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What is the minimum number of bulbs, each marked $60W$ , $40V$ that can work safely when connected in series with a 240V mains supply?
A. $2$
B. $4$
C. $6$
D. $8$

Last updated date: 13th Jun 2024
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Hint Firstly, we will calculate the resistance of each bulb and maximum current that can flow through the bulb without damaging it. Then we will find the minimum resistance required in the circuit. We will compare the two results to get the required number of bulbs.

Complete Step by step solution
Given: voltage of supply= ${V_s} = 240V$
            Voltage required for bulb= $V = 40V$
            Power of each bulb= $P = 60W$
Now resistance of each bulb is given by,
$\because R = \dfrac{{{V^2}}}{P}$
On Putting value, we get,
$R = \dfrac{{{{40}^2}}}{{60}} = 26.67\Omega ......(1)$
Now maximum current that can flow through bulb without damaging it is given by,
  {I_{\max }} = \dfrac{P}{V} \\
  {I_{\max }} = \dfrac{{60}}{{40}} \\
  {\operatorname{I} _{\max }} = 1.5A \\
  \because {I_{\max }} = 1.5A \\
Therefore, minimum resistance required in the circuit is,
${R_{\min }} = \dfrac{{{V_s}}}{{{I_{\max }}}}$
$\therefore {R_{\min }} = \dfrac{{240}}{{1.5}} = 160\Omega ......(2)$
Let n number of bulbs be connected in series.
Then resistance of n bulbs is equal to ${R_{eq}} = n26.67......(3)$
Using (2) and (3) we get,
$n26.67 \leqslant 160$
From above,
$n = 6$
Hence the minimum number of bulbs required is 6.

Option (C) is correct.

Note Above we have calculated the maximum value of current that flows through each bulb without damaging it. If we provide more than that value of current that would cause heating effects and would damage the bulbs.