Question

Two bulbs with different powers are first connected in series and then in parallel across the same voltage supply. The ratio of powers produced in the two respective cases is 2 : 9. If resistance of one bulb is 8 ohm, then the resistance of the other is​

Answer 1

Answer:

Explanation:

Let the individual resistances be r1 and r2.

Then

In series          r1+r2=9               ...(i)

In parallel        r1r2r1+r2=2

or,                   r1r2=18                ...(ii)

Solving these equations (9−r2)r2=18

or                    r22−9r2+18=0

or                    r22−6r2−3r2+18=0

or             r2(r2−6)−3(r2−6)=0

or             (r2−6)−(r2−3)=0

∴   If          r2=6,r1=3

And if                     r2=3,r1=6

Hence the resistance are of 3 and  6ohm.