Question

Please provide correct explanation for ques 51.
(The answer is (2) - decreases)

Answer 1

Answer:

Hi !

Because all the bulbs are identical, we can suppose their resistance to be R each.

Now, if the switch was closed initially, so that combination of bulb 2 and 3 in parallel was in series with B1

We would have had 3R/2 as the equivalent resistance

Which means the current in circuit given by the battery would have been V/3R/2 = 2V/3R

And, hence the power supplied to the bulb 1 will be (2V/3R)^2 * R = 4V^2/9R

Ok,

Now, if the switch was opened then just B1 and B2 would have remained in the circuit which are in series and hence the net resistance of the circuit would be 2R

so, current supplied then by the battery = V/2RAnd, hence the new power supplied to B1 will be (V/2R)^2 * R = V^2/4R which is for obvious reasons less than the power supplied initially

(Since, 4/9> 1/4)

And, hence because the net power supplied to the Bulb 1 in second scenario is less than the first one , the incandescence that is the brightness of the bulb will decrease.