In: Physics
Here for the first case the a bulb(B1) is connected to the battery of potential difference as shown in the figure,
Let the resistance offered by the bulb (B1) is
So,For the circuit as shown,the current in circuit,
So,The current through bulb(B1),
the votage drop at the bulb,
Since there is no other bulbs or resistors in the circuit,as per the kirchoff's voltage law,the entire voltage applied by the battery is dissipated in the bulb B1 itself making it glows brighter.
But in the 2nd case,When another light bulb B2 is also added in series with this bulb B1,cannected to the same battery of voltage ,
Now conider the circuit as shown,
Here let the resistance of the bulb B2 be
So,The net resistance in circuit, (Connected in series)
So,The current in circuit,
SInce bulbs connected in eries,thee currets through both bulbs are same.
The voltage drop in bulb B1, ie,
Also voltage drop in bulb B2,
Also by Kirchoff's voltage law for the circuit, or,
It shows that the voltage applied by the battery is divided among the bulbs in the ratio of thier resistors.And the net voltage applied by battery is the sum of the voltage drops in the both bulbs.
So,In this case the voltage drop in the bulb B1 is less than the applied battery voltage or less than the voltage drop in the same bulb B1 in the previous case where only one bulb B1 is connected.
ie, (in this case)
So,The voltage drop in B1 when only B1 is connected is greater than the voltage drop in B1 when both B1 and B2 connected in series.
Also the brightness of the bulb is determined by the voltage drop across the bulb.
So,The bulb with more voltage drop will glows brighter than the bulb with low voltage drop.
So,Since the bulb B1 has more voltage drop in case 1 when connected alone will glows brighter than the 2nd case where B2 is also connected in series reduces te voltage drop in B1.
So,when an second bulb is connected in series with the first bulb,the brightness of the first bulb B1 will decrease than the previous case.