Question

Two electric bulbs whose resistances are in the ratio of $1:2$ are connected in series. The powers dissipated in them have the ratio -

• A. $1:2$
• B. $2:3$
• C. $1:1$
• D. $1:4$

Answer 1

The correct option is A $1:2$

In series combination of resistors, the current is the same through all resistors.

So,

$\frac{P_1}{P_2}=\frac{i^2{R}_1}{i^2{R}_2}=\frac{1}{2}=1:2$

Hence, option $\left(A\right)$ is the correct answer.