Two electric bulbs, P and Q, have their resistances in the ratio of $1:2$. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.

Hint: In the case of series combination, we know that the power dissipation in the bulb is directly proportional to the resistance of the bulb. So, we can say the ratio of the power dissipation in the two bulbs will be the same as the ratio of their resistance. Here, the ratio of the resistances of the bulbs is given. So, we can calculate the ratio of their power dissipation also.

Complete step by step solution:
Let, ${R_P}$ and ${R_Q}$ be the resistances of the two bulbs P and Q.
Here, it is given that ${R_P}:{R_Q} = 1:2$ .
Now, we know that the bulbs are connected in a series combination; the current flowing through both must be the same. But the voltage required to force the current through the two bulbs is not the same; it must be proportional to their resistances.
So, the power dissipation is given by, $P = {I^2}R$ .
$\Rightarrow P \propto R$ .
Now, if we consider, ${P_P}$ and ${P_Q}$ be the power dissipation in the bulbs P and Q,
Then, we can say, $\dfrac{{{P_P}}}{{{P_Q}}} = \dfrac{{{R_P}}}{{{R_Q}}}$
Since, $\dfrac{{{R_P}}}{{{R_Q}}} = \dfrac{1}{2}$ is given,
So, we have, $\dfrac{{{P_P}}}{{{P_Q}}} = \dfrac{1}{2}$
Therefore, if the two bulbs P and Q having resistances as a ratio of $1:2$ , are connected in series across a battery, then the ratio of the power dissipation in these bulbs is also $1:2$ .

Note: A resistor in any circuit that has a voltage drop across it wastes electrical power. We know all resistors have a power rating. This is because the dissipated electrical power is converted into heat energy. This power is the maximum that can be dissipated from the resistor without it burning out. The power of dissipation is the conversion rate.