Question

The above configuration shows three identical bulbs, grade them in order of their brightness.

• A. $B_3>B_2=B_1$
• B. $B_1=B_2>B_3$
• C. $B_3>B_1>B_2$
• D. $B_2>B_1>B_3$

Answer 1

The correct option is A $B_3>B_2=B_1$

Power consumed, $P=i^{2}R$

Since all three bulbs are identical, so their resistance is same. So power consumed is directly proportional to the square of the current.

$\therefore P\propto i^2$

$B_1$ & $B_2$ withdraw less current as compared to $B_3$ because in series they have $2R$ resistance whereas $R$ is the resistance due to $B_3$. Moreover, current through $B_1$ & $B_2$ will be equal.

$\therefore i_3>i_1=i_2$

And we know that brightness is directly proportional to the power consumed.

$\therefore$ Brightness order: $B_3>B_2=B_1$.

Hence, option $\left(a\right)$ is correct.
$$\begin{array}{c}\text{Why this question?}\\ \text{Tip: When identical bulbs are connected then}\\ \text{brightness of that bulb will be more, where}\\ \text{current is more or potential difference is more.}\end{array}$$