B1, B2 and B3 are the three identical bulbs connected to a battery of steady emf with key K closed. What happens to the brightness of the bulbs B1 and B2 when the key is opened ?
Answer & explanation
Correct answer: option 3
Let EMF of the Source is E and resistance of each bulb is R.
When key is closed then Equivelent resistance is $R_{eq} = R + \frac{R}{2} = \frac{3R}{2}$
Total current is $I = \frac{E}{R_{eq}}= \frac{2E}{3R}$
Current through B1 is $ \frac{2E}{3R}$ , current through B2 and B3 is $\frac{I}{2} =\frac{E}{3R}$
After Key is opened
$R_{eq} = 2R$
Total current is $ I' = \frac{E}{2R}$ through both B1 and B2
Current through the bulb B1 decreases and B2 increases
Hence Brightness of the bulb B1 decreases and B2 increases.