Question

30 bulbs are connected in a series. If one bulb is fused and the remaining 29 bulbs are joined in a series and connected to the same supply, the light in the rom will be:
A) Increased
B) Decreased
C) Remain same
D) None

Answer 1

Hint: Using the formula of power,$P = \dfrac{{{V^2}}}{R}$ to find the power consumed by bulbs in both cases. Also sing the formula of equivalent resistance in series so that we could determine the total resistance by the bulbs on the whole system.

Complete step by step answer:
Let a single bulb with resistance be $R$and the voltage be $V$.
We know that equivalent resistance in series is given by ${R_{equivalent}} = {R_1} + {R_2} + {R_3} + ............ + {R_n}$, where $n$is the natural number. Therefore, resistance of 30 bulbs is $30 \times R = 30R$.
Hence, power consumed by one bulb$ = \dfrac{{{V^2}}}{R}$. Therefore, power consumed by 30 bulb, ${P_1} = \dfrac{{{V^2}}}{{30R}}$ ---------------- ( I )
Now, resistance of 29 bulbs is $29 \times R = 29R$.
Therefore, power confused by 29 bulb, ${P_2} = \dfrac{{{V^2}}}{{29R}}$. -----------------( II )
Hence, from (I) and (II)
${P_2} > {P_1}$ that is power consumed by 29 bulbs is more than power consumed by 30 bulbs.
So light in the room will increase since less bulbs consume more power and glow more.
So the correct answer is A.

Note: Remember greater the value in the denominator , lesser the total value of function. Also note that there are two cases – power consumed by 30 bulbs and power consumed by 29 bulbs. Also always use the power formula in the terms given, here we can’t use $P={I}^2{R}$ as we are given voltage is constant hence we should use formula in terms of voltage and resistance.