Question

Consider the diagram in which $B_1, B_2$ 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?

• 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 bulbs $B_1$ decreases and $B_2$ increases
• D. Brightness of the bulbs $B_1$ and $B_2$ decreases

Answer 1

Answer: (C)

When key, $K$ is opened, bulb $B_2$ will not draw any current form the source, so that terminal voltage source increases. Hence, power consumed by bulb increases, so light of the bulb becomes more. The brightness of bulb $B_1$, decreases.