Question

Two bulbs 40 W, 220 V and 110 W, 220 V are connected in parallel with 200 volts supply, then
(A) 110 W bulbs glow with more intensity than 40 W.
(B) a 40W bulb glows with more intensity than 110 W bulb.
(C) Both 110W and 40 W glow with equal intensity
(D) No bulb will glow

Answer 1

Hint: In a rated device, the voltage rating means that the device will consume the mentioned amount of power if the device is connected across that particular amount of voltage. So from the formula \[P = IV\] we will get the answer.
Formula used: In this solution we will be using the following formulae;
\[P = IV\] where \[I\] is the current flowing through the device, and \[V\] is the voltage.
\[I = \dfrac{V}{R}\] where \[R\] is the resistance of the device.

Complete Step-by-Step solution:
Generally, every device will consume a particular amount of energy per unit time when connected across a voltage source. This power consumed is very dependent on the voltage source it is connected to. A particular device consuming a particular amount of power across a particular voltage source may not consume the same amount on another voltage source.
Generally, the power consumed of a device having a particular resistance to electrical current will be given as
\[P = IV\] where \[I\] is the current flowing through the device, and \[V\] is the voltage.
But since according to ohm's law, we have
\[I = \dfrac{V}{R}\] where \[R\] is the resistance of the device.
hence,
\[P = V\left( {\dfrac{V}{R}} \right) = \dfrac{{{V^2}}}{R}\]
Hence, the power consumed is proportional to the square of the voltage across it.
This is why power ratings must have standards at which it consumes said amount. If a device is rated 40 W 220 volts consumes 40 W of power if connected across a 220 Volts. However this doesn’t mean that the device will not operate at a lower voltage. It would only consume a lower amount of power.
Hence, the 110 W will glow with more intensity than the 40 W bulb.

Thus, the correct option is A.

Note: For clarity, the intensity is higher at the 110 W because power consumed is more at the 110 W and power is directly proportional to intensity.
However, on real bulbs, the efficiency may be so low, that only a tiny fraction of the electrical power will be converted to light energy hence, this has to be taken into consideration.