Question

When two resistances $ {{R}_{1}} $ and $ {{R}_{2}} $ are connected in series, they consume $ 12 \,W $ power. When they are connected in parallel, they consume $ 50\, W $ power. What the ratio of the powers of $ {{R}_{1}} $ and $ {{R}_{2}} $ ?

• A. 1/4
• B. 4
• C. 3/2
• D. 3
• E. 1

Answer 1

Answer: (C)

In series, $ {{p}_{s}}=\frac{{{P}_{1}}\times {{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}=12W $
In parallel, $ {{P}_{p}}={{P}_{1}}+{{P}_{2}}=50\,W $
$ \therefore $ $ {{P}_{1}}{{P}_{2}}=12\times 50=600 $
Now, $ {{({{p}_{1}}-{{p}_{2}})}^{2}}={{({{p}_{1}}+{{p}_{2}})}^{2}}-4{{p}_{1}}{{p}_{2}} $
$ ={{(50)}^{2}}-4\times 600 $
$ =2500-2400=100 $
$ \therefore $ $ {{P}_{1}}-{{P}_{2}}=10 $
or $ {{P}_{1}}=30W,{{P}_{2}}=20\,W $
$ \therefore $ $ \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{3}{2} $