Question

two bulbs have the ratings (a) 40W at 220V (b) 40W at 110V .the fillament of which bulb has a higher resistence.what is the ratio of their resistences.

Answer 1

hello friend ,

here is your answer

✍✍✍. answer. ✍✍✍✍

$we \: know \: thar \\ p = \frac{ {v}^{2} }{r} \\ the \: first \: lamp \: resistor \: is \\ r = \frac{ {v}^{2} }{p} oham \\ = > r = \frac{ {200}^{2} }{40} \: oham \\ = > r = \frac{40000}{40} \: oham\\ = > r = 1000 \: oham\\ the \: resistor \: of \: second \: bulb \\ r = \frac{ {v}^{2} }{p} \\ = > r = \frac{ {100}^{2} }{40} \: oham \\ = > r = \frac{10000}{40} \: oham \\ = > r = 250 \: oham\\ so \: the \: filament \: of \: 1000 \: oham \\ \: bulb \: has \: higher \: recitence$

$the \: ratio \: of \: the \: recitence \: is \\ = 1000 : 250 \\ = 4 : 1$

✨✨hope it helps✨✨

Answer 2

$\bold{FOR \: A}$

Power P = 40W

Potential difference = 220V

Resistance $R_a$ =?

$\huge \boxed{R =\frac{{V}^{2}}{P}}$

$R_a=\frac{{220}^{2}}{40}$

$R_a=\frac{220 \times 220}{40}$

$R_a= 1210 ohm$

$\bold{FOR \: B}$

Power P = 40W

Potential difference = 110V

Resistance $R_b$ =?

$\huge \boxed{R =\frac{{V}^{2}}{P}}$

$R_b=\frac{{110}^{2}}{40}$

$R_b=\frac{110 \times 110}{40}$

$R_b=302.5 ohm$

◼The filament of bulb " a" has higher resistance which is 1210 ohm.

On comparing both resistance,

$\frac{R_a}{R_b} = \frac{1210}{302.5}$

$\frac{R_a}{R_b} = \frac{242}{605}$

So, the ratio between their resistances is 242:605.