Question

HEY PLZ ANSWER MY THIS QUESTION IT IS IMPORTANT FOR MY BOARD..plz explain the answer also​

Answer 1

Answer:

bulb 1

refer to attach image for answer

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

${\huge\tt\bf{GIVEN}}\begin{cases}\sf{Potential\:difference\:of\:both\:the\:bulbs:- 220\:V}\\\sf{First\:bulb:-60\:W}\\ \sf{Second\:bulb:-100\:W}\\\sf{Connected\:in\:series\:and\:Potential\:difference=440\:V} \end{cases}$

$\huge\tt\bf {TO\:FIND:-}$

• Which one will fuse and why?

$\huge\tt\bf {SOLUTION:-}$

We know that,

$\huge\tt {\boxed {Power=V\times I}}$

$\tt {Power=V\times \dfrac{V}{R}}$

By this,

$\huge\tt{\boxed {R=\dfrac {{V}^{2}}{Power}}}$

Case 1 :- 25 W bulb,

$\leadsto\tt {R=\dfrac{220\times 220}{25}}$

$\leadsto\tt\huge{\boxed {\pink {R=1936\:\Omega}}}$

Case 2 :- 100 W bulb,

$\leadsto\tt {R=\dfrac {220\times 220}{100}}$

$\leadsto\huge\tt {\boxed{\pink{R=484\: \Omega}}}$

We know that,

$\tt\huge{\boxed{I = \dfrac{V}{R}}}$

Now,when the bulbs are connected in series along with supply of 440 V, then,

$\leadsto\tt{\dfrac{440}{1936+484}}$

$\leadsto \tt\huge{\boxed{\purple {I = 0.1819\:A}}}$

We know,

$\tt\huge{\boxed {V= I \times R}}$

Now,Let us find the voltage drop in 25 W bulb :-

$\leadsto \tt {0.1819 \times 1936}$

$\leadsto\huge\tt{\green{\boxed {352\:Volts}}}$

As, the voltage drop is more than 220 volts, the fuse will melt.

Now, finding the voltage drop in 100 W bulb :-

$\leadsto\tt {0.1819 \times 100}$

$\leadsto\huge\tt{\green{\boxed {88\:Volts}}}$

As, the voltage drop is less than 220 volts, the fuse will not melt.

So, $\huge\tt{\red{100\:W\:bulb\:will\:not\:fuse.}}$

AND

$\huge\tt {\red {25\:W\:bulb\:will\:fuse.}}$

$\rule {200}2$