Series Connected Bulbs – Rankers Physics
Topic: Current Electricity
Subtopic: Power of Electrical Circuit

Series Connected Bulbs

Assertion (A): \(100 \text{ W}\), \(60 \text{ W}\) and \(20 \text{ W}\) bulbs, each marked \(220 \text{ volt}\), are connected in series with a voltage source, then \(20 text{ W}\) bulb gives maximum illumination.
Reason (R): Resistance of filament \(20 \text{ W}\) bulb is maximum.
 
(1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are false

Solution:

The power rating of a bulb is given by \(P = V^2 / R\). For bulbs rated at the same voltage \(V\) (here \(220 text{ V}\), resistance \(R = V^2 / P\). A lower power rating implies higher resistance. Thus, the \(20 \text{ W}\) bulb has the highest resistance (Reason R is true). When bulbs are connected in series, the same current \(I\) flows through each. The power dissipated by each bulb is \(P_{actual} = I^2 R\). Since \(I\) is common, the bulb with the highest resistance will dissipate the most power and therefore glow brightest. Hence, the \(20 \text{ W}\) bulb will provide maximum illumination (Assertion A is true). Reason (R) correctly explains Assertion (A).

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