Question

A $50W$ bulb connected in series with a heater coil is put to an AC mains. Now the bulb is replaced by a $100W$ bulb. The heater output will
(A) Double
(B) Halve
(C) Increase
(D) Decrease

Answer 1

Hint We will take into account the resistance of the two bulbs. Then relate it with their power. Finally we will conclude with the change in the overall action.
Formulae Used: \[P = \mathop I\nolimits^2 R\]
Where, $P$ is the power rating of the bulb, $I$ is the current flowing through and $R$ is the resistance of the bulb.
For resistances \[{R_1}\] and \[{R_2}\] in series, the equivalent resistance \[{R_{eq}}\] turns out to be
\[{R_{eq}} = {R_1} + {R_2}\]

Step By Step Solution
We take $I$ to be the current flowing through both the cases which remains the same because the mains voltage remains the same as they are connected in series.
Now,
Resistance of the heater is \[{R_H}\]
\[{R_{eq}} = {\text{ }}{R_1} + {R_H}\]
Also,
\[P = \mathop I\nolimits^2 R\]
Now,
When the power of the bulb increases, the resistance decreases and due to this the current in the circuit increases.
Now,
We know from Joule’s law of heating that,
Heat, \[H = {I^2}Rt\]
Thus,
\[H \propto I\]
Hence,
The heater output will increase.

Hence, going through all the options, we can say that the correct option is (3).

Note When the current increases in the bulb, it will increase for the whole circuit as the circuit resistances are in series with each other. In other words, the current flowing through a series circuit remains constant for each and every resistance placed in the circuit. And hence the current increases in the heater. As a result, the heater output also increases.