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**Hint:**This question is based on the concept of electrical power consumed by an electrical bulb. This power consumed can be related to the voltage applied across the bulb, the current drawn by the bulb & the resistance of the bulb.

**Formula used:**

Power consumed by the bulb, $P = V \times I$

Where, $V = $ Voltage across the bulbs

$I = $ current drawn by the bulbs.

**Complete step by step solution:**

It is given that the bulb is rated as $40W - 220V$ this means that the bulb when put into use across the voltage of $220V$ it will be produce a power of $40W$

If suppose the bulb is put across a voltage $V$ which is not equal to $220V$ then power generated by the bulb can be calculated by using Ohm’s law.

For example, the same bulb that is $40W - 220V$is put across a supply voltage of $110V$ than, the power generated by the bulb can be calculated as follow:

Given- Voltage across the bulb- $110V$

Rated voltage $ = 220V$

Power rated $ = 40W$

Now considering Ohm’s Law which says $V = IR$

Where,

$V = $ Voltage.

$I = $ Current.

$R = $ Resistance.

Power \[ = VI = {I^2}R = \dfrac{{{V^2}}}{R}\]

Using power $ = \dfrac{{{V^2}}}{R}$

We know that $R' = \dfrac{{{V^2}}}{{P'}}$

And the resistance of the bulb remains constant does not change with either current or voltage, so equating the resistance for the rated case and for the case, voltage $V = 110V$

$ \Rightarrow R' = \dfrac{{{{(110)}^2}}}{{P'}}$

Equating the resistances we can calculate the value of

We know that both the resistances are equal.

$ \Rightarrow R = R'$

$ \Rightarrow \dfrac{{{{(220)}^2}}}{{40}} = \dfrac{{{{(110)}^2}}}{{P'}}$

$P' = 10W$

Now using the same logic, $P = VI$

Given voltage, $V = 220V$

Power $ = 40W$

$I = \dfrac{{40}}{{220}}A = 0.1818A$

**So option (D) is correct.**

**Additional Information:**

The electrical power \[ = VI = {I^2}R = \dfrac{{{V^2}}}{R}\]

Where,

$V = $ Voltage.

$I = $ Current.

$R = $ Resistance.

Proper formula should be chosen according to the data given in the question. If the data given does not match exactly with the rated quantities of the given bulb, then calculations can be done by taking resistance of the bulb to be constant.

**Note:**Electrical power is defined as the amount of work done by the electrical forces per unit time. It's S.I. The unit is Watt.

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