is brighter when they are connected in parallel - when they both have the same voltage of 240V across them. But when they are connected in series they have the same current flowing through them; the voltage across each is different. The power dissipated in each bulb is $I^2R$ where $I$ is current and $R$ is resistance. So when the current is the same the bulb with greater resistance emits more light power and is brighter.

Which bulb has more resistance? You can get the resistance of the bulb from the rating using formula $P=\frac{V^2}{R}$. $V$ is the same for both bulbs so the resistances are inversely proportional to the power ratings : $R_{60}/R_{100}=P_{100}/P_{60}=100/60$. The 60W bulb has the higher resistance. Therefore when the same current flows through both (eg when connected in series) the 60W bulb is brighter.

sammy gerbil
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    Nice and concise. One important thing relevant for real light bulbs: their resistance changes a LOT with temperature. So the bulb that starts out with the higher resistance will get hotter, become more resistive, and "steal" even more of the power than you might have thought. This makes the difference in brightness larger than you'd expect. – Floris Nov 28 '16 at 19:39
  • A simple question just came to my mind, u said the voltage will be different in the series circuits, but why? Isn't it supposed to be constant?? – loody Nov 29 '16 at 18:14
  • No. In parallel the voltage across each is the same. In series the current through each is the same. If the same current $I$ passes through the 2 resistors in series, the voltages across them are $V_1=IR_1$ and $V_2=IR_2$. If $R_1 \ne R_2$ then $V_1 \ne V_2$. – sammy gerbil Nov 29 '16 at 18:34
  • What if the two resistors have the same resistance value, then v1=v2, is that possible? – loody Nov 29 '16 at 19:25
  • Yes, of course. If the two lamps are identical (eg both 100W) then the voltage across each will be the same when in series; they will be equally bright when in series and when in parallel. (But notice that if 240V is placed across them in parallel, they will shine brighter than when 240V is placed across them in series, because in the latter case the voltage across each lamp is only 120V.) However, if they have the same voltage rating (240V) but different power ratings (100w, 60W) then they cannot have the same resistance. – sammy gerbil Nov 29 '16 at 19:43
  • so far I understand what you say, but just to make it clearer for me, you said if the current is the same because its series then the bulb with the higher resistance is brighter. lets assume the current isn't the same or its parallel circuit how will this apply on the two formula you wrote in the answer and how you will come to the fact of higher power rating means more brightness in parallel circuit? – loody Nov 29 '16 at 20:07
  • If different currents flow though each bulb, the brighter bulb will be the one for which $P=I^2R$ is higher. In my example the 100W bulb has $R=V^2/P=240^2/100=576$ Ohm and the 60W bulb has resistance $R=100*576/60=960$ Ohms. If instead you know the voltages across each, then compare $P=V^2/R$ to find which is higher. The 2 formulas are really the same if you substitute $V=IR$. (However, note the comment by Floris. You cannot rely on these formulas to tell you the correct power output - except at the rated voltage - but they will tell you which should be brighter.) – sammy gerbil Nov 29 '16 at 20:32