Question

Three identical light bulbs are connected to a 30V battery in a circuit. Bulb A is on main loop and closest to battery. Bulbs B and C are in parallel and the junction is after Bulb A S1 is in the loop with bulb B and S2 is in loop with bulb C

Initially switch 1 is closed and switch 2 is opened. When switch 2 is closed what happens to the brightness of bulbs A,B and C Please explain

Answer 1

When switch 1 is closed and S2 is opened.

then no current will flow through bulb B.
and now bulb A and C are in series with battery,

hence curremt through both = 30 / 2R = 15/R

so brightness of bulb B will be zero and brightness of A and C will be same.

when switch 2 also closed.

now bulb B and C are in parllel.

1/R' = 1/R + 1/R

R' = 0.5 R

now R' and A are in series.

Req = R + 0.5R = 1.5R

current through bulb A = 30 / (1.5R) = 20/R

current through B and C = (20/R) / 2 = 10/R

SO now brightness of bulb A will increases.

Bulb B was will have brightness now .

Bulb C brightness will decrease.

A - Increase

B - 0 to increase

C - decrease.