question_answer

                Two 220 V, 100 W bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 V AC supply line. The power drawn by the combination in each case respectively will be:

A)                                                                                                                                                                                            200 W, 150 W    

B)                 50 W, 200 W      

C)                 50 W, 100 W                      

D)                 100 W, 50 W

Correct Answer: B

Solution :

                Let, \[{{P}_{1}}=100\,W,\,\,{{P}_{2}}=100\,W,\,V=220\,volt\]                 \[{{P}_{1}}=\frac{{{V}^{2}}}{{{R}_{1}}}\,and\,\,{{P}_{2}}=\frac{{{V}^{2}}}{{{R}_{2}}}\]                 \[\therefore {{R}_{1}}=\frac{{{V}^{2}}}{{{P}_{1}}}=\frac{{{(220)}^{2}}}{100}=\frac{220\times 220}{100}\Omega \]                 and        \[{{R}_{2}}=\frac{{{V}^{2}}}{{{P}_{2}}}=\frac{{{(220)}^{2}}}{100}=\frac{220\times 220}{100}\Omega \]                 Case I: When two bulbs are connected in series.                                 In series, \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}\]                 \[=\left( \frac{220\times 220}{100} \right)\times 2\]                 Hence, \[{{P}_{eq}}=\frac{{{V}^{2}}}{{{R}_{eq}}}=\frac{220\times 220}{\left( \frac{220\times 220}{100}\times 2 \right)}\]                 \[=\frac{100}{2}=50\,W\]                 Case II: When two bulbs are connected in parallel.                                 In parallel,                 \[{{R}_{eq}}=\frac{{{R}_{1}}\,{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\]                 \[=\frac{{{\left( \frac{220\times 220}{100} \right)}^{2}}}{\frac{220\times 220}{100}\times 2}\]                 \[{{R}_{eq}}=\frac{220\times 220}{100}\times \frac{1}{2}\]                          Hence, \[{{P}_{eq}}=\frac{{{V}^{2}}}{{{R}_{eq}}}=\frac{220\times 220}{\frac{220\times 220}{100}\times \frac{1}{2}}\]                 = 200 W                 Alternative:                 For series \[{{P}_{eq}}=\frac{{{P}_{1}}{{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}=\frac{100\times 100}{200}=50\,W\]                 For parallel \[{{P}_{eq}}={{P}_{1}}+{{P}_{2}}=100+100=200\,W\]