Question

Two bulbs have ratings 100W, 220 V and 60W, 220V. Which one has the greater resistance?​

Answer 1

P=VI

= V²/R

For same V,R is inversely proportional to P

For 1st case,

power =100W

voltage = 220V

power =V²/R

100 = (220)²/R

R=(220)²/100

R=48400/100

R=484 Ohm

In 2nd case,

power= 60W

voltage = 220V

power =V²/R

60=(220)²/R

R=(220)²/60

R=48400/60

R=806.6 Ohm

∴in second case the resistance is more.

We can say that less the power more will be the resistance.

Answer 2

Given :-

$\:$

• Two bulbs have ratings 100W, 220 V and 60W, 220V.

$\:$

To Find :-

$\:$

• Which one has the greater resistance ?

$\:$

Solution :-

$\:$

We know that,

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: P \: = \: \frac{ {V}^{2} }{R}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ {V}^{2} }{P}$

$\:$

Here in the first bulb,

➠ P = 100 W

➠ V = 220 V

$\:$

According to the question by using formula we get,

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ {V}^{2} }{P}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ {(220)}^{2} }{100}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \cancel\frac{ 48400 }{100}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ 484 }{1}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: \underline{\boxed{ \bf \: R\: = \: 484 \: Ω}}$

$\:$

In the second bulb,

➠ P = 60 W

➠ V = 220 V

$\:$

According to the question by using formula we get,

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ {V}^{2} }{P}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ {(220)}^{2} }{60}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \cancel\frac{ {48400}}{60}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: R\: = \: \frac{ {806.6 }}{1}$

$\:$

$\displaystyle \bf \: \: : \: \implies \: \: \underline{\boxed{ \bf \: R\: = \: 806.6 \: Ω}}$

$\:$

∴ The second bulb possessing 60 W, 220 V has the greater resistance.

$\:$