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Question: An electric bulb is rated \[40W - 220V\], when it is connected to a \(220V\) source. Find the curren...

An electric bulb is rated 40W220V40W - 220V, when it is connected to a 220V220V source. Find the current drawn by the bulb?
A) 0.1666A0.1666A
B) 0.1333A0.1333A
C) 0.1623A0.1623A
D) 0.1818A0.1818A

Explanation

Solution

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×IP = V \times I
Where, V=V = Voltage across the bulbs
I=I = current drawn by the bulbs.

Complete step by step solution:
It is given that the bulb is rated as 40W220V40W - 220V this means that the bulb when put into use across the voltage of 220V220V it will be produce a power of 40W40W
If suppose the bulb is put across a voltage VV which is not equal to 220V220V then power generated by the bulb can be calculated by using Ohm’s law.
For example, the same bulb that is 40W220V40W - 220Vis put across a supply voltage of 110V110V than, the power generated by the bulb can be calculated as follow:
Given- Voltage across the bulb- 110V110V
Rated voltage =220V = 220V
Power rated =40W = 40W
Now considering Ohm’s Law which says V=IRV = IR
Where,
V=V = Voltage.
I=I = Current.
R=R = Resistance.
Power =VI=I2R=V2R = VI = {I^2}R = \dfrac{{{V^2}}}{R}
Using power =V2R = \dfrac{{{V^2}}}{R}
We know that R=V2PR' = \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=110VV = 110V
R=(110)2P\Rightarrow R' = \dfrac{{{{(110)}^2}}}{{P'}}
Equating the resistances we can calculate the value of
We know that both the resistances are equal.
R=R\Rightarrow R = R'
(220)240=(110)2P\Rightarrow \dfrac{{{{(220)}^2}}}{{40}} = \dfrac{{{{(110)}^2}}}{{P'}}
P=10WP' = 10W
Now using the same logic, P=VIP = VI
Given voltage, V=220VV = 220V
Power =40W = 40W
I=40220A=0.1818AI = \dfrac{{40}}{{220}}A = 0.1818A

So option (D) is correct.

Additional Information:
The electrical power =VI=I2R=V2R = VI = {I^2}R = \dfrac{{{V^2}}}{R}
Where,
V=V = Voltage.
I=I = Current.
R=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.