Q147 / 180 · KCET 2010 · Physics

$\mathrm{B}_{1}, \mathrm{~B}_{2}$ and $\mathrm{B}_{3}$ are the three identical bulbs connected to a battery of steady emf with key $\mathrm{K}$ closed. What happens to the brightness of the bulbs $B_{1}$ and $B_{2}$ when the key is opened?
Figure

  • A
    Brightness of the bulb $\mathrm{B}_{1}$ increases and that of $\mathrm{B}_{2}$ decreases
  • B
    Brightness of the bulbs $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ increase
  • Correct answer
    C
    Brightness of the bulb $\mathrm{B}_{1}$ decreases and $\mathrm{B}_{2}$ increases
  • D
    Brightness of the bulbs $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ decrease

Solution

When key $\mathrm{K}$ is opened, bulb $\mathrm{B}_{2}$ will not draw any current from the source, so that terminal voltage of source increases. Hence, power consumed by bulb increases, so light of the bulb becomes more. The brightness of bulb $\mathrm{B}_{1}$ decreases.

Hence, the correct answer is option: 'C'.

Relevant Questions