Question

The coil of an electric bulb takes $ 40W $ to start glowing. If more than $ 40W $ is applied, $ 60\% $ of the extra power is converted into light and rest into heat. The bulb is rated $ 100W $ $ 220V. $ Find the percentage drop in light intensity at a point if the supply voltage is changed from $ 220V $ to $ 200V $ .
(A) $ 16\% $
(B) $ 22\% $
(C) $ 28\% $
(D) $ 34\% $

Answer 1

Hint
The rate at which work is done by the source of e.m.f in maintaining the current in an electric circuit is called electric power of the circuit. Electrical power can be written as the product of potential difference and current. Here $ 40W $ power is required to heat up the coil of the electric bulb.
 $ P = \dfrac{{{V^2}}}{R} $ (where $ P $ are the power in the electrical circuit, $ V $ stands for the potential difference of the circuit, $ R $ stands for the resistance of the circuit)

Complete step by step answer
Let us first consider the case of $ 220V $
Given that the rating on the bulb is $ 100W $ $ 220V $
It is said that $ 40W $ is used to heat the coil, the remaining power is given by,
Excess power $ = 100 - 40 = 60W $
It is said that $ 60\% $ of the extra power is converted into light energy, this can be written as,
Power converted into light, $ = \dfrac{{60 \times 60}}{{100}} = 36W $
With this, the bulb will glow completely.
Now, let us consider the case of $ 200V $
We have to find the power corresponding to $ 200V $
To find the power we use the formula, $ P = \dfrac{{{V^2}}}{R} $
For that, we have to find the total resistance of the circuit.
From the expression of power we can write the resistance as,
 $ R = \dfrac{{{V^2}}}{P} $
From the given rating, $ V = 220V $ and $ P = 100W $
 $ \therefore R = \dfrac{{{V^2}}}{P} = \dfrac{{{{\left( {220} \right)}^2}}}{{100}} = 484\Omega $
Now we can find the new power using the formula , $ P = \dfrac{{{V^2}}}{R} $
Substituting the values we get,
 $ P = \dfrac{{{V^2}}}{R} = \dfrac{{{{(200)}^2}}}{{484}} = 82.64 $
We know that $ 40W $ is used to heat the coil of the light,
The excess power can be written as,
Excess power = $ 82.64 - 40 = 42.64W $
Out of this power $ 60\% $ is converted into light, this can be written as
Power converted into light $ = \dfrac{{60 \times 42.64}}{{100}} = 25.584W $
We know that the intensity of the light is directly proportional to the power.
Initially, the power that was converted into the light was, $ 36W $ after the voltage is changed, the power that is converted into light became $ 25.584W $
The difference in power can be written as, $ 36 - 25.584 = 10.416W $
The percentage drop in light intensity can be written as,
Percentage drop $ = \dfrac{{\Delta P}}{P} \times 100 = \dfrac{{10.416}}{{36}} \times 100 = 28.9\% $
The answer is Option (C): $ 28\% $ .

Note
In most devices, the power rating will be given. It is the maximum power that is allowed to pass through a device In most cases devices convert electrical energy into another form of energy. Electrical motors convert electrical energy into mechanical energy.