Question

Assertion : If three identical bulbs are connected in series as shown in figure then on closing the switchs. Bulb `C` short circuited and hence illumination of bulbs `A` and `B` decreases.
Reason : Voltage on `A` and `B` decreases

A. both A and B will glow more brightly
B. both A and B will glow less brightly than before
C. A will glow less brightly and B more brightly
D. None of the bulbs will glow

Answer 1

Correct Answer - C
Bulbs B and C are connected in parallel. Suppose, voltage across A and B is same. If bulb C is fused, bulb B will work
`P_(B)=(V^(2))/(R_(B))impliesP_(A)=(V^(2))/(R_(A))`
`R=(R_(B)R_(C))/(R_(B)+R_(C))`
`because " " R_(B) lt R_(A) therefore P_(B) gt P_(A)`