Question

A 100 W bulb B1 and two 60 W bulb B2 and B3 are connected to a 250 V source as shown in the figure. Now W1, W2 and W3 are the out-put powers of the bulbs B1, B2 and B3 respectively ​

Answer 1

Answer:

A 100 W Bulb B1 and two 60 W bulbs B2 and B3... Solution: Let the given power ratings be defined at an operating voltage V. The resistances of the three bulbs are given by R1=V2/100,R2=V2/60,R3=V2/60. R 1 = V 2 / 100 , R 2 = V 2 / 60 , R 3 = V 2 / 60 .

Explanation:

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Answer 2

Solution -

Given : A 100 W bulb B1 and two 60 W bulb B2 and B3 are connected to a 250 V source .W1, W2 and W3 are the out-put powers of the bulbs B1, B2 and B3 respectively

To find : To compare the output powers of these three bulbs

Solution : We know that

$\sf{P\:=\:\dfrac{V^2}{R}}$

As power is proportional to the square of voltage and is directly proportional to resistance .

Let the resistance of bulb $\sf{B_1\:,\:B_2\:and\:B_3}$ be $\sf{R_1\:,\:R_2\:and\:R_3}$ respectively .

Now , finding out their resistance one by one

$\sf{R_1\:=\:\dfrac{V^2}{P_1}}$

$\sf{R_1\:=\:\dfrac{V^2}{100}}$

Similarly ,

$\sf{R_2\:=\:\dfrac{V^2}{P_2}}$

$\sf{R_2\:=\:\dfrac{V^2}{60}}$

And as given in the Question $\sf{R_2\:=\:R_3}$

Also , ${i_1\:=\:\dfrac{250}{R_1\:+\:R_2}}$

${i_1\:=\:\dfrac{250}{\frac{V^2}{100}\:+\:\frac{V^2}{60}}}$

${i_2\:=\:\dfrac{250}{R_3}}$

${i_2\:=\:\dfrac{250}{\frac{V^2}{60}}}$

${i_2\:=\:\dfrac{250\:\times\:60}{V^2}}$

Now , we have

$\sf{W_1\:=\:{i_1}^2{R_1}\:=\:{\dfrac{250}{V^2\:\times\frac{1}{100}\:+\:\frac{1}{60}}}^2\:\times\:R_1}$

$\sf{W_2\:=\:{i_2}^2{R_2}\:=\:{\dfrac{250}{V^2\:\times\frac{1}{100}\:+\:\frac{1}{60}}}^2\:\times\:R_2}$

$\sf{W_2\:=\:{i_2}^2{R_2}\:=\:\dfrac{250\:\times\:60}{V^2}\:\times\:R_3}$

Thus ,

$\sf{W_1\:=\:{\dfrac{250}{V^2\:\times\frac{1}{100}\:+\:\frac{1}{60}}}^2\:\times\:\frac{V^2}{100}}$

$\sf{W_1\:=\:\dfrac{(250)^2}{{V^2\:\times\:\frac{160}{100\:\times\:60}^2}}\:\times\:\frac{1}{100}}$

$\sf{W_2\:=\:\dfrac{(250)^2}{{V^2\:\times\:\frac{160}{100\:\times\:60}^2}}\:\times\:\frac{1}{60}}$

$\sf{W_3\:=\:\dfrac{(250)^2}{V^2\:\times\:{\frac{1}{60}^2}}\:\times\:\dfrac{1}{60}}$

Now , after solving these equations we get

$\sf{W_1\::\:W_2\::\:W_3\:=\:15\::\:25\::\:64}$

Therefore

$\sf{W_1\:<\:W_2\:<\:W_3}$