In the circuit shown, the total power output is P when S is closed. As the switch S is opened, the power output becomes (all bulbs are identical)

Question

In the circuit shown, the total power output is P when S is closed. As the switch S is opened, the power output becomes (all bulbs are identical) (A) (2/3) F (B) (3/2) F (C) (4/9) P (D) (9/4) P

Tardigrade Admin · Accepted Answer

When switch S is closed Re f f=2 R(R is resistance of one bulb) P=(V2/2 R) When switch S is open Re f f=3 R P prime=(V2/3 R) ⇒ (P prime/P)=(2/3)

Q. In the circuit shown, the total power output is $P$ when $S$ is closed. As the switch $S$ is opened, the power output becomes (all bulbs are identical)

$\frac{2}{3} F$

$\frac{3}{2} F$

$\frac{4}{9} P$

$\frac{9}{4} P$

When switch $S$ is closed $R_{e ff} = 2 R (R$ is resistance of one bulb)
$P = \frac{V ^{2}}{2 R}$
When switch $S$ is open $R_{e ff} = 3 R$
$P^{'} = \frac{V ^{2}}{3 R}$
$\Rightarrow \frac{P ^{'}}{P} = \frac{2}{3}$

Q. In the circuit shown, the total power output is P when S is closed. As the switch S is opened, the power output becomes (all bulbs are identical)

32​F

23​F

94​P

49​P