Question

Forty electric bulbs are connected in series across 220V supply. After one bulb is fused the remaining 39 are connected in series across the same potential. The illumination will be
a) more with 40 bulbs than with 39 bulbs
b) more with 39 bulbs than with 40 bulbs
c) equal in both cases
d) in ratio ${{40}^{2}}:{{39}^{2}}$

Answer 1

Hint: Illumination of a bulb is given by the heat released which is: \[H=\dfrac{{{V}^{2}}t}{{{\operatorname{R}}_{eq}}}\]
Since the bulbs are connected in series, the voltage across each bulb remains the same and the effective resistance is the cumulative sum of all the resistances.So, use this formula to find the heat released by 40 bulbs and 39 bulbs and compare the results.

Complete step by step answer:
We have:
Resistance of a bulb = R
So, effective resistance of 40 bulbs connected in series is: ${{\operatorname{R}}_{eq}}=40R$.
So, effective resistance of 39 bulbs connected in series is:${{\operatorname{R}}_{eq}}=39R$.
So, heat released by 40 bulbs is:\[{{H}_{40}}=\dfrac{{{V}^{2}}t}{40R}\]
So, heat released by 39 bulbs is: \[{{H}_{39}}=\dfrac{{{V}^{2}}t}{39R}\]
Now, by comparing both the heat released, we get:
\[{{H}_{40}}<{{H}_{39}}\]
Therefore, illumination of 39 bulbs will be greater than 40 bulbs.
Hence, option (b) is the correct answer.

Note: Since the voltage is the same for the two connections, the resistance will be less for 39 bulbs. And since there is an inverse relation between the current flowing through a conductor and the resistance, less current means more current and vice-versa. The relation between current and resistance is given by Ohm’s law.