Answer
77.7k+ views
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.
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)