A circuit consists of three identical lamps connected to a battery as shown in the figure. When the switch S is closed then the intensities of lamps A and B

Explanation

When the switch is open,
`V_(A) = V_(B) = V_(C) = V//3`
and `P_(A) = P_(B) = P_(C) = ((V/3)^(2))/(R) = (V^(2))/(9R) ` = P (say)
(a) When the switch is closed, the bulb C is short circuited and hence there will be no current through C.So `P_(C) = 0`
`V_(A) = V_(B) = (V)/(2)`
So , `P_(A) = P_(B) = ((V/2)^(2))/(R) = (v^(2))/(4R) = (9)/(4) P` .
Therefore , the intensity of illumination of each of the bulb A and B becomes `(9)/(4)` times the intital value but the intensity of the bulb C becomes zero.
(b) The power dissipated in the circuit before closing the switch is `P_(i) = P_(A) + P_(B) +P_(C) = 3P`
The power dissipated after closing the switch is
`P_(f) = P_(A) + P_(B) + P_(C) = (9)/(4)P + (9)/(4)P + 0 = (9)/(2)P = (3)/(2)P`
Thus , power dissipated in the circuit becomes `(3)/(2)` times the intial value.