Bharat takes five identical bulbs `B_(1), B_(2), B_(3), B_(4)` and `B_(5)` and connects them to a cell as shown in the figure. He compares
(i) The current passing through the different bulbs and
(ii) the potential difference across different bulbs when the circuit is closed. What does he conclude?

Explanation

When the circuit is closed, current starts flowing from the positive electrode of the cell. At 'A' the total current splits, a part passes through `'B_(1)'` and other part passes through `'B_(4)'` . As all the bulbs are identical , the resistance offered by them is the same. If the total current flowing in the circuit is i, at A it splits into two equal halves, and a current of '1/2' passes through `B_(1)` and `B_(4)`. Here, `B_(1)` and `B_(4)` are connected in parallel between A and B. As `B_(2)` is connected in series the current that passes through it is 'i'. Similarly, `B_(3)` and `B_(5)` are connected in parallel between C and D. Hence, '1/2' current passes through both of them. Their potential difference depends on the current that passes through the different bulbs. As shown, the potential drop across `B_(1),B_(4),B_(3)` and `B_(5)` is equal and the potential drop across `B_(2)` is twice that of the others.