Question

Bharat takes $5$ identical bulbs $B1,B2,B3,B4,andB5$ and connects them to a cell as shown in the figure

He compares,

1. The current passing through the different bulbs
2. The potential difference across different bulbs when the circuit is closed. What does he conclude?

Answer 1

Step 1: Given data:

the bulbs $B1,B2,B3,B4,andB5$ are identical and the circuit

Step 2: Finding the current and potential:

1. Now, as the current starts flowing, the current$\left(i\right)$ splits at $A$.

. $\Rightarrow B_1$ and $B_4$ are parallel, the total amount of current $\left(i\right)$ through these two will split as $\frac{i}{2}$.

In $B_2$ the value of the current will be $i$.

again in bulbs, $B_{3}and{B}_5$ the value of the current will be $\frac{i}{2}$in each as they are in parallel

ii. $potentialdifference\left(V\right)=iR(R=resis\tan ce)$

As the bulbs are identical, their resistance is the same i.e. $R$

$\Rightarrow potentialdifference\left(V\right)$depends on $i$

As the current across $B_{1,}{B}_4,B_{3}and{B}_5$ is same, this means the potential across them is also the same and the current across $B_2$ was double of $B_{1,}{B}_4,B_{3}and{B}_5$ , similarly potential drop across $B_2$ will also be double.

So, if the potential across $B_{1,}{B}_4,B_{3}and{B}_5$ is $V$ then across $B_2$ it will be $2V$ .