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.
Answer
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.
No comments:
Post a Comment