Consider a simple circuit containing a battery and three identical incandescent bulbs A, B and C. Bulb A iswired in parallel with bulb B and this combination is wired in series with bulb C. What would happen to thebrightness of the other two bulbs if bulb A were to burn out?(1) Only bulb B would get brighter.(2) Both A and B would get brighter.(3) Bulb B would get brighter and bulb C would get dimmer.
(4) There would be no change in the brightness of either bulb B or bulb C

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Solution

$In the given question there are 2 cases Case 1 when the bulb didnt burn out Two of the resistance are in parallel with a single in series so the effective resistance will be for the two resistance in paraller the eff . resistance is \frac{r}{2} and with when they are connected in series effective resistance becomes \frac{r}{2} + r = \frac{3 r}{2}, so the current through C is \frac{Vx 2}{3 r} = \frac{2 V}{3 r} Current though A and B \frac{V}{3 r} Case 2 When the bulb A burns, there are only 2 bulbs in series, so the eff resistance is 2 R Current trough each bulb is \frac{V}{2 r} Finally we can conclude that the current will be directly proportional to the power which is the brightness . so the answer will be option 3$

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