Question

which one have lesser resistance A 220V , 60 W bulb or 220V , 40W bulb​

Answer 1

$\mathfrak{\large{\underline{\underline{Given:-}}}}$

1 st bulb have :-

Power = 60w

2 nd bulb have :-

Power = 40w

Potential difference = 220 v

$\mathfrak{\large{\underline{\underline{</strong><strong>T</strong><strong>o</strong><strong> </strong><strong>find</strong><strong>:-}}}}$

Which bulb has lesser resistance=?

$\mathfrak{\large{\underline{\underline{</strong><strong>S</strong><strong>o</strong><strong>l</strong><strong>u</strong><strong>t</strong><strong>i</strong><strong>o</strong><strong>n</strong><strong> </strong><strong>:-}}}}$

For finding the resistance of both bulbs, we have to use the relation between power,resistance and voltage.

It is given by formula:-

★ $\boxed{\sf{ </strong><strong>R</strong><strong> </strong><strong>=</strong><strong>\frac{ {v}^{2} }{p} </strong><strong> </strong><strong>}}$

Now, we can put the value of given quantities.

Resistance of 1st bulb = $\bold{</strong><strong>R</strong><strong>1</strong><strong> </strong><strong>=</strong><strong> </strong><strong>\frac{ {(220)}^{2} }{60} </strong><strong>}$

$\implies$ $\bold{</strong><strong>R</strong><strong>1</strong><strong> </strong><strong>=</strong><strong>\frac{220 \times 220}{60} </strong><strong> </strong><strong>}$

$\implies$ $\bold{</strong><strong>R</strong><strong>1</strong><strong> </strong><strong>=</strong><strong> </strong><strong>8</strong><strong>0</strong><strong>6</strong><strong>.</strong><strong>6</strong><strong>7</strong><strong> Ω</strong><strong>}$

Now,resistance of 2 nd bulb = $\bold{</strong><strong>R</strong><strong>2</strong><strong> </strong><strong>=</strong><strong> \frac{ {(220)}^{2} }{40} </strong><strong> }$

$\implies$ $\bold{</strong><strong>R</strong><strong>2</strong><strong> </strong><strong>=</strong><strong> </strong><strong> \frac{220 \times 220}{40} </strong><strong> }$

$\implies$ $\bold{</strong><strong>R</strong><strong>2</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>2</strong><strong>1</strong><strong>0</strong><strong> </strong><strong> Ω</strong><strong>}$

from above we can say that :-

★ $\boxed{\sf{ </strong><strong>R1</strong><strong> </strong><strong><</strong><strong> </strong><strong>R2</strong><strong>}}$

therefore, bulb 1 st has lesser resistance than bulb 2 nd.