Question

If two bulbs, whose resistance are in the ratio $1:2$, are connected in series. The power dissipated in them has the ratio of:
$A)\text{ }1:1$
$B)\text{ }1:2$
$C)\text{ 2}:1$
$D)\text{ }1:4$

Answer 1

This problem can be solved by using the power dissipated by a resistor in terms of the current passing through it and the resistance. Since the two bulbs are in series, the same amount of current passes through both of them. Therefore, we can find the power dissipated by both of them and get the required ratio.

Formula used: $P={{I}^{2}}R$

Complete step by step answer:
The bulbs can be considered to be resistors. Since they are in series, the same current will flow through them. We will use the formula for the power dissipated by a resistor in terms of the current through it and its resistance.
The power $P$ dissipated by a resistor of resistance $R$ is given by
$P={{I}^{2}}R$ --(1)
where $I$ is the current passing through the resistor.
Now, let us analyze the question.
Let the resistances of the two bulbs be ${{R}_{1}}$ and ${{R}_{2}}$ respectively.
According to the question,
${{R}_{1}}:{{R}_{2}}=1:2$
$\therefore \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{1}{2}$ --(2)
Let the current flowing through each of them be $I$.
Let the power dissipated by the two bulbs be ${{P}_{1}}$ and ${{P}_{2}}$ respectively.
We have to find out ${{P}_{1}}:{{P}_{2}}$.
Now using (1), we get
${{P}_{1}}={{I}^{2}}{{R}_{1}}$ --(3)
${{P}_{2}}={{I}^{2}}{{R}_{2}}$ --(4)
Therefore, dividing (3) by (4), we get
$\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{{{I}^{2}}{{R}_{1}}}{{{I}^{2}}{{R}_{2}}}=\dfrac{{{R}_{1}}}{{{R}_{2}}}$
Using (2) in the above equation, we get
$\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{1}{2}$
$\therefore {{P}_{1}}:{{P}_{2}}=1:2$
Hence, we have got the required ratio of the powers dissipated by the two bulbs.

So, the correct answer is “Option B”.

Note: This problem could have also been solved by realizing the fact that for bulbs in series, the power dissipated by each bulb will be directly proportional to its resistance (as the current remains constant). Thus, we could have directly used this proportionality to get the ratio and could have done away with the unnecessary variable of the current. Ultimately we reached the same result in our calculations as the current got cancelled out in the calculation process.