Question

Figure shows three identical bulbs A B and C connected to a battery of supply voltage V when the switch S is closed discuss the change in
a) The illuminations of the three bulbs
b) The power dissipated in the circuit. ​

Answer 1

$\huge\fcolorbox{red}{black}{ANSWER}$

$\mathbb\green{REFERS \: TO \: ATTACHMENT}$

Answer 2

Q]___?

When the switch is Open,

Vᴀ = Vʙ = Vᴄ = V/3

and

Pᴀ = Pʙ = Pᴄ = (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, Pc = 0

Vᴀ = Vʙ = V/2

Pᴀ = Pʙ = (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 initial value but the intensity of the bulb C becomes Zero.

B)_____

The power dissipated in the circuit before closing the switch is

Pi = Pᴀ + Pʙ + Pc = 3P

The power dissipated in the circuit before closing the switch is

Pf = Pᴀ + Pʙ + Pc = 3P

The power dissipated after closing the switch is,

Pf = Pᴀ + Pʙ + Pc

= 9/4 P + 9/4 P + 0

= 9/2 P = 3/2 Pi

Thus, the power dissipated in the circuit becomes 3/2 times the initial value :)

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