Question

When two resistances R1 and R2 are connected in series, they consume 12W power. When they
are connected in parallel, they consume 50 W power. What is the ratio of the powers of R1 and R2?

Answer 1

so for series connection equivalent resistance(R) = R1 + R2

for parallel connection equivalent resistance(r) = $\frac{R1*R2}{R1+R2}$

so we know P = VI ⇒ P = I²R 

so for series connection 12 = V²/R   ..... {i}

for parallel connection 50 = V²/r .............. {ii}

dividing i from ii

so $\frac{50}{12}= \frac{R}{r}$

⇒ $\frac{25}{6}= \frac{(R1+R2)^2}{R1R2}$

⇒ 6R1² + 6R2² - 13R1R2 = 0

⇒  6R1² - 9R1R2 - 4R1R2 + 6R2²= 0

⇒ 3R1(2R1 - 3R2) - 2R2(2R1 - 3R2) = 0

⇒ (2R1 - 3R2)(2R1 - 3R2) = 0

so  $\frac{R1}{R2}= \frac{3}{2},\frac{2}{3}$   ANSWER