`B_(1), B_2 and B_3` are the three identical bulbs connected to a battery of steady emf with key `K` closed. What happens to the brightness of the bulbs `B_(1)` and `B_(2)` when the key is opened?

Options

A. Brightness of the bulb `B_1` increases and that of `B_2` decreases
B. Brightness of the bulbs `B_1` and `B_2` increases
C. Brightness of the bulb `B_1` decreases and that of `B_2` increases
D. Brightness of the bulbs `B_1` and `B_2` decreases

Answer

C

Explanation

When swith is closed
`i=V/(R_(1) + (R_(2)R_(3))/(R_(2) + R_(3)))`
When switched is open:
`i^(.) =V/(R_(1) + R_(2))`
If `R lt R.` then `I. lt i`
Hence, Brightness of the bulb `B_1` decreases and `B_2` increases.